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September 2007 Thesis supervisor: Arno van den Essen Second reader: Michiel de Bondt Radboud University of Nijmegen Department of Mathematics 9011145

Voor Els, Zoë, Ella en Kaat.

/012341 This thesis is special. It is mine.

How to make squares of numbers with ‘magic’ patterns? How do you know if they are impossible? These are the main questions I tried to answer in this thesis. The first chapter is an introduction to what magic squares are about and is intended to motivate and activate the reader. It introduces some broad and useful notions found throughout the magic square literature in a, I hope, clear way. As every mathematician knows, Mathematics is not easily understood without ‘doing.’ So throughout this thesis I added exercises. Some of the more innocent looking questions have given rise to several published papers in peer-reviewed mathematical magazines. In a few cases I have pointed these out, either in a footnote, or, in the answers section (page 77). Do not let it hold you back from trying! A short remark on the language used in this thesis. For people not used to reading mathematical literature, the language used can occasionally come across as arcane. For me this presents sundry challenges like does anal-retentive have a hyphen? For you, the reader, it might seem somewhat odd. In the text it will be mostly apparent in the later, more technical chapters. There are also several customs that might come across as odd. Mathematics has such a long history, that I expect they are here to stay. For instance, even though we are not of royalty, we will use mostly we instead of I. Since this thesis, by using linear algebra and other mathematical constructs, builds on the work of others it seems highly appropriate. We still hope this thesis will be read by all and sundry! The second chapter, ‘Making magic squares’, is really amalgamated history. I have reformulated several well-known ways to make magic squares into one framework. The main innovation is that it is build in a logical and elementary fashion from one simple observation. It is extremely difficult to attribute the original algorithms to persons. The amount of literature on magic squares is staggering and scattered across the globe. Attributions in different papers not seldom contradict each other, or are so vague they are useless. Often methods and theorems in various guises are being reinvented. It is not uncommon to find an algorithm from before 1100 presented as a ‘new’ method in this century (cf. the references in ‘Almost the same %

repeated’, page 44). The first discoverers or inventors of the methods used in the second chapter are, as far as I can tell, lost in time. The next chapter, ‘Generalizing’, generalizes the framework of the second chapter greatly. It is formulated in more current mathematical language, namely that of linear algebra over finite rings. This results in an extremely powerful tool to prove existence in general. Several ‘easy’ and ‘obvious’ extensions, like making Sudoku’s (page 40), are shown. A common occurrence in the magic square literature is to state results without providing proof. In ‘Mission impossible’ (page 28) I prove one of the most often repeated phrases. The chapter ‘Multimagic squares’ is the main application of the ideas presented in the previous chapters. Large parts will appear in the American Mathematical Monthly october 2007 as an article by Harm Derksen, Arno van den Essen and myself. The preprint [DEEa] and internal report [DEEb] are much more technically formulated. All are far less general than the methods considered here. I want to point out that section ‘Universal existence’ (page 59) gives a general answer to the first question posed, i.e. it shows a very general way to prove the existence of squares (and cubes) with special properties. The only real requirement being that the properties can be expressed in terms of linear vector spaces over a finite field. In ‘Magic Zoo’ I showcase a number of nice examples. All are applications of the two main theorems (page 34 and 55), occasionally with some extensions as hinted at in ‘Explaining Dudeney’.

It is a great pleasure for me to thank everyone who directly or indirectly has supported me in my mathematical endeavours. Often unwittingly, people around me have contributed to this thesis. It is wonderful to me, for instance, that a sideremark of a cameraman from TV Gelderland has made a significant difference to how I decided to present parts of this thesis. I want to thank one person in particular. His everlasting enthusiasm, positive attitude, hospitality and friendship made every meeting a joy. His human side taught me how to be a better person. Thanks Arno! Christian Eggermont Nijmegen September 2007

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Back to the Future Associative squares A fine magic key Variations and Sudoku’s Almost the same repeated

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Introduction

Magic squares are remarkable squares of numbers. They have been found around the globe and some are ancient. A few are even famous and have legends associated with them. most famous is the 3 ×3 square to the left. If you can believe 3 8 1 the The stories about this square, it was found by the mythical Chinese 2 4 6 emperor Yu (circa 2200BC) on the back of a turtle climbing out of the yellow river. Historians have been able to trace it back to the third 7 0 5 century before Christ, which makes this square somewhere between 2300 and 4200 years old! It is commonly called the Lo Shu. The square at the right has been dated around 1100. 6 11 0 13 It has been found inscribed above a gate at Khajuraho in central India. In Khajuraho the rulers of the Chandela 1 12 7 10 dynasty build 85 temples around the end of the first mil15 2 9 4 lennium, of which some 20 remain. These temples have decorations of gods, kings and mythological animals. Most 8 5 14 3 famous are however the detailed images of mostly women in various sexual positions. In Europe the famous astronomer Henrici Cornelii Agrippæ (1486–1535) related magic squares to the seven ‘planets’ of its time, which included the Sun and the Moon. Manuscripts from Picatrix from the thirteenth century and Cardano from the sixteenth century, relate the planets to the exact same magic squares with sides from three through nine, but in the opposite sequence! For ‘a bunch of numbers’, they have made their way to rather prominent places in some ancient cultures and believe systems. Besides cultural, religious and historical significance there is also an interesting mathematical aspect. But first the A

obvious (mathematical) question:

What makes a number square a magic square? You might want to study the two examples given, to see what might be so remarkable about them. Note that, even after all these years, it is certainly possible that you might find properties of the square(s) that nobody has found before. If later on, I do not mention your properties; you might have something new. So please write anything you find remarkable about the squares down. For those that need some help, a cryptic kind of hint is supplied below1 .

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❛ It is not surprising that the Lo Shu square should exert a powerful fascination in an age when, to most people, even the simplest arithmetic was a matter for amazement. N. L. Biggs, The roots of combinatorics, Historia Mathematics 6 (1979), p.119.

❜

Defining magic squares 6 1 15 8

3 8 1 2 4 6 7 0 5 (a) Lo Shu

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0 7 9 14

13 10 4 3

(b) Khajuraho

Figure 0.1: Some properties and patterns you might have found in the Lo shu: (i). the numbers in each row, column and diagonal have the same sum, i.e. (rows) 3+8+1 = 2+4+6 = 7+0+5 = (columns) 3+2+7 = 8+4+0 = 1+6+5 = (diagonals) 3 + 4 + 5 = 1 + 4 + 7 = 12. (ii). the even numbers form a cross and the odd numbers are in the corners. Opposite numbers next to the center (4) have the same sum (8). (iii). by concatenating the numbers in a row respectively column you get the following sums: 381 + 246 + 705 = 327 + 840 + 165 (iv). 32 + 82 + 12 = 72 + 02 + 52 and 32 + 22 + 72 = 12 + 62 + 52 (v). 3812 + 2462 + 7052 = 1832 + 6422 + 5072 (vi). raising it to an odd power as a matrix results in a square with property (i)

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Some properties and patterns you might have found in the square from Khajuraho: (i). the numbers in each row, column and diagonal have the same sum; (ii). if you would tile the plane with copies of this square and take a diagonal of any 4 × 4 square it will have the same constant sum. (iii). the numbers in any 2 × 2 square have the same sum. Defining magic squar@

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(iv). (rows) 62 +112 +02 +132 = 152 +22 +92 +42 = (columns) 62 +12 +152 + 82 = 02 +72 +92 +142 and (rows) 12 +122 +72 +102 = 82 +52 +142 +32 = (columns) 112 + 122 + 22 + 52 = 132 + 102 + 42 + 32

So which properties has a ‘magic square’? It is difficult to say what to call remarkable and what not, let alone ‘magical.’ Somewhere, somehow there came a consensus on when to call a number square a magic square. To kill the suspense: it is property (i), so A magic square is a square consisting of numbers for which the sum of the numbers in any row, column or diagonal will result in the same number. This number will be called the magic sum. A square with only rows and columns magic, possibly without magic diagonals will be called a semi-magic square. Something you might have taken for granted is the use of the consecutive integers. All old magic squares are either made with 1, 2 . . . or 0, 1, . . .. In the first case such a square will be called classical, and in the latter normal. A square where two entries are equal is deemed trivial. The length of the side of the square is called its order. So the Lo Shu has order 3, and the square of Khajuraho is of order 4. They are both normal and (hence) not trivial. We will see later that most of the properties as observed above, have been studied in more generality.

Exercise 0 If you turn a magic square 90◦ , ‘another’ magic square arises2 . What other transformations can you think of?

Exercise 1 Can you see that if you can make a magic square with any series of consecutive numbers, you can also make one with 0, 1, . . .? What happens if each number in a magic square is multiplied by a constant? Or if you add to each entry a constant?

Exercise 2 What is the magic sum for a normal magic square of order n? What is the sum of 0, 1, . . . , n2 −1? If you replace each number x in a normal magic square with n2 − 1 − x will a normal magic square result? Can you explain your result?

Exercise 3 How many normal magic squares are there of order 1, 2, or 3?

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Often two magic squares will be regarded as the same square if they differ only by some ‘triviality’, like a rotation. I will not make these distinctions at this time. L

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Exercise 4 Complete the following magic squares: 24 16 13 6

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Exercise 5 Continuation of exercise 4. i) How many numbers have to be in a square to make a normal solution unique? ii) Can a square have a unique normal and a unique classical solution?

Exercise 6 Can you figure out the meaning of figure 0.2(b) and figure 0.2(d)? They are so-called connection figures.

3 8 1 2 4 6 7 0 5 (a) Lo Shu

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Figure 0.2: connection figures

Proper properties  Most of the patterns and structures that were men-  a0,0 a0,1 a0,2 . . . tioned at page 3 suggest interesting questions and  a1,0 a1,1 a1,2 . . .   generalizations. We shortly introduce the most well-   a2,0 a2,1 a2,2 . . .   known properties with an example and a short de-  . .. .. .. .. . . . scription. Notions that are useful in a more general setting will, if possible, be noted. To ease the descriptions we establish some notations. A square of numbers, which we will always denote with a capital letter, e.g. A, can have all kinds of properties, it can be magic for instance. We will denote the number at the r-th row (counting from the top and starting with 0) and the c-th column (counting from Proper propewi@

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the left and starting with 0) as ar,c. So the whole square can be denoted as A = (ar,c)r,c=0,...,n−1. In case the order is known or obvious we will write (ar,c).

Associative squares Suppose you are asked the sum of 0,1, …, 9999. The long solution would consist in typing it into a computer. The trick is to realise that to sum a consecutive series of numbers, it is much easier to sum it twice! Do this as follows: 0+9999 + 1+9998 + 2+9997 + . . . + 9999+0. Since each pair (a+9999−a) has the same sum and you know how many pairs you have to add, the total of the original sum ( 12 10000 · 9999) is readily obtained3 . This pairing of numbers that are equidistant to the middle of a sequence, has become a useful trick. The main use seems to be, given a semi-magic square to obtain correct sums on the main diagonals by re-ordering rows and columns—operations that do not destroy the constant row- and column-sums. Another place this idea pops up is when we replace each number i in a normal magic square of order n with n2 − 1 − i, normally called it’s complement, or ‘the complement of i.’ The result is of course a normal magic square (do you see this?). In some cases the resulting square will be a rotation or mirrored version of the original. If the resulting square will be the same square 180◦ rotated it is called associative. More explicitly, the sum ar,c + an−1−r,n−1−c is the same constant for every 0 ≤ r, c ≤ n − 1. If the square is normal, this constant is automagically n2 − 1.

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There have been various studies of, and classifications based on, the " " " patterns of the pairs of complementary numbers, also called connection figures. Most well-known is the classification of H. E. Dudeney " " " (1857–1931) in 12 ‘types’ (see [Dud17, p.120], and, cf. [And17, Chap.8] and [Abe94]). To the side a connection figure from the Lo Shu.

Pandiagonal squares Property (ii) of the Khajuraho square is rather special. Such a square is a pandiagonal square and it is useful to see several equivalent descriptions. The first is as on page 3 that of tiling the plane with copies of the square. Suppose you take a paper and cut out holes such that you can only see the numbers in a diagonal. In this viewpoint a square is pandiagonal if you put the paper aligned with the grid anywhere and the numbers visible will have the same sum. The second equivalent hands-on definition is that if you remove any number of columns from the left (resp. rows of the top) of the square and stick them after the last column (resp. row) then the resulting square has again magic diagonals. By the way, the diagonals that occur are called the broken diagonals. 3

Note I have not mentioned the story as told by Euler of the nine year old Gauß. Euh or does this count as such… J

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The underlying notion is really that of ‘wrap-around’. Most people are familiar with computer games where 5 6 7 8 9 if you move off the right side of the screen you appear 10 11 12 13 14 at the left side, and if you move upwards off the screen 15 16 17 18 19 you’ll come up from the bottom of the screen. In a sense the same is happening here with your paper with holes. 20 21 22 23 24 In the next chapter this notion will be crucial. In a more formal language4 , the last equivalent definition is: A square A is called pandiagonal if for all c = 0, 1, . . . , n − 1, where n is the order of the square, the sums a0,c + a1,c+1 + · · · + an−1−c,n−1 + an−c,0 + · · · + an−1,c−1 and 1

0

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a0,c + a1,c−1 + · · · + ac,0 + ac+1,n−1 + · · · + an−1,c+1

are the same constant. The numbers in the first (respectively second) sum constitute what we call a downward (respectively upward) broken diagonal. A few are shown in the last picture.

Exercise 7 Show that there exists normal associative squares of any order. Show there are normal associative pandiagonal squares of any order. Can you see there are no associative or pandiagonal normal magic squares of order 4k + 2?

Exercise 8 An addition table is the result from adding numbers in a table. More concretely, given two sequences r0, r1, . . . , rn−1 and k0, . . . , kn−1, the addition table of these two sequences is an n × n square (ai,j) of numbers where ai,j = ri + kj for all i,j. i) Show that any addition table is pandiagonal. ii) When is an addition table associative? iii) What are the sequences corresponding to the following squares:

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11 13 9 15 0 10 2 8 iv) Show that if ai,j +as,t = ai,t +as,j for all i#= s, j#= t, then it is an addition table. Can you tell what the sequences are? Addition tables, given by the property in item iv), are sometimes called primitive squares, but I find ‘addition table’ more appropriate. Addition tables have been the basis of very interesting work into pandiagonal squares of Rosser and Walker ([RW38], [RW39]). They also played a major role in obtaining all most-perfect squares by Ollerenshaw and Brée ([Oll86],[OB98]). 4

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Exercise 9 Show that in any pandiagonal square of order 5 the sum of the 5 numbers in the form of ‘a cross’ (anywhere in the square) is a constant5 . Show the same holds for the other patterns.

Vice versa, is a square that admits ‘magic crosses‘ also necessarily pandiagonal?

Ply squares Property (iii) of the Khajuraho square of order 4 suggests the following generalization. A square A is called r × c-ply if every r × c block (height r ‘rows,’ width c ‘columns’) of numbers in the square, including wrap-around (!), has the same sum. Every pandiagonal magic square of order 4 happens to be also 2 × 2-ply. This is not true for larger orders. It is a most interesting property required for squares I will barely mention: Franklin squares, most-perfect squares.

Exercise 10 Can you see that if the order of a square is even and each 2 × 2 square has the same sum, that also all ‘wrapped’ squares have this same sum? What if the square has odd order?

Bimagic squares Property (iv) suggests the following question which I will quote from Britain’s greatest puzzlist, Henry Ernest Dudeney (1847-1930) in [Dud17]:

❛ 408.—MAGIC SQUARES OF TWO DEGREES

While reading a French mathematical work I happened to come across the following statement: “A very remarkable magic square of 8, in two degrees, has been constructed by M. Pfeffermann. In other words, he has managed to dispose the sixty-four first numbers on the squares of a chessboard in such a way that the sum of the numbers in every line, every column, and in each of the two diagonals, shall be the same; and more, that if one substitutes for all the numbers their squares, the square still remains magic.” I at once set to work to solve this problem, and, although it proved a very hard nut, one was rewarded by the discovery of some curious and beautiful laws that govern it. The reader may like to try his hand at the puzzle.

❜

His solution shown in figure 0.3(a) was not the first, but is very nice and also pandiagonal. Such a special property deserves a name: A square A that is a magic square and where also (a2i,j) is a magic square, is called bimagic. 5

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cf. [And60, p.147-148]. K. Magic squar@

According to M. Frolov ([Fro92, p.78-79]) the first to publish bimagic squares was G. Pfeffermann early 1891 ([Pfe91]). He published it in a periodical as a puzzle: figure 0.3(b). Its solution is a classical bimagic square. Can you solve the puzzle?

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For other interesting properties see [Abe94], [And17] and [Ful74]. We must stress that because research into properties has been undertaken by people of all walks of life, there is no real consensus on terminology.

Relative relatives The most interesting about magic squares for most people is the (almost) hidden structures and patterns. Recognizing a pattern and applying it to good use are different things.

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Figure 0.3: Close cousins of the Lo Shu Relative relativ@

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Here you see a few magic squares that are intimately related to the Lo Shu. I will not explain how they were made, can you figure it out? 30 75 12 35 80 17 28 73 10

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(d) magic squares of order 9 some 600 years apart

Figure 0.3: (continued) more close cousins of the Lo Shu The left square from figure 0.3(d) ([Kar94, pp.148-149, fig. 6]) was written down by Yang Hui, a scholar-antiquarian that published a small collection of diagrams ‘to illustrate the wonders of shu in the form of shushu’ ([Yok91, p. 510]) in 1275. Most diagrams where magic squares. Yang Hui did not know much about the construction method of the magic squares, but copied them from older sources (see [Cam69b, footnote 23, p.280]). The right square of order 9 from figure 0.3(d) is taken from [Bar88, fig.22]. So it is from 1888, but probably much older since the two squares of order 9 are made by the same process. Note that both squares are divided in nine squares of order three. If, as is here the case, these squares are all magic squares, I call such a square 3-partitioned. In general, if a square can be divided, in the same way as the squares of order nine, in order n magic squares, then the square is called n-partitioned.

❛ Pretend that you’re Hercule Poirot: Examine all clues, and deduce the truth by order and method. ❜ Helpmessage from TEX

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Y3x[8Z y3Z[4 :\]30@ How, how, how and also why?

In the previous chapter we have hopefully gotten a taste of what magic squares are and seen various (additional) properties. But how to go about making magic squares? And if you want additional properties you have just thought of? In this chapter we come to how to go about making magic squares. An ideal way of making magic squares should, if possible, also entail a way to predict and/or instill other magical properties—like pandiagonality or any other property we fancy.

Square turning

We will start with a specific case, while keeping the more general situation in mind. So let’s try order 5, while keeping in mind other (larger) orders.

Do what you can If no brilliant flash happened yet1 , start by doing what you can: write down the numbers in a square. How well did we do? The magic sum is the sum of 0 1 2 3 4 all the numbers in the square divided by the number of 5 6 7 8 9 rows so 15 (0+1+· · ·+52 −1) = 12 5(52 −1) = 60. Only the third row and third column have the correct sum. But 10 11 12 13 14 maybe more surprising is that all broken diagonals have the 15 16 17 18 19 correct sum!

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The mind keeps turning So we will try to make the (broken) diagonals into rows and columns. For a young child a square turned 45◦ is no longer a square, but it will call it a kite. In the same way, the numbers on a (broken) diagonal are after turning the square 45◦ no longer 1

if it did, please stop reading and write it down! AA

on a diagonal, but in a column or a row. Vice versa the columns and rows have after turning ‘turned into’ diagonals. This sounds like a viable route. It is clear though, that we have to carefully keep track of which situation / interpretation we are talking about. Lets call the rotated square a kite and prepend ‘kite,’ respectively ‘square,’ to the corresponding terms, so that we can speak of a kite-row and broken upward square-diagonal. 0 1

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Note that when we take a paper and write down such a square, we have to add some spacing to keep it square. It also seems like in the kite we have more columns and rows!? But wait a moment, for square-diagonals we had ‘wrap-around’, so for the kite we should have by analogy ‘wrap-around’ for the kite-rows and kitecolumns! So if we consider wrap-around for the kite-rows and kite-columns, we see that we have, by turning a square into a kite, a one-to-one correspondence from downward broken square-diagonals (from top-left to bottom-right) into kitecolumns and the upward broken square-diagonals (from top-right to bottom-left) into kite-rows. In particular all kite-rows and kite-columns have (with wrap-around) the same sum. Similarly we have only one downward and one upward kite-diagonal with the correct sum, since we have only one square-row and one square-column with the correct sum. But that sounds like a magic square. ‘Magic kite really,’ any young child would say.

Back into shape So how to get the numbers back into the square-shape without disturbing the correct sums for all the kite-rows, all the kite-columns and the two kite-diagonals?

AI

A. Making magic squar@

Clearly the two kite-diagonals are the nat0 ural choice to becoming the new diagonals 5 1 for the somehow resulting target square-shape 10 . 6 . 2 we are looking for. So lets put a square tar15 . 11 . 7 . 3 get in the picture with those as diagonals and see if we can get the numbers into that 20 16 . 12 . 8 4 target square-shape and also keep them in 21 . 17 . 13 . 9 the same kite-diagonals. 22 . 18 . 14 Any number belongs precisely to two square23 19 diagonals (an upward and a downward), or 24 equivalently precisely one kite-row and one kite-column. But note that by the extra spacing we will have empty spots in our target! Maybe we can move the numbers outside the target into the empty spots? To avoid confusion we use the same tactic as before and use target-row for a row in the target etcetera. So lets start with the number 0. It must stay in the main downward square-diagonal, 5 1 which turned into the middle kite-column. 10 . 6 . 2 And 0 must stay in the main upward square15 . 11 . 7 . 3 diagonal, which turned into the kite-row with numbers 21, 17, 13 and 9. So to keep 0 in its 20 16 . 12 . 8 4 original square-diagonals, or equivalently 21 . 17 . 13 . 9 in its kite-row and kite-column, and hence 22 . 18 . 14 keep the sums the same, there is only one 23 19 place available for it in the target: we have 24 to place 0 in the intersection of the middle target-column and the fourth target-row. Similarly 1 must stay in the kite-column given by 7, 13, 19, 20 and kite-row 5, (1,) 22, 18, 14. Again only one space in the target remains possible. We have to place 1 at the intersection of the fourth target-column and the fifth target-row. 0

In the same way there is, after having chosen the tar10 23 6 19 2 get, for each number only one possible spot. This can eas3 11 24 7 15 ily be seen by noting that the added spaces / empty spots 16 4 12 20 8 are placed between each number in a kite-row and kite9 17 0 13 21 column. Since each kite-row and kite-column intersect each other precisely twice, once in a place already hav22 5 18 1 14 ing a number, and once where is an empty spot, there is for each number outside the target precisely one empty spot inside the target. The resulting square is of course a magic square! Square turning

AB

Re-evolving Let’s turn back and take a closer look at what we have found.

Diagonal sums At first we noted that each diagonal, the middle row and the middle column already had the correct sum. To see this for any order n, note that each broken diagonal contains precisely 0,1,2, …, n − 1 (in some order) in both righthand squares, when we split the square as in figure 1.1. Note that the reasoning to

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=

+

Figure 1.1: note the broken diagonals see that the diagonals each has the same sum, does not depend on the specific set of numbers chosen but on the fact that the square we started with is an addition table (cf. page 7). We will later see that this is useful to note. If the order is odd it is easy to see that the middle row and middle column have the correct sum2 , and, these precisely turn into the main diagonals.

Calendar trick As a nice application of the same reasoning take the following trick. Let a person take any square of numbers from any month on a calendar with side length four. Let her tell you the smallest number in it. Let her choose from her square any number. Tell her to cross out all the numbers on the same row and column, except her chosen number. Let her choose another number, that is not crossed out. As before, let her cross out all the other numbers on the same row and column. Let her do that two more times, so precisely four numbers are selected and all others are crossed out. You should now be able to give her the sum of her freely chosen numbers!

Only the odd From the previous subsection it is clear that turning 45◦ of any odd order square results in a full square where the numbers in a column (respectively row) turn (reordered) into the numbers of a upward (respectively downward) 2

The middle row (respectively column) of the right-hand squares in figure 1.1 contain, either n times the number 12 (n − 1), or the numbers 0, 1, . . . (n − 1). Note, like little schoolboy Gauß, that 0 + 1 + . . . + (n − 1) = n · 12 (n − 1). AL

A. Making magic squar@

broken diagonal. And the numbers of a upward (respectively downward) diagonal turn into a row (respectively column). If, however, we try turning a square of an even order, we see that the intersection of the diagonals are already filled in! So turning 45◦ only works for squares of odd order.

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15 We have in short proven the following lemma.

Lemma 1 Let M be a square of numbers of odd order. The turning transformation as described above, exchanges rows with upward diagonals and columns with downward diagonals. If M is pandiagonal, then the transformation, will result in a semi-magic square. If also the middle row and column have the magic sum then the resulting square will be a magic square. Vice versa, any semi-magic square M can be turned into a pandiagonal square.

0 0 1 2 3 4 5

3 . 1 !

6 7 8

6

. 4 .

3 8 1 !

2

7 . 5 8

2 4 6 7 0 5

Figure 1.2: Turning to the Lo Shu

One small step … After writing out a few examples with larger orders, it is clear that you waste a lot of paper while turning squares and redrawing. It would be better if we think of a alternate way to generate the same magic squares. So take a good look at a larger order square, made by the turning 45◦ process, and see if you can make it in a more direct way. One smaz hick-up …

AC

You might want to ponder about the structure of this square before reading the next paragraph…

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65

If you follow the sequence 0,1,… in the square, a clear pattern emerges. It looks as if someone has been ‘stepping’ on a different size chessboard3 and kept track by writing down the number of the steps along the way. Starting at 0, we ‘step’ one column to the right and one row down, and place 1. We continue stepping one to the right and one down, till we hit an edge. When we arrive at the edge of the square, we just ‘wrap-around’ to the opposite side. For instance from 4 to 5 and again to 6. When after a step we stand on an already occupied spot, we take a ‘cross-step,’ one step to the left and one step down, once (e.g. from 10 via 0 to 11) and then go along stepping like before (e.g. from 11 to 12 et-cetera). One can easily check that continuing this stepping process completes the square. Please check the steps taken in this square of order 11 and the square of figure 1.3 on page 17. Notice that the cross-steps precisely take place at the multiples of the order of the square. This is true for all magic squares made by turning. So instead of ‘the turning process,’ we could have made the same magic square of order 5 by stepping with step (1, 1) and cross-step (1, −1). Here the step and cross-step are denoted by a vector giving the next row and column relative to its previous location. The starting position was (3, 2), i.e. row 3 and column 2.4 We could of course try other step/cross-step values with possibly different starting positions. This gets somewhat complex so we need more abstraction to cope. In other words, we need ‘to leap’ to a model. 3 4

AJ

of course without touching any edges… remember we start counting rows and columns at 0. A. Making magic squar@

!

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(a) the first steps

10!

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(d) a spectacular cross-step "

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(b) note the first cross-step from 4 to 5

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(c) more steps

10 23 6 19 2 3 11 24 7 15 16 4 12 20 8 9 17 0 13 21 22 5 18 1 14 (e) finished

Figure 1.3: stepping to make the turning square of order 5

A model step We use the previous remarks as inspiration, and model the stepping process in a more formal and general setting.

Going in circles The first thing to note in the previous section is the ‘wrap-around’ of the row-indices and column-indices5 . This means that any row-index r and columnindex c will be taken modulo the order of the square, say n. Cf. ‘Pandiagonal squares’ at page 6. So, in our language from page 5, with A = (ar,c), the wrap-around amounts to postulating ar,c = ar+n,c = ar,c+n.

Taking steps We will now look at the same kind of steps as before. The procedure to fill the square is as follows. See figure 1.4. We start by writing 0 at (p, q), i.e. row p and column q. We take a step (a, b) and write 1 at (p + a, q + b). And so 2 at (p + 2a, q + 2b), 3 at (p + 3a, q + 3b), and so on. We continue 5

Strictly speaking, ‘Going toroidal’ would be a more accurate heading. A model {ep

AF

a

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# n ! −1! !

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a

"

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# n ! +2 a

"

Figure 1.4: step (a, b), cross-step (a , b ), start (p, q)

b

#

stepping and writing until at step n, we get to (p + na, q + nb) = (p, q) which is already occupied (by 0)6 . So we also take a cross-step and put n at (p + na + a " , q + nb + b " ) = (p + a " , q + b " ). And we merrily continue stepping again with (a, b), putting n + 1 at (p + a + a " , q + b + b " ) and so on. At every multiple of n we not only take a step but also a cross-step. In other words we write the number x + x " n with x " , x ∈ {0, 1, . . . , n − 1} at position7 (r, c) where

'

r c

(

≡

'

p q

(

+

'

a a" b b"

('

x x"

(

(1.1)

The notation ≡ is used to denote the fact that the numbers are taken modulo n. A square as produced by this process will be called a uniform step square. If it is not completely filled it will be ill-defined since some positions have more than one possible number and some positions will be left empty. But since we are not interested in these degenerate cases, we will ignore these subtleties for the moment.

Exercise 11 Can you figure out if the following are uniform step squares? If it is, then what step, cross-step and starting position are needed to generate it? If it is not, can you prove it? i) the Lo Shu, see page 3 ii) the square of order 5 in figure 0.3(a) at page 9 iii) the squares of order 9 in figure 0.3(d) at page 10 6

Yes, this could happen sooner, say after m < n steps. Then (p + ma, q + mb) = (p, q), so it is clear that if gcd(a, n) = gcd(b, n) = 1 then this will not happen before step n. But for the moment we will ignore this. 7 remember our convention r, c ∈ {0, 1, . . . , n − 1} AM

A. Making magic squar@

Example 0 Some uniform step squares: 0 1

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step (0, 1) cross-step (1, 0)

start (1, 2), step (0, 1)

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cross-step (1, 1)

start (0, 0), step (0, 2) cross-step (1, 0)

Exercise 12 Try the following step, cross-step and starting positions for some odd orders, like n=7 or 9. What happens? For convenience write h = 12 (n − 1). i) step (1, 3), cross-step (2, 3) at starting position (0, 0) ii) step (1, 1), cross-step (1, −1) at starting position (n, h) iii) step(1, 1), cross-step (1, −1) at starting position (h + 1, h) iv) step (−h, h), cross-step (−h, −h) at starting position (0, h)

The obvious questions are of course to determine sufficient (or necessary) requirements for an uniform step square to be completely filled, semi-magic et-cetera. We will postpone these questions till we have made yet another translation into more common mathematical ground.

This? A fine map! With ‘turning squares’ we can make a magic square for any odd order. A small step led us to uniform step squares. Uniform step squares are not always magic. The equation (1.1) at page 18 which in a sense defines a uniform step square, is not to my liking. To easily check, or possibly predict, properties of an uniform step square, it is useful to have a formula which specifies, given the row and column, the number at that location. So we rather have the inverse of the map given by equation (1.1)! If we identify a number written in base n as a pair of numbers (x, x " ), where each number of the pair is taken from {0, . . . , n − 1}, then what we are looking for can easily be distilled out of equation (1.1). The value of ar,c at row r and column c is then given by

1 ar,c ≡ · " (ab − a " b)

'

b " −a " −b a

('

r−p c−q

(

where we always reduce r, c, a, b, a " , b " , p and q modulo the order n. Vice versa each map above corresponds exactly to a uniform step square. In other words, we were looking at an affine map from (Z/nZ)2 → (Z/nZ)2! |}? A fine map!

AD

Before we try to answer obvious questions, like when will a uniform step square be completely filled, we better take a step back and check our assumptions yet again. This, and answers to more questions, can be found in the next chapter.

IK

A. Making magic squar@

2

~18103•[€[8Z More, more, most?

At first, in ‘Back to the Future’, we will give the general framework. The previous chapter will stand model. Getting acquainted with the notations will come first. In ‘Associative squares’ we show the ‘direct calculation’ approach. Linear algebra will in ‘A fine magic key’ provide insight in how to make a large class of patterns magic. Several other applications, including Sudoku’s, that highlight other aspects of the method used, will be discussed. In ‘Almost the same repeated’ we discuss an ancient technique, which assures us that many properties from small orders can, in a way, be transfered to larger order squares.

Back to the Future We have seen in the previous chapter that the magic squares made by ‘turning,’ turned out to be uniform step squares. These where no longer always magic but gave a lot more room to experiment and more importantly suggested interesting questions naturally. The uniform step squares turned out to be affine maps over finite rings. Since this translates the questions surrounding (uniform step) magic squares in questions in the realm of ‘Linear Algebra’, it seems worthwhile to look closer at what this means. Time to see how all this translates in a familiar example.

The familiar example Consider the first magic square made ‘by turning’ of page 13. It is also a uniform step square (of course): use step (1, 1), cross-step (1, −1) and starting position (3, 2). IA

The translation as suggested in ‘This? A fine map!’ (page 19) amounts to: 00 01 02 03 04

20 43 11 34 02

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03 21 44 12 30 L

3 T

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The left square contains the row and column indices (concatenated together by removing the visual clutter of (, )’s for pairs). We view these as pairs of elements of R = Z/5Z. These pairs are then mapped by an affine map L : R2 → R2 to the center square. The pairs in the center square are translated back using the usual base 5 representation to the final square at the right. Since we will be using the sets {0, 1, . . . , s} often, we will denote it by s.

So what are these maps and identifications exactly? The row and column indices were mapped in the most simple way. Call the identifying map for the rows ρ and for the columns κ. In the familiar example we used ρ : R → 5 with ρ(i) = i, where for each i ∈ 5 we denote by i the equivalence class of i in R. We also took κ to be the same map. The ‘affine map’ L : R2 → R2 we found for this example is:

L:

'

r c

(

&→

'

3 2 3 3

('

r c

(

+

'

2 0

(

In general, for any finite ring R, an R-affine map L : Re → Rd, or simply affine map if the ‘base ring’ R is obvious, is a map for which there is a t ∈ Rd and an Rlinear map (or equivalently R-module homomorphism) L0 : Re → Rd such that L(a) = L0(a) + t for all a ∈ Re. The last identification ‘back’ is a bijection T : R2 → 52. It mimics the base 5 representation: T ((a, b)) = ρ(a)·5+ρ(b). Note that all maps so far are bijections. We see that a uniform step square A = (ai,j) could have been defined as aρ(r),κ(c) = T (L((r, c))) for all r, c ∈ R. Compare this to the formula in ‘This? A fine map!’ at page 19. In short we are considering the following diagram: (ρ×κ)−1

L

T

m × m −−−−→ R2 −−−−→ R2 −−−−→ m2 So the matrix M that corresponds to this diagram is defined by

mρ(a),κ(b) = T (L((a, b))), II

I. Generalizing

where a, b ∈ R.

(2.1)

This is the ‘framework’ in which we will study magic squares. There will be one small subtlety added later. Note that we have taking notational liberty by denoting the vectors from R2 as both row- and column-vectors. Unless ambiguity problems might be probable we will take that liberty for obvious layout reasons.

The choices The finite ring R = Z/mZ was choosen since we wanted to ‘step around’ without having to worry about ‘falling of the edge’ — the ‘wrap-around’. In a sense this meant that we needed to add and multiply the row and column indices modulo the order. But really any finite ring R with the number of elements of R equal to the order of the square, i.e. |R| = m, will do. We will however in this thesis only consider commutative associative finite rings R with identity (1 #= 0). The identifying maps ρ, κ and T , for the square to be completely filled, have to be bijections. They will most often be the ‘usual’ bijections, i.e. similar to the maps in ‘The familiar example’ (page 21). Nothing else in general will be required from these maps. Note that a change in ρ (resp. κ) constitutes a permutation of rows (resp. columns). The affine map L obviously also has to be a bijection. In particular, the determinant of the corresponding matrix of L0 has to be a unit in the finite ring. Not a relation you might immediately have noticed from the equivalent parameters (step, cross-step and starting position) in the uniform stepping process! What are the benefits of all this? Being more exact (and abstract) makes the choices made more transparent. It also makes abundantly clear, the choices one can make. So more ‘freedom’ if you will, at the expense of more ‘administration.’ The abstraction also provides insight and inspiration: if you fancy cubes more than squares, it is now sooo much easier to see how to make completely filled uniform step cubes using the relation with affine maps (cf. [And17, p.64-76] and [BC87, p.216221]). It also makes it completely obvious what to do for even orders, something that could not be handled by the uniform step process directly. Try a different ring or identification!

Other flavours of T The following makes good use of the possibility of choosing a different T . We keep using the same notations as in (2.1) (page 22). Let’s make a magic square of order 4. We have at least three rings to choose from but stick for the moment to the most well known Z/4Z, and denote it again by R. Let ρ and κ be standard or the ‘usual’ bijections, i.e. analogues to the maps in the familiar example above. We leave T unspecified for the moment. Take L : R2 → R2 Back to • Future

IB

to be the R-linear map

'

r c

(

&→

'

1 2 2 1

('

r c

(

Now take a look at the square (L((ρ−1(r), κ−1(c)))), so with elements in R2. We write again (a, b) ∈ R2 as ab for clarity (less clutter) in the square below.

00 21 02 23 12 33 10 31 20 01 22 03 32 13 30 11 Since there are repetitions in the first column in the second coordinate (only 0’s and 2’s) it is clear this square will not be automatically magic for all T (really?). If all 2’s would have been 3’s and vice versa, the sums for all the rows and columns for both coordinates would all be 2, as it should ( 0+1+2+3 = 2 in R!). In a way this can be accomplished by adjusting the standard T . Simply use T ((a, b)) = τ(a) · 4 + τ(b) with τ : R → 4 the bijection that maps the sequence 0, 1, 2, 3 (in R) to the numbers 0, 1, 3, 2. The resulting magic square is even pandiagonal and 2 × 2-ply!

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Exercise 13 What are the step, cross-step and startposition used above before the adjustment of T ? (Hint: look at L and page 19) After the adjustment to T , is the resulting square as shown above still a uniform step square in the sense of the previous chapter?

Exercise 14 Show that Z/4Z, F2[x]/(x2), F2 ⊕ F2 and F2[x]/(x2 + 1) are all the finite (associative) rings with identity of 4 elements. Which one is isomorphic to the finite field F4 of four elements? IL

I. Generalizing

A subtlety Instead of changing T to make the magic square above, we could also have used the ring R = F22 (with some added subtleties, look for them). 0000 0001 0010 0011

···

(ρ×κ)−1

−→

0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

0000 1101 0011 1110 L

−→

0111 1010 0100 1001 1100 0001 1111 0010 1011 0110 1000 0101

T

−→ · · ·

We use the binary number representation as inspiration for the identifications. So use τ : F2 → 2 with τ(0) = 0 and τ(1) = 1 to build the identifying maps; ρ((a, b)) = κ((a, b)) = τ(a) · 2 + τ(b) and T ((a0, a1, a2, a3)) = τ(a0)23 + τ(a1)22 + τ(a2)2 + τ(a3). Then define the linear map L : F42 → F42 using the matrix  

1 0  1 1   0 1 0 1

0 0 1 1

1 1   0  1

the resulting square defined by Mρ(a0 ,a1 ),κ(a2 ,a3 ) = T (L((a0, a1, a2, a3))) will be exactly the same as the magic square of ‘Other flavours of T’. In diagram-form the maps look like: (ρ×κ)−1

L

T

4 × 4 −−−−→ F42 −−−−→ F42 −−−−→ 42 Do you see what is different from the diagram on page 22? We have silently used the fact that (F22)2 is isomorphic to F42 ! There is no need to make the isomorphism explicit and in the rest of this text we will simply use it without comment. In other words, for squares of order m = qn we get the following diagram: (ρ×κ)−1

L

T

qn × qn −−−−→ R2n −−−−→ R2n −−−−→ q2n

(2.2)

where R is a finite commutative ring with q elements. The matrix M that corresponds to this diagram is defined by

mρ(a),κ(b) = T (L((a, b))),

where a, b ∈ Rn.

This is the ‘extended framework’.

These notations will be used in the following chapters, if possible, for ease of understanding. Back to • Future

IC

Exercise 15 How would a affine map from (F2)2 → (F2)2 look like? Can you see the benefits of using F42 instead of (F22)2 ?

Exercise 16 Can you write the following famous magic squares as an affine map? i) the Lo Shu (page 3) ii) the square from Khajuraho (page 1) iii) Dudeney’s bimagic square of order 8 (page 9) iv) Any (famous) square by Benjamin Franklin? (see Internet).

Associative squares For some reason I decided to make associative squares. In the setting outlined a square A of order m is associative precisely if

aρ(r),κ(c) = m2 −1 − am−1−ρ(r),m−1−κ(c). So how to proceed? A key observation is that the usual bijection ρ(i) = i (with R = Z/mZ, i ∈ m) has the property m − 1 − ρ(a) = ρ(−a − 1) for all a in R (check this). In the general setting, with R any finite ring with |R| = m, we consider bijections τ : R → m with the property that there exists an element c in R such that τ(a)+τ(−a+c) = m−1 for all a in R. We call such a bijection a bijection of type c. Note that by this definition the standard bijection is a bijection of type −1. One nice thing about these bijections is that it is possible to combine them like a base q representation. It inspired the following definition. Suppose for all i ∈ n we have a number ti ∈ Z and a bijection τ(i) : R → q n n n such " that T : R → q is a bijection such that for all r ∈ R we have T (r) = ti · τ(i)(ri). I call such a bijection T a weighted sum of the τ(i). If it is immaterial which specific bijections τ(i) are meant, we will simply refer to T as a weighted sum bijection. Note that, since Rn is not only an R-module but also a ring, T can be of type c ∈ R2n as in the definition above.

Lemma 2 Let R be a finite ring with q elements. Suppose T : Rt → qt is a weighted

sum of ρ(i) : R → q. If each ρ(i) is a bijection of type ci, then T is a bijection of type (c0, c1, . . . , ct−1) ∈ Rt.

Exercise 17 Prove lemma 2.

Exercise 18 Show that all bijections from F2 → 2 are of type 1. Show that 8 (of the 24 possible) bijections Z/4Z → 4 are of type c ∈ {1, 3}.

Exercise 19 Show which maps in the previous section are of type c for some c ∈ R. Are there any maps used that are not of type c? IJ

I. Generalizing

Now we are in a position to answer the question. Notations are as at the end of ‘Other flavours of T’ (diagram (2.2) of page 25)

Proposition 3 Let M be a square of order qn, defined by mρ(a),κ(b) = T (L((a, b))). Suppose ρ is a bijection of type c, κ a bijection of type d, and T a bijection of type e. Lets denote the translation vector of L by t and its matrix by A/ . Then 0

M is associative

if and only if

2t = −A

c d

+e

(in R2n).

Proof: Since ρ (resp. κ) is of type c ( resp. d) we have for any a, b ∈ Rn that q2n−1− mqn −1−ρ(a),qn −1−κ(b) is equal to q2n−1−mρ(−a+c),κ(−b+d). By the definition of M this is equal to q2n −1−T (L((−a+c, −b+d))). Since T is of type e, this in turn is equal to T (−L((−a+c, −b+d))+e). In other words, M is associative if and only if T (L(a, b)) = T (−L((−a+c, −b+d))+e) for all a, b ∈ Rn. Since T is a bijection, the last condition is equivalent with L((a, b)) = −L((−a + c, −b + d)) + e. In other words M is associative if and only if

/ 0 / 0 / 0 / 0 −a + c a −c A ab + t = −A −b − t + e = A + A +d b −d − t + e

Reordering the terms gives the result.

"

This ‘direct calculation’ worked out nicely. But do we even know if the bijections exist? Lucky me, the following lemma shows they do.

Lemma 4 If 2 is a unit in R, then for each c in R there exists a bijection N of type c. If 2 is not a unit in R, then for each unit c in R there exists a bijection N of type c. Proof: For c in R define ϕ = ϕc : R → R by ϕ(a) = −a + c. Then ϕ2 is equal to the identity, so all orbits of ϕ have length one or two and an element a is a fixed point of ϕ if and only if 2a = c. For a in R denote its orbit under ϕ by O(a). i) Let 2 be a unit in R. Then ϕ has exactly one fixed point, namely, a0 := 2−1c. The orbit O(a0) of a0 consists of one element. Let O(a1), . . . , O(as) be the other orbits. They all have two elements. In particular, q = 2s + 1. Define N(a0) = s, N(ai) = i − 1, and N(ϕ(ai)) = q − 1 − N(ai) (= q − i) for i = 1, 2, . . . , s. Then N is the desired bijection. ii) Let 2 not be a unit in R. Then for c in R∗ , the group of units in R, the map ϕ = ϕc has no fixed point: indeed, if ϕ(a) = a for some a, then 2a = c. This implies that 2 is a unit in R, a contradiction. Consequently, we can write R as the disjoint union of orbits O(a1), . . . , O(as), each of which consists of two elements. In particular, q = 2s. Finally define N(ai) = i−1 and N(ϕ(ai)) = q−1−N(ai) (= q − i) for all i. Again N is a bijection of type c. "

Exercise 20 The proof of lemma 4 shows more bijections are possible than the statement of the lemma concludes. Show this by constructing a bijection of type 3 from Z/6Z to 6. A‚ociative squar@

IF

Exercise 21 Can you use proposition 3 to make the following square associative? i) the pandiagonal square of order 4 from ‘Other flavours of T’ (page 23) ii) the square from Khajuraho (page 1)

Exercise 22 In an associative square of order m any two rows or columns, symmetrically located from the center, can be exchanged without compromising associativity. For example exchange row i and m − 1 − i. What is the relationship with bijections of type c?

Intermezzo – Mission Impossible Since we are discussing associative squares it might be a nice break to discuss one of the often repeated phrases (in various forms) in the magic square literature:

❛ It is believed by many that it is impossible to construct an associative or pandiagonal magic square of order 2(2k + 1) ❜

e.g. see [Ful74, p.60] or [And17, p.263.] But since I believe you should put your proof where your mouth is, I will provide a proof. In fact a lot of the fun for me is in proving (and beware, also in some cases disproving!) these kind of seemingly bold statements. In other words the proof of the pudding is in the proving1 . The following lemma is inspired by [And17, p.277]

Lemma 5 Let A be a associative square of order 2m. Let A be the square obtained from A by re-ordering the columns and rows in the order 0, 1, …, m-2, m-1, 2m-1, 2m-2, …, m. Then for all r and c the sum ar,c + ar+m,c+m, where the indices are taken modulo 2m, is a constant. Proof: Denote S for the constant sum of any two skew related cells in the associative square. Write x! = S − x for the complement of x, so in particular S = a!r,c + ar,c. Look at A as partitioned in 4 sub-squares of order m. Since A is associative, the top two squares (bi,j) and (ci,j), determine the bottom two squares. The reordering of the rows and columns, as stated above, has the following effect:

A=

1

bi,j

ci,j

c!m−1−i,m−1−j b!m−1−i,m−1−j

1

2

!

1

bi,j c!i,m−1−j

ci,m−1−j b!i,j

2

=A

The real proverb is the proof of the pudding is in the eating and proof has the sense of ‘test’ (as it also has, or used to have, in phrases such as proving-ground and printer’s proof ). The proverb literally says that you won’t know whether food has been cooked properly until you try it. Or, putting it figuratively, don’t assume that something is in order or believe what you are told, but judge the matter by testing it; it’s much the same philosophy as in seeing is believing and actions speak louder than words. The proverb is ancient—it has been traced back to 1300 and was popularised by Cervantes in his Don Quixote of 1605. IM

I. Generalizing

Clearly ar,c + ar+m,c+m is constant since a!r,c = ar+m,c+m, i.e. they are complements. " Note that in particular A is pandiagonal, since for any number in a broken diagonal, also its complement is in the same broken diagonal.

Exercise 23 Check the following statements. In any semi-magic square the rows (respectively columns) can be permuted, the result will still be semi-magic. In particular, the process of lemma 5 can change an associative semi-magic square of even order in a pandiagonal magic square. In a general magic square of order m one can exchange row i and m − 1 − i and also exchange column i with m − 1 − i and the result will still be magic.

17 30

1

32 11 14

17 30

1

15

35

2

15

35 19 27

7

27 19

13 10 29 31

9

12

23 26

4

6

25 22

16

8

33

0

21 24

3

34

7

14 11 32 2

13 10 29 12

9

31

21 24

3

5

34

28 20

16

33 20 28

0

5

23 26

4

6

18

!

8

18

22 25

An example of the reordering and its effect on an associative square of order 6 It is interesting to note that A. L. Candy claimed that a square of order 2(2k+1) can be made associative except for two pairs of numbers, symmetric with respect to the horizontal or vertical axis (see [Can34], [Can35]). Which seems to partially answer problem 2.20 from Gakuho Abe in [Abe94], when using lemma 5.

Exercise 24 In general, you can not use lemma 5 to transform a pandiagonal square into an associative square. Provide such a counter example. And now for the pudding:

Lemma 6 Let M be a semi-magic square of order 2m using consecutive numbers. If M is associative or pandiagonal then m is even. Proof: By lemma 5 it is sufficient to consider a normal pandiagonal square of order 2m. We need to deduce that m has to be even. Write S for the magic constant of M (= m(4m2 − 1)) and partition M in 2 × 2 squares with ‘colors’ a, b, c and d. Write A (respectively B,C,D) for the sum of the numbers with color a (respectively b,c,d). The sum of the first, third, fifth, … columns is equal to A + C = mS . The sum of the first, third, fifth, … rows is equal to A + B = mS . The sum of the second, fourth, sixth, … downward diagonals is A‚ociative squar@

ID

Ö

1 2

equal to C + B = mS . So A = B = C = a integer so m has to be even.

a c a c a .. .

b d b d b .. .

a c a c a .. .

b d b d b .. .

a c a c a .. .

· mS = 12 m2(4m2 − 1). Clearly A is b d b d b .. .

a c a c a .. .

... ... ... ... ... ..

.

" The proof given above can also easily be adapted to rule out other sets of numbers. There do exist non-normal associative pandiagonal magic squares of order 6, e.g. see figure 2.1 on page 30.

34 16 29 20 30 15

34 16 29 15 30 20

35 11 40

7

39 12

35 11 40 12 39

7

6

48

2

43

6

44

1

43

2

48

28 18 33 14 32 19

0

46

5

47

4

42

41

9

9

36

8

37 13

0

46

44

1

36 13 37

8

41

5

47

28 18 33 19 32 14

42

4

(a) 3 × 3-ply

(b) associative

Figure 2.1: A non-normal 2 × 2- and 3 × 3-ply pandiagonal magic square of order 6, and its’companion 2 × 2-ply associative pandiagonal magic square by the transformation of lemma 5. For the first square also holds that ar,c + ar+3,c+3 is constant.

A fine magic key We have seen how various magic squares nicely fit into the framework. So the obvious question is, can we see why they are magic, when seen within the framework?

The familiar example (again) Consider the sets of rows, columns and diagonals of the familiar example (page 21). Clearly a row consists of the entries aρ(r),κ(c), where r, c ∈ Rn and r fixed (for a BK

I. Generalizing

column it is c that is fixed). Why do the rows in the resulting square have the same sums?

20 43 11 34 02 03 21 44 12 30

2 4 1 3 0

0 3 1 4 2

0 2 4 1 3

3 1 4 2 0

T

L

· · · −→ 31 04 22 40 13 ! 3 0 2 4 1 · 5 + 1 4 2 0 3 14 32 00 23 41

1 3 0 2 4

4 2 0 3 1

42 10 33 01 24

4 1 3 0 2

2 0 3 1 4

Well of course! We already saw the argument needed in ‘Re-evolving’ (page 14)2 . The same is of course happening here, since those diagonals have turned into rows (and columns) but that is not the point. Look at the first row of the left square (the result of L before applying T ). In the first coordinate we have the sequence 2, 4, 1, 3, 0 and in the second coordinate 0, 3, 1, 4, 2. So both consist of all the elements of F5. As you can easily check, this is also true for the other rows. As can also be noted at the right, T (a, b) = ρ(a)5 + ρ(b). So any sum of the entries in a row is the same, since we sum the same elements. This observation will be fruitful (again). The sums of the columns are constant in the same way. These sequences are the result of an affine map. So why has this affine map this property for rows? Take any row, say the ρ(r)-th, for some r ∈ R = F5. The ρ(r)-th row consists of the entries aρ(r),κ(s), where s runs over R and r is fixed. So its sum is equal to the sum of the entries T (L((r, s))), where the sum is taken over s ∈ R. Since T (a, b) = ρ(a)5 + ρ(b), the sum is equal to

3 #

4

ρ(3r + 2s + 2)

s∈R

·5+

#

ρ(3r + 3s)

s∈R

The ‘coordinate maps’ from R → R given by s &→ 2s + c and s &→ 3s + c are surjective, and so in this case isomorphisms (2 and 3 are units in R) for any constant c ∈ R. So the sum is equal to

3 # s∈R

4

ρ(s)

·5+

#

ρ(s).

s∈R

This sum is clearly independent of r, so each row has the same sum.

Exercise 25 Write for this square the sums for the ρ(c)-th column as above. Are the resulting ‘coordinate maps’ also isomorphisms? What about the diagonals? 2

And probably also in your answer for exercise 8 at page 7. It is also why the calendar trick (page 14) works. A fine magic key

BA

The argumentation above can, after some modifications, be reused for other finite rings and fields, but to understand the argument better, and get more out of it, it is useful to look at it in the context of ‘Linear Algebra.’ So how does this turn into ‘Linear Algebra’? The main observation is that the set of pairs of row and column indices are identified (using (ρ × κ)−1) with elements of the vector space F25 (since Z/5Z = F5). In this light the rows and columns are affine subspaces! We use the common notation {e0, e1} for the natural basis of F25. A notation for the vector subspace spanned by several vectors will be that list of vectors, enclosed in angle brackets )*. So since the first row is the linear subspace spanned by the vector e1, the first row is )e1*. The affine subspaces parallel to this correspond to the other rows. Similarly, the columns of the square are of the form r + )e0* for suitable choices of r in F25. The main downward diagonal is the (linear) subspace )e0 + e1* and the main upward diagonal is the affine subspace −e1 + )e0 − e1*. How does this work out in the ‘extended framework’ from page 25 (with fields)?

Getting even In ‘A subtlety’ (page 25) the square of order 4 also used a vector space: F42. Likewise, the rows are affine subspaces of )e2, e3* and the columns of )e0, e1*. As before we look at the square resulting from L before applying T . If we look at the coordinates separately (figure 2.2), we see the same thing happening here: in each row, column, and diagonal there are as many 0’s as 1’s. By the form of the T used, this results in constant sums.

0 0 1 1

1 1 0 0

0 0 1 1

1 1 0 0

0 1 1 0

1 0 0 1

0 1 1 0

1 0 0 1

0 1 0 1

0 1 0 1

1 0 1 0

1 0 1 0

0 1 0 1

1 0 1 0

1 0 1 0

0 1 0 1

Figure 2.2: ‘subtle’ square in coordinates

So how do we relate this again to the affine map L? Since L is an affine map, each of the coordinate maps are affine maps from F42 to F2. Note that an affine map Fs2 → Ft2 takes affine subspaces to affine subspaces. Since d-dimensional subspaces 4 are isomorphic to Fd 2 , we see that for any linear subspace of F2 such a map is either constant, or it is surjective and hence takes the values 0 and 1 equally often. If the latter is true on a subspace, it is equally true on any parallel affine subspace. So how do we use these ideas in the more general setting? BI

I. Generalizing

A fine sum In the framework of page 25 we are considering finite rings, not just fields. Here R2n is not a vector space but it is an R-module. The fact that it is also a ring is only used sparingly. Subspaces correspond to (R-) submodules S ⊂ R2n. If S is an Rsubmodule of R2n then for x ∈ R2n we will call the set S + x an affine submodule. If also x #∈ S then S + x will be called parallel to S. Let S be a submodule of R2n. Suppose it is generated by s0, s1, . . . , sd−1 ∈ R2n as an R-module, which we again denote by S = )s0, s1, . . . , sd−1*. Now we want to get a better understanding of a sum taken over the elements in a parallel affine submodule S + x, which is by definition:

#

Mρ(a),κ(b) =

(a,b)∈S+x

#

T (L((a, b)))

(a,b)∈S+x

The reasoning above depended on the ‘shape’ of T . It used the fact that T is a weighted sum bijection. If T is of this form, then, loosely speaking, we can try using surjectivity of the coordinate maps of L on S + x to create constant sums. So assume T is a weighted sum bijection. Then, as in the previous examples, we consider the i-th coordinate part of the sum (without the constant factor):

#

(2.3)

ρ(i)(L((a, b))i)

(a,b)∈S+x

So what is L((a, b))i, the i-the coordinate of the vector L((a, b))? Note that each " (a, b) ∈ S + x can be written as d−1 s r j=0 j j + x with rj ∈ R, since S is generated 2n by the sj. If we denote " the i-th row of the matrix of L by rowi ∈ R then we see that with (a, b) = sjrj + x we get

L((a, b))i = ti + )rowi | x* + "

d−1 # )rowi | sj*rj j=0

where )u | v* = k ukvk is the usual bilinear form3 . As in the previous paragraph we assume surjectivity for the ‘coordinate map’ L(i) : Rd → R defined by d−1 # L(i) : r &→ )rowi | sj*rj . j=0

3

I am well aware that rowi really is a row-vector and I use it as a column-vector, so it should be rowi . But, as before, I trust the reader in reading through these ambiguities, and in knowing when to transpose a vector if needed. t

A fine magic key

BB

In defining the coordinate map we left out the constant factor since it is not relevant. We want to plug-in this map in the sum. So at this point, as in the finite fields case, we like to sum over Rd in equation (2.3). However this is only possible if we assume the R-module S to be free, i.e. S , Rd. If R is a finite field, all modules are free. So assume S to be a free module and the sj to be a basis of S. Then the sum (2.3) is equal to

#

r∈Rd

5 6 ρ(i) L(i)(r) + )rowi | x* + ti

(2.4)

Since the linear map L(i) is assumed surjective, we have for each a ∈ R a d non-empty collection L−1 (i) (a) = {r ∈ R ; L(i)(r) = a}. Say L(i)(a0) = a. Then

−1 d d−1 L−1 for (i) (a) = a0 + ker L(i). Since R / ker L(i) , R we see that |L(i) (a)| = |R| each a ∈ R. Note that Rd is a disjoint union of the collections L−1 (i) (a) where a runs through −1 R. Since the collections L(i) (a) are all the same size we see the sum (2.4) is equal to

d−1

|R|

#

a∈R

|R|−1 d−1

ρ(i) (a + )rowi | x* + ti) = |R|

#

j

j=0

Since each of the coordinate parts has a sum independent of x, the total sum is independent of x. So in short, we have proven

Theorem 1 Let M be a square defined as on page 25. Let T be a weighted sum bijection and S ⊆ R2n a free R-module. If all of the (2n) maps L(i) : Rd → R as defined above are surjective, then the sum of the entries of M taken over S, and taken over any parallel affine module of S, is constant.

Note that besides the shape of T , there is considerable freedom. For instance the choice of the translation vector of L, is completely irrelevant in this theorem. So proposition 3 from page 27 can often be used (not always if characteristic 2 cf. exercise 21, page 28). Also note that the order of the rows of the matrix of L is irrelevant. Ofcourse, so are ρ and κ.

Exercise 26 In theorem 1 there is no reference to the specific basis of S chosen. It is however used to define the coordinate maps L(i). To show this is not a problem, prove that if b0, b1, …, bd−1 is another basis of S as R-module, then Im(L(i)) = "d−1 Im(B(i)) where B(i) : r &→ j=0 )rowi | bj*rj. The ideas of the proof of theorem 1 are certainly applicable in other situations, and often point to the right direction in case they are not directly applicable. In ‘Variations and Sudoku’s’ (page 40) we give some examples of this statement. BL

I. Generalizing

Read before use The theorem suggests a lot of questions. What do these (free) affine submodules of R2n look like? What are appropriate submodules for rows, diagonals or any pattern we fancy? And how do you use them to give wonderful (new?) magic patterns in a square?

What do (parallel) affine submodules look like? First of all, the elements in any affine submodule correspond to places that are given by the bijections ρ and κ. So a collection of an affine submodules can ‘look’ different, if different ‘identifications’ are used. Example 2 on page 37 shows this most clearly. There are two collections of affine submodules that are clear exceptions to this. Whatever finite ring R, or ρ and κ are used, the columns are always the parallel affine submodules of )e0, e1, . . . , en−1*, and the rows always of )en, en+1, . . . , e2n−1*.

(a) overlap

(b) parallel?

Figure 2.3: Second, there are some small things to watch out for. For instance, rather trivial, the two shaded portions in figure 2.3(a) can not be parallel to the same submodule, since the parallel affine submodules form a partition of R2n and so have no overlap, i.e. they are disjunct. Slightly less trivial, with the ‘usual’ bijections ρ and κ, the two broken diagonals in figure 2.3(b) are parallel affine submodules in (Z/4Z)2 but not in F42. To see R-submodules as something like vector spaces is useful, within limits. In figure 2.4 we show a five collections of parallel affine submodules, the generator for the ‘linear’ submodule is given in the caption. The bijections chosen are the standard bijections of type −1 = 4. Each affine submodule has its own ‘color.’

Exercise 27 In figure 2.5 a number of collections of parallel affine submodules are shown. Can you give generator(s) for each. The square is identified with F42 as in ‘A subtlety’ (page 25).

Exercise 28 For a submodule S denote by S⊥ ⊆ R2n the R-submodule of the vectors v such that )v | s* = 0 for all s ∈ S. When can the parallel affine subspaces of a free submodule S and of S⊥ have constant sums? A fine magic key

BC

(a) )(0, 1)*

(b) )(1, 1)*

(c) )(1, −1)*

(d) )(1, 2)*

(e) )(1, 3)*

Figure 2.4: A few parallel affine submodules in (Z/5Z)2

(a)

(b)

(c)

(d)

(e)

Figure 2.5: Some parallel affine submodules in F42

What to do with submodules you fancy? Suppose you found a collection of free submodules, such that when the parallel affine submodules have a constant sum, the property you somehow crave is a necessary consequence. Then what? In other (longer) words, when we have a set of free submodules, how to find possible rows for L that makes them all ‘magic’? Note each row is an element of R2n. It is a simple matter to list all elements of R2n, since R is finite. You can reduce this list by noting that it is a disjunct union of sets vR! , where R! is the group of units in R, for some collection of elements v. Only one representative of each vR! suffices4 . To see if an element v of the remaining list might be a row of L, we use theorem 1. Simply remove the vector v from the list if the ‘coordinate map’ L(i) as defined in the proof of theorem 1 with v instead of rowi, is not surjective. Remember that a linear map Rt → Rs is surjective when the columns of the matrix generate Rs. The remaining elements in the list can be used as rows for the matrix of L, if the determinant is a unit in R. And by theorem 1, any resulting L gives these collections of parallel affine submodules constant sums. When R is a field, to check for surjectivity of the ‘coordinate map’ of v, it is enough to check for each submodule that there is a generator g such that )g | v* is not zero. If this fails, remove it. When R is for instance Z/mZ a variation on 4

Check surjectivity of corresponding coordinate maps, and do not forget the possible effect on the determinant of L! BJ

I. Generalizing

fraction free gauss-elimination can work fast. The following exercises show a few nuances of using ‘non-fields.’

Exercise 29 Let R = Z/6Z. Check which of the linear maps corresponding to the following matrices'are surjective. (

3 2 2 3

i)

ii)

5

2 3

6

iii) iv)

'

'

2 3 0 4 1 2

'

2 3 0 5 1 2 (

(

(

6 15 10 . 10 6 15 i) Show that no two by two minor of A has a unit in R as determinant. ii) Show the map R3 → R2 defined by r &→ Ar is surjective anyway. Note that there is no unit in the matrix. iii) Can you see how A was constructed?

Exercise 30 Let R = Z / 30 Z and A =

An example of the before mentioned process is in order:

Example 1 A family of squares of order 9. Let R = F3. We take the standard bijec-

tions equivalent to base three representation. We take an (almost) random set of free submodules: )e0, e1*, )e0, e2*, )e2, e3*, )e0 + e1, e3*, )e1, e2 + e3*, )e0 + e1, e2 + e3*, )e0 − e2, e1 − e3*, )e0 + e1, e2 + e3*, )e0 − e1*, and )e2 − e3*. The process above results in a short list of only 8 remaining possible rows:

(1, 0, 0, 1), (2, 1, 0, 1), (0, 1, 1, 0), (0, 1, 2, 0) (1, 0, 0, 2), (2, 1, 0, 2), (0, 1, 1, 2), (0, 1, 2, 1) From the 70 possible choices5 of 4 rows, there are 58 that have a unit as determinant. The happy coincidence is that all of the resulting squares are 3 × 3-ply pandiagonal magic squares. We have not checked for other consequences of the choice of submodules. An impression of what the collections of parallel affine submodule look like, is given in figure 2.6. We have given a ‘color’ to each of the parallel affine submodules to show which partitions will have equal sums. We left out the columns, we assume you can guess how they would look.

Example 2 In ‘Mission impossible’ we proved the well-known transformation from associative squares of even order to a special kind of pandiagonal squares in lemma 5 (page 28). These transformed squares of even order (2m) are pandiagonal since ar,c + ar+m,c+m is constant for all r and c. Note this only depended on a reordering of the rows and columns. This can be seen in a different way. 5

There are more; but I have taken the rows of L in the order they are in the list. Any reordering of the rows or multiplying rows with units is also allowed. And more bijections L result from adding any translation vector, of course. A fine magic key

BF

(a) )(1, 0, 0, 0), (0, 0, 1, 0)*

(b) )(0, 0, 1, 0), (0, 0, 0, 1)*

(d) )(0, 1, 0, 0), (0, 0, 1, 1)*

(e) )(1, 1, 0, 0), (0, 0, 1, 1)* (f) )(1, 0, −1, 0), (0, 1, 0, −1)*

(g) )(1, 0, 1, 0), (0, 1, 0, 1)*

(h) )(1, −1, 0, 0)*

(c) )(1, 1, 0, 0), (0, 0, 0, 1)*

(i) )(0, 0, 1, −1)*

Figure 2.6: Nine collections of parallel affine submodules of F43

Take F2n 2 , some n > 1. Use the binary representation for the identification by ρ = κ. Then suppose we have an affine bijection L : 2n F2n 2 → F2 such that the coordinate maps are surjective on the parallel affine submodules of )e0 + en*. In other words if the sum of the entries ar,c + ar+2n−1 ,c+2n−1 is constant! By taking different ρ and κ as suggested by lemma 5, we see that associativity of even order squares can be acquired using theorem 1! BM

I. Generalizing

Exercise 31 What submodule could you use instead of the submodule )e0 + en* in the above example, when R = F2 × Z/mZ, with the ‘usual bijections’?

Example 3 A family of normal pandiagonal associative magic squares. Let R = Z/qZ with gcd(6, q) = 1, ρ = κ of type c ∈ R, T a weighted bijection of type e ∈ R2. Define L : R2 → R2 by ' ( ' (' ( ' ' (( r0 1 2 r0 c −1 L: &→ +2 e−3 r1 2 1 r1 c Note L is a bijection, and )e0*, )e1*, )e0 +e1* and )e0 −e1* all are ‘magic’, as well as their parallel affine submodules, since ±1, 2, ±3 are units in R. The choice of the translation-vector is an application of proposition 3.

Example 4 Let R = F2[x]/(x2 + x + 1) (, F4) and" ρ = κ given by ρ(0) = 0, ρ(1) = 1, ρ(x + 1) = 2, ρ(x) = 3. Let T (r) = notation). And define



  r0  r1   &  L:   r2  →  r3

0 x 1 0

x 0 0 1

1 0 0 1

3−i i ρ(ri)2

 0 r0  r1 1   1   r2 0 r3

(as the binary

   

Then M is a normal pandiagonal 4 × 4-ply6 magic square of order 16, with all ‘bent diagonals’ (of length 16) the same sum. A bent diagonal is, loosely speaking, like an ordinary diagonal except that if you follow it from a corner in the direction of the center, it changes direction half way through, and follows ‘another diagonal’.

❛ Computers are useless, they can only give you answers 6

❜

Pablo Picasso (1881-1973)

This is not exactly trivial. Good luck :-) A fine magic key

BD

Variations and Sudoku’s In the proof of theorem 1 the reason that an affine submodule had constant sums was because it contained for each of the coordinate sums the same elements the same number of times. This can be used to our advantage in at least two ways.

Variations by variables By the reason just mentioned, the precise numbers are irrelevant. This suggests the possible use of variables. If T maps to variables instead of numbers, we can more easily control specific sums or perhaps get other properties. We first show a few special (non-normal) squares of order 6, often considered difficult to create.

Just not normal Example 5 Take for ρ, κ the standard bijections from R = Z/6Z to 6 and take

T ((r0, r1)) = τ(0)(r0) + τ(1)(r1) where τ(0), τ(1) are bijections from R to the set of variables {a, b, c, A, B, C} respectively {x, y, z, X, Y, Z}. Define the linear map L using the matrix ' ( 3 2 2 3 The resulting square is:

x+a z+A Y+a x+A z+a Y+A

X+c Z+C y+c X+C Z+c y+C

x+B z+b Y+B x+b z+B Y+b

X+a Z+A y+a X+A Z+a y+A

x+c z+C Y+c x+C z+c Y+C

X+B Z+b y+B X+b Z+B y+b

Note it is already pandiagonal. It is only magic if a + A = b + B = c + C = x + X = y + Y = z + Z and 2(x + Y + z) = 2(X + y + Z) = 2(a + B + c) = 2(A + b + C) = 3(a + A). In other words, the square is magic precisely if the following two rectangles are magic rectangles.

a B c

x Y z

A b C

X y Z

Since the square is already pandiagonal, we know, by lemma 6 (page 29), that it cannot be made normal. If we want the square to be 2×2-ply, then the requirements become X + x = Y + y = Z + z and A + a = B + b = C + c. If we want the LK

I. Generalizing

square to be 3 × 3-ply, X + y + Z = x + Y + z and A + b + C = a + B + c. In other words this square is magic if and only if it is 2x2-ply and 3 × 3-ply! Can you figure out which values are chosen in figure 2.1 at page 30?

Example 6 A 2 × 3-ply square of order 6, constructed similar to the previous example.

A+a C+b E+c A+d C+e B+d D+e

F+f

B+a D+b F+c

C+a E+b A+c C+d D+d

F+e

E+f

E+e

A+f

B+f D+a F+b

B+c

E+a A+b C+c E+d

A+e C+f

F+d

B+b D+c

B+e D+f F+a

It will also be semi-magic as long as a+d = b+e = c+f and A+C+E = B+D+F . It cannot be normal and semi-magic because of the second requirement, since 0 + 1 + . . . + 5 = 15 is odd. If a + c + e = b + d + f and 3(A + D) = A + B + C + D + E + F the diagonals will be magic.

Exercise 32 What is the matrix from the map L used in the previous example? Being bi A bimagic square of order 9, whatever we fill in for the variables! X+x+A+a Z+z+A+c Y+y+A+b Z+x+B+b Y+z+B+a X+y+B+c Y+x+C+c X+z+C+b Z+y+C+a

Y+x+B+c X+z+B+b Z+y+B+a X+x+C+a Z+z+C+c Y+y+C+b Z+x+A+b Y+z+A+a X+y+A+c

Z+x+C+b Y+z+C+a X+y+C+c Y+x+A+c X+z+A+b Z+y+A+a X+x+B+a Z+z+B+c Y+y+B+b

Z+y+A+c Y+x+A+b X+z+A+a Y+y+B+a X+x+B+c Z+z+B+b X+y+C+b Z+x+C+a Y+z+C+c

X+y+B+b Z+x+B+a Y+z+B+c Z+y+C+c Y+x+C+b X+z+C+a Y+y+A+a X+x+A+c Z+z+A+b

Y+y+C+a X+x+C+c Z+z+C+b X+y+A+b Z+x+A+a Y+z+A+c Z+y+B+c Y+x+B+b X+z+B+a

Y+z+A+b X+y+A+a Z+x+A+c X+z+B+c Z+y+B+b Y+x+B+a Z+z+C+a Y+y+C+c X+x+C+b

Z+z+B+a Y+y+B+c X+x+B+b Y+z+C+b X+y+C+a Z+x+C+c X+z+A+c Z+y+A+b Y+x+A+a

X+z+C+c Z+y+C+b Y+x+C+a Z+z+A+a Y+y+A+c X+x+A+b Y+z+B+b X+y+B+a Z+x+B+c

I constructed this one using F43, standard ρ and κ (of type 2 of course) and L defined by the matrix  

1 2  1 0   0 2 2 2

2 0 1 2

2 1   0  1

As before, take T ((r0, r1, r2, r3)) = τ(0)(r0) + . . . + τ(3)(r3). With τ(0) (resp. ρ(1),τ(2),τ(3)) sending F3 to {a, b, c} (resp. {A, B, C}, {x, y, z}, {X, Y, Z}). The choice of matrix for the affine map will be made clear in ‘Existence’ (page 55). Vaƒations and Sudoku’s

LA

Exercise 33 Can you make a bimagic square with your birthdate (dd-mm-19-yy) somewhere?

Sudoku’s By the same reason as noted at the beginning of this section (‘Variations and Sudoku’s’), we can make Sudoku’s. It is ironical to note that most western people know the puzzle by the Japanese name Sudoku (or Su Doku), while most Japanese know it by the english name ‘Number place’. Anyway, we only need to choose ρ and κ somewhat more carefully, to be able to make the well-known variants with rectangular blocks. But first the ‘regular’ Sudoku’s.

Sudoku’s of order n2 A Sudoku is a square of order n2 for which every row, column and ‘standard’ square block of order n contains exactly the numbers from the set n2. Order 9 are still most popular. Take as finite ring R = Z/nZ. Write down the affine map L : R4 → R2 as



   r0 ' ( r0 ' (  r1   r1  a a a a t 0,0 0,1 0,2 0,3 0 &  + L:   r2  → a1,0 a1,1 a1,2 a1,3  r2  t1 r3 r3

Take ρ, κ and T to be the standard bijections. The rows ()e0, e1*), columns ()e2, e3*) and blocks ()e1, e3*) need to contain the same elements. By theorem 1, if the following minor’s have determinant a unit in R, then the resulting square will be a Sudoku.

7 7 a0,0 a0,1 7 7 a1,0 a1,1

Sudoku’s of order mn

7 7 7 7

7 7 a0,2 a0,3 7 7 a1,2 a1,3

7 7 7 7

7 7 a0,1 a0,3 7 7 a1,1 a1,3

7 7 7 7

Any number from 0 to nm − 1 can be uniquely written as am + b with a ∈ n and b ∈ m. This is also called the mixed radix representation. We make a square of order nm with each row, column and standard n × m-block containing nm, an Sudoku with n × m blocks. Use R = Z/nZ × Z/mZ, κ : (a, b) &→ a + bn, and ρ : (a, b) &→ am + b where a ∈ n and b ∈ m. Then with the assumption that a0,0, a1,0 ∈ Z/nZ, and a0,1, a1,1 ∈ Z/mZ, take T and L as follows;

Mρ((r0 ,r1 )),κ((c0 ,c1 )) = ρ((a0,0r0 + a1,0c0 + t0, a0,1r1 + a1,1c1 + t1))

LI

I. Generalizing

The requirements on the rows, columns and blocks result by theorem 1 in the requirements on the ai,j. It is a nice exercise (see below) to work these out. A few examples with (1, 1), (1, 1) ∈ Z/2Z × Z/2Z, (1, 2), (1, 1) ∈ Z/2Z × Z/3Z, resp. (1, 1), (1, 1) ∈ Z/4Z×Z/2Z. Since it is customary to use consecutive numbers starting from 1, we added 1 to each entry.

1 2 3 4

3 4 1 2

2 1 4 3

4 3 2 1

1 2 5 6 3 4

3 4 1 2 5 6

5 6 3 4 1 2

2 1 6 5 4 3

4 3 2 1 6 5

6 5 4 3 2 1

1 2 3 4 5 6 7 8

5 6 7 8 1 2 3 4

2 3 4 1 6 7 8 5

6 7 8 5 2 3 4 1

3 4 1 2 7 8 5 6

7 8 5 6 3 4 1 2

4 1 2 3 8 5 6 7

8 5 6 7 4 1 2 3

Exercise 34 Figure out what T and L are in the usual notation. Write down a basis for the rows. Do you see what the requirements are? Do the same for the columns and n × m blocks. Check your findings with a few examples, like (2, 1), (1, 1) ∈ Z/4Z × Z/3Z.

Exercise 35 Make a partially filled Sudoku with exactly 2 solutions. What about one with exactly 4 solutions? 5?

In the same way one can make so called ‘detrivializers’ (see ‘Compound intrest’, page 48). In a somewhat similar way it is possible to, as a rule, make r×c-ply squares of order rc, but that will be put to paper another time.

❛ If dealing with trifles like magic squares is worthy of an Euler or a Cayley; if in short it is legitimate, it is because amusement has value. ❜ [McC97, p.100]

Vaƒations and Sudoku’s

LB

The same repeated but ever so slightly different The square of order 9 at the right of figure 0.3(d) at page 10 has a rather clear relation with the Lo Shu. If you have not yet tried to determine this relation, now would be a good time to spend a few minutes on it. It is a nice example of one of the oldest known ways of using small magic squares to obtain larger magic squares and I call it compounding. Interestingly, compounding makes the set of all (n-dimensional) magic squares with this operation into a free monoid (see [Adl97]). Compounding, in various guises, is very old and has been often reinvented (cf. [Ful74, chap.6], [WR89], [ASY77]). Strangely enough, I was unable to find any references or publications of proofs on how various properties (besides magic) can ‘propagate by compounding.’ It is certainly a simple way to furnish existence proofs. That is what this section is all about. First an example:

7 0 5 2 4 6 ⊗ 3 8 1

0

13

3

14

7

10

4

9

12

1

15

2

11

6

8

5

=

7

0

5

124 117 122

34

27

32

133 126 131

2

4

6

119 121 123

29

31

33

128 130 132

3

8

1

120 125 118

30

35

28

129 134 127

70

63

68

97

90

95

43

36

41

88

81

86

65

67

69

92

94

96

38

40

42

83

85

87

66

71

64

93

98

91

39

44

37

84

89

82

115 108 113

16

9

14

142 135 140

25

18

23

110 112 114

11

13

15

137 139 141

20

22

24

111 116 109

12

17

10

138 143 136

21

26

19

106

104

61

54

59

79

72

77

52

45

50

101 103 105

56

58

60

74

76

78

47

49

51

102 107 100

57

62

55

75

80

73

48

53

46

99

To be more precise: Let A be a square of order m and B a square of order n. We define a square A ⊗ B of order mn by

(A ⊗ B)i+mk,j+ml = ai,j + m2bk,l where 0 ≤ i, j ≤ m − 1 and 0 ≤ k, l ≤ n − 1. LL

I. Generalizing

A few basic examples:

0 2 0 0 0 0 1 0 ⊗ 0 0 0 = 2 3 2 0 0 0 0 2

1 3 1 3 1 3

0 2 0 2 0 2

1 3 1 3 1 3

0 2 0 2 0 2

0 0 0 9 9 9 0 0 0 9 9 9 0 0 0 0 0 0 9 9 9 0 1 0 0 0 ⊗ = = 32 · 2 3 18 18 18 27 27 27 0 0 0 18 18 18 27 27 27 18 18 18 27 27 27

1 3 1 3 1 3 0 0 0 2 2 2

0 0 0 2 2 2

0 0 0 2 2 2

1 1 1 3 3 3

1 1 1 3 3 3

1 1 1 3 3 3

Note that almost always A ⊗ B #= B ⊗ A !

Exercise 36 I have said before that the squares of order 9 from figure 0.3(d) at page 10 have a clear connection with the Lo Shu. Can you figure out for both squares what relation I meant? The following proposition shows the construction’s main significance.

Proposition 7 Let A and B be squares with A of order m and B of order n. i) ii) iii) iv) v) vi) vii)

If A and B are normal, then so is A ⊗ B If A and B are associative, so is A ⊗ B If A and B are pandiagonal, so is A ⊗ B if A and B are (semi-)magic, then so is A ⊗ B If A is s-partitioned, then so is A ⊗ B If A is magic and B is s-partitioned, then A ⊗ B is sm-partitioned If A is semi-magic and B is c × r-ply, then A ⊗ B is cm × rm-ply

Proof: I use the same notations as in the definition of A ⊗ B above. Most proofs follow the following guideline: show that the sum of two squares that both possess property P is also P . Then use A ⊗ B = A ⊗ (0)n + (0)m ⊗ B where we denote with (0)s the zero-matrix of order s. Since both summands are particular simple to imagine, the checking of the property P is much less of a problem. With this in mind most items are straightforward administrative chores (try a few yourself) and below I only show the most interesting/difficult steps. ii) Since both squares are associative, there are constants A, B such that

A = ai,j + am−1−i,m−1−j B = bk,l + bn−1−k,n−1−l

for all for all

0 ≤ i, j ≤ m − 1 and 0 ≤ k, l ≤ n − 1

|e same |- .'&-

|f gbdf

LC

So take any i, j, k, l and again with notation α = mk + i and β = ml + j, and since mn − 1 − α = m(n − 1 − k) + m − 1 − i we see that

(A ⊗ B)α,β + (A ⊗ B)mn−1−α,mn−1−β = 5 6 5 6 = ai,j + m2bk,l + am−1−i,m−1−j + m2bn−1−k,n−1−l = (ai,j + am−1−i,m−1−j) + m2 (bk,l + bn−1−k,n−1−l) = A + m2B

which is clearly constant. iii) Note that the sum of two pandiagonal squares is pandiagonal. Since any broken diagonal in A ⊗ (0)n is precisely n times a broken diagonal of A, it is clear that A ⊗ (0) is pandiagonal if A is. For (0)m ⊗ B note that any broken diagonal is either, m times the (down- or upward) main diagonal of B, or either the sum of b times the numbers from one broken diagonal of B and m − b times the numbers from the broken diagonal of B right next to it (1 ≤ b ≤ m − 1). So (0)m ⊗ B is pandiagonal if B is. " The remarks at the start of the proof of the previous proposition are so useful that we will make definitions accordingly. It also makes it much easier to talk about various compounds. A property P (like being semi-magic, or normal or…) is called additionable if for any two squares A and B of the same order that have property P , the sum A + B also has property P . A property is called compoundable or propagates by compounding when for any squares A and B with the property, also A ⊗ B necessarily has the property. Two special cases deserve their own name. A property is called tilable if for any positive number s and any square A with property P , also A ⊗ (0)s has property P . You can also say that the property propagates by compounding from the right with (0)s. Similarly a property can be blown up if it propagates by compounding from the left with (0)s for any s.

Exercise 37 Show that if A and B are magic then A⊗B is m-partitioned and magic where the order of A is m.

Exercise 38 Continuation of exercise 36. Which items of proposition 7 would still be true if you used the way the left square of figure 0.3(d) was made as definition for compounding? Most compounded squares show the structure of sub-squares very clearly and look like a stack of distinct squares. This will addressed shortly, but the way they are stacked when compounding makes that the properties propagate. The following exercise shows this clearly.

Exercise 39 Show that a random square stack of pandiagonal squares will not automatically be pandiagonal. LJ

I. Generalizing

Magical powers Exercise 40 Show that the sum of two bimagic squares is in general not bimagic. In ‘Proper properties’ we have seen bimagic squares. As the previous exercise shows, being bimagic is not an additionable property.

Exercise 41 Show that the property of a square A which asserts that (a3ij) is magic, does not propagate by compounding. Does it propagate when compounding from left or right with a ‘special’ matrix?

The straightforward generalized property of requiring M!d, the matrix obtained from M by raising each entry of M to the power d, to be magic for some specific number d (e.g. 3 as above) is not something that might easily benefit from compounding. However consider the following definition: a square A is called pmultimagic if for each integer e from 1 to p the square A!e is a magic square. So in particular, being a 2-multimagic square is equivalent to being a bimagic square. The perhaps most surprising fact is the following

Proposition 8 If A and B are p-multimagic, then so is A ⊗ B Proof: I only show that the column-sum is independent of the column-index. The other sums be shown " to be constant in a similar way. Fix 1 ≤ e ≤ p. We denote " can x Sx = i Ai,j and Tx = i Bxi,j for 1 ≤ x ≤ p. Since A and B are p-multimagic squares these Sx and Tx are constants. Fix the column-index β = ml + j and write α = mk + i. mn−1 # α=0

(A ⊗

B)eα,β

=

m−1 n−1 ##

e ' ###

k=0 i=0

5 6e ai,j + m2bk,l

( # # 'e( # e 5 2 6e−x x = m bk,l ai,j = m2(e−x)be−x axi,j k,l x x x k i x=0 k i ' ( ' ( ## e # e # = m2(e−x)Sxbe−x m2(e−x)Sx be−x k,l = k,l x x x x k k # 'e( = m2(e−x)SxTe−x x x

So the column-sum is independent of it’s index β, hence a constant.

"

For me this suggests that p-multimagic squares are a ‘nice’ generalization of bimagic squares. |e same |- .'&-

|f gbdf

LF

Compound interest If you want to make a magic square of order 6, then you can not do this directly by compounding (really?). But take a look at figure 0.3(c) at page 9. It is purported to have a clear relation with the Lo Shu. Yet again. How was it made? If we look at the main reasoning of proposition 7, as outlined at the start of its proof, then it is obvious how to proceed. We start with Lo Shu ⊗(0)2 or (0)2 ⊗ Lo Shu and have to add a square of order 6 to make it into a normal square. I will call such a square a detrivializer. A more explicit definition later.

12 14 35 34

4

6

3 3 8 8 1 1

0 2 3 2 0 2

15 13 32 33

5

7

3 3 8 8 1 1

3 1 0 1 1 3

10

8

18 19 26 24

11

9

16 17 27 25

− 22 ·

2 2 4 4 6 6 2 2 4 4 6 6

=

2 0 2 3 2 0 3 1 0 1 3 1

28 30

3

2

22 20

7 7 0 0 5 5

0 2 3 2 2 0

29 31

1

0

21 23

7 7 0 0 5 5

1 3 1 0 1 3

I reconstructed above the detrivializer used in the construction of figure 0.3(c) by subtracting (0)2 ⊗ Lo Shu. Can you see why the detrivializer does its work? You might want to take some time to figure it out before proceeding. The detrivializer works because at all the positions that contain the same number, the detrivializer contains precisely the numbers 0, 1, 2, 3. In general, to ‘detrivialize’ a trivial square as A " = (0)n ⊗ A, for each number in A the set of positions in A " that contain that number, should correspond in the detrivializer to the set n2. To detrivialize a trivial square as A " = A ⊗ (0)n, with A of order m, we proceed similar, the only difference being that we have to use the set m2n2, or equivalently we have to add the detrivializer m2 times. This is the only requirement to make these trivial squares normal when A is normal. However if we want the resulting square to be e.g. (semi-)magic, then the detrivializer has to have the same properties. This only works of course if the property is additionable. As we have seen in the proof of proposition 7 most properties are additionable. Notable exception is p-multimagic-ness.

Example 7 A most simple case. A sample detrivializer for

0

1

2

3

like this:

0 1 2 3

0 1 0 1 2 3 2 3 0 1 0 1

+ 22 ·

2 3 2 3 LM

I. Generalizing

3 2 1 0 1 0 3 2 2 3 0 1

0 =

⊗ (0)2 wil work

5

8

13

14 11

6

3

4

1

12

9

10 15

2

7

And indeed it is normal. Note it is not semi-magic or associative (which

0

1

2

3

is),

since the detrivializer is not. It is pandiagonal though.

Example 8 To the left a detrivializer of Lo Shu ⊗ (0)3. The resulting magic square

is at the right. It certainly does not look as ‘regular’ as the two squares from figure 0.3(d) at page 10!

8 3 5 3 1 0 1 8 7

75 35 46 30 17

4 1 2 2 8 4 6 3 6

38 13 24 20 76 42 56 31 60

6 1 3 4 3 2 2 8 7

61

4 2 1 2 6 8 6 4 3

39 26 10 21 62 73 57 44 28

8 6 7 3 4 0 1 2 5

74 58 69 29 40

5 6 1 0 2 6 7 4 5

52 54 14

0 7 6 7 5 4 5 0 2

3

71 55 66 53 37 48

0 5 3 7 0 8 5 7 1

2

49 33 65

1 5 8 8 7 4 3 0 0

16 45 77 79 63 41 34

9

1

12 80 64

32 43 27 23 25 72 68

7

6

11 22 51

18 59 70 36 50 4

8

19

78 47 67 15 0

5

So how to make detrivializers with nice properties? And what do we gain?

Exercise 42 It is possible to make a detrivializer for (0)2 ⊗ Lo Shu such that the result will be magic. Can you?

Example 9 The square of Khajuraho (see page 1) is rather remarkable for being pandiagonal and 2 × 2-ply. By ordinary compounding we lose the 2 × 2-ply-ness. Using detrivializers this is not a property that is difficult to keep. Use for instance the following detrivializer for Khajuraho ⊗ (0)2:

2 3 2 3 0 1 0 1

1 0 1 0 3 2 3 2

2 3 2 3 0 1 0 1

1 0 1 0 3 2 3 2

3 2 3 2 1 0 1 0

0 1 0 1 2 3 2 3

3 2 3 2 1 0 1 0

0 1 0 1 2 3 2 3

Can you see how this easily extends to orders a multiple of four?

Lemma 9 There exists magic detrivializers for any s and any magic square A compounded from left or right with (0)s Proof: By proposition 7 i) and iv) and the fact that there exists magic squares for all orders larger than two, we see s ≥ 3 is proven. Since s = 1 is trivial, we only need |e same |- .'&-

|f gbdf

LD

to look at s = 2. The case that the order of A is three has already been disposed of in the example and in the not-too-hard exercise 42. First A⊗(0)2 with A order 4 or more. Look at the right square. The quadrants of the detrivializer of example 2 can be put in the corresponding empty spots if we let the empty spots have order 2. The result will be a magic detrivializer of order 8. The quadrants of the square of exercise 42 can be used if the empty spots have order 3. You get a magic detrivializer of order 10. In precisely the same way, by applying the template recursively, we get detrivializers for all even orders. The case (0)2 ⊗ A is left as an easy exercise (43) to the reader.

0

3

0

0

1

2

3

3

0 0

···

···

0

1

2

3

3

3

0

3

0

0

···

0

3

1

2

3

1

2

. . .

. . .

. . .

. . .

0

3

1

2

2

1

1

1

1

3

0

2

2

···

2

3

0

2

2

2

2

1

1

1

3

0

2

1

3

0

2

1

. . .

. . .

. . .

. . .

3

0

2

1

···

··· ···

···

3

1

"

In a sense it is rather silly that I have used direct combinatorial proofs. It is just as easy to apply the same ideas as e.g. used to make the various Sudoku’s to these detrivializers.

Exercise 43 Complete the proof of lemma 9. Prove this also using affine maps. Hint: F2 × Z/nZ.

Exercise 44 Make a detrivializer for Khajuraho ⊗ (0)2 (see page 1) that is 2 × 2ply pandiagonal magic and ‘4-partitioned’. Show that the detrivializer can be used to generate 2 × 2-ply pandiagonal 4-partitioned (Franklin) magic squares with an order a multiple of 8.

Exercise 45 Is it possible to make pandiagonal detrivializers for (0)2 ⊗ A with A pandiagonal? And for A ⊗ (0)2? And if they also have to be semi-magic?

Exercise 46 There have been many methods in the past to make magic squares of order 2n when given a magic square of order n, most notably by Candy ([Can34, Can35]) and Strachey ([BC87, p.196-199]). The last has found its way in almost all recent recreational books. Compare the known methods to this one with respect to flexibility, ease of use, and possible propagation of properties.

Other blocks I am mostly interested in squares with certain patterns having constant sums, but proposition 7 and the forgoing generalizes readily to rectangles, cubes, and higher dimensional figures. See for instance figure 2.7 and the next (tricky but possible) exercise. CK

I. Generalizing

Exercise 47 Below is a normal magic cube of order 3 in ‘slices’ as seen in [And17, p.66, fig.158].

9

25

5

22

2

23

0

16

6

13 20

7

14 18

11 24

15

8

4

12 19

10 26

3

21

17

1

The question is: can you make a normal magic cube of order 6 out of it?

6

13

0

8

13

9

3

4

9

2

10

0

2

7

12 14 ⊗ 3

7

11 =

8

10 11

5

4

12

5

6

14

1

1

208 204 198 199 201

13

9

3

4

6

133 129 123 124 126

195 197 202 207 209

0

2

7

12 14

120 122 127 132 134

203 205 206 200 196

8

10 11

5

1

128 130 131 125 121

148 144 138 139 141

43 39 33 34 36

163 159 153 154 156

135 137 142 147 149

30 32 37 42 44

150 152 157 162 164

143 145 146 140 136

38 40 41 35 31

158 160 161 155 151

58 54 48 49 51

118 114 108 109 111 178 174 168 169 171

45 47 52 57 59

105 107 112 117 119 165 167 172 177 179

53 55 56 50 46

113 115 116 110 106 173 175 176 170 166

73 69 63 64 66

193 189 183 184 186

88 84 78 79 81

60 62 67 72 74

180 182 187 192 194

75 77 82 87 89

68 70 71 65 61

188 190 191 185 181

83 85 86 80 76

103 99 93 94 96

223 219 213 214 216

28 24 18 19 21

90 92 97 102 104 210 212 217 222 224

15 17 22 27 29

98 100 101 95 91

23 25 26 20 16

218 220 221 215 211

Figure 2.7: A normal associative ‘3 × 5-partitioned’ magic square of order 15

|e same |- .'&-

|f gbdf

CA

❛ Die Mathmatiker sind eine Art Franzosen: redet man zu ihnen, so übersetzen sie es in ihre Sprache, und dann ist es alsobald ganz etwas Anderes. ❜ Goethe (1893)

❛ Le secret d’être ennuyeux c’est de dire tout.

CI

❜

Voltaire

3

Y]•9[y3Z[4 „\]30@ Want to raise powers? Well, superman, eat (one of) your heart(s) out!

This chapter is the main application of the ideas presented in the previous chapters. Large parts will appear in the American Mathematical Monthly october 2007 as an article by Harm Derksen, Arno van den Essen and myself. The preprint [DEEa] and internal report [DEEb] are much more technically formulated. All are far less general than the methods considered here.

Introduction

As noted in ‘Bimagic squares’ at page 8, the first 2-multimagic (or bimagic) square was published early 1891. In 1905 the first 3-multimagic square, a square of order 128, was constructed by Tarry. In 2001 both 4- and a 5-multimagic squares were constructed by Boyer and Viricel, squares of order 512 and 1024, respectively (see [Boy01] and [Boy], where a nice history of the subject is given). The record was a 6multimagic square of order 4096 constructed by Pan Fengchu in October 2003 [Fen]. However, the following question remained open:

Do there exist n-multimagic squares when n ≥ 7?

A fine sum of powers In ‘A fine magic key’ (page 30) we looked at sums of entries of rows, or as it turned out, parallel affine submodules. This resulted in theorem 1, which is a rather flexible tool to obtain constant sums. How to do the same with sums of powers of entries? Since the sums of the entries to the power one have to be constant too, we will assume that we are in the same setting as in theorem 1. So as before, let S = )s0, s1, . . . , sd−1* be a free submodule of R2n, and let T be a weighted sum bijection of ρ(i)’s (i ∈ 2n). Consider the sum of entries of M raised to the power p, where CB

the entries range over the elements in a affine (or ‘linear’) submodule S + x:

#

Mpρ(a),κ(b) =

#

T (L(r + x))p

r∈S

(a,b)∈S+x

Note that T is a linear combination of ρ(i)’s. So we are looking at a sum (x0 + x1 + k−1 . . . + x2n−1)p. The terms of this sum are of the form αxej00 · · · xejk−1 , in which α is a positive integer, 1 ≤ j0 < j1 < . . . < jk−1 ≤ 2n, e0, . . . , ek−1 ≥ 1, and e0 + · · · + ek−1 = p. As in the proof of theorem 1, we look at the individual terms. By the remark above, such a term looks like (without the leading constant factor):

# r∈S

ρ(j0 )(L(r + x)j0 )e0 · · · ρ(jk−1 )(L(r + x)jk−1 )ek−1

Rewrite this by denoting cj = )rowj | x* + tj and using the fact that the L(i)’s are surjective:

#

r∈Rd

5 6e 5 6e ρ(j0 ) L(j0 )(r) + cj0 0 · · · ρ(jk−1 ) L(jk )(r) + cjk−1 k−1

(3.1)

Note the sum is now over r ∈ Rd. At this point in the proof of theorem 1, we used the surjectivity of L(i) to sum over fibers of L(i). But note that in this case, a term is not related to just one ‘coordinate map’ L(i), but to a combination of several! Therefore, put J = {j0, . . . , jk−1} and define the linear map L(J) : Rd → Rk by

L(J)(r) =

5 6 L(j0 )(r), . . . , L(jk−1 )(r)

and assume it is surjective. In particular, this means that k ≤ p ≤ d. Since the linear map L(J) is assumed to be surjective, we have for each a ∈ Rk −1 a non-empty collection L−1 (J) (a). Say it contains b, so L(J)(b) = a. Then L(J) (a) = d−k b+ker L(J). Since Rd/ ker L(J) , Rk we see that |L−1 for each a ∈ Rk. (J) (a)| = |R| Note that Rd is a disjoint union of the collections L−1 (J) (a) where a runs through −1 Rk. Since the collections L(J) (a) are all of the same size |R|d−k, we see that the sum

in (3.1) is equal to

|R|d−k

#

a∈Rk

5 6e ρ(j0 ) (a0 + cj0 )e0 · · · ρ(jk−1 ) ak−1 + cjk−1 k−1

which in turn is equal to

|R|d−k

3

#

a0 ∈R

CL

ρ(j0 ) (a0 + cj0 )e0

4

B. Multimagic Squar@



···

#

ak−1 ∈R

 5 6ek−1  ρ(jk−1 ) ak−1 + cjk−1

Since each ρ(jm ) is a bijection, we get

"

ρ(jm )(a + cjm )ejm =

a∈R

|R|−1 "

iejm . So the

i=0

sum of each term is independent of x, and hence also the total sum we started with. Note that if L(J) is surjective then L(K) is surjective for any non empty K ⊆ J. In other words, I have proven

Theorem 2 Let M be a square defined as on page 25. Let T be a weighted sum bijection and S ⊆ R2n a free R-module of rank1 d. Let p ≤ d. If for each subset J of 2n with |J| = p, the map L(J) : Rd → R|J| as defined above is surjective, then the sum of the entries of M!e taken over S, and of any parallel affine module of S, is constant, for each 1 ≤ e ≤ p. As with theorem 1 there is considerable freedom of choice w.r.t. the translationvector of L and the bijections T, ρ and κ. Note that the highest power p for an affine submodule is bounded by its rank d. It is, as is to be expected, considerable 52n6 more work to make an affine submodule ‘p-multimagic’. You need to prove p maps L(J) to be surjective! In the next sections we shall see how to use this theorem to prove the existence of n-multimagic squares for all n.

Existence To prove the existence of n-multimagic squares for all n we translate the problem into a problem concerning the existence of some special matrix. This special matrix should of course correspond to the map L from an (n-multimagic) square M in the setting of diagram (2.2). The key is theorem 2. So assume T to be a weighted sum bijection. Consider the affine submodules corresponding to the rows, columns and both diagonals. Since the highest power p, in the notation of theorem 2, is bounded by the rank d of the affine submodule under consideration, we need to have at least rank n for the rows, columns and diagonals. Let say the rank for each is n. How to translate the requirement of the maps as mentioned in theorem 2 to requirements for the matrix A of L? Consider the columns, these are parallel affine submodules of the free submodule )e0, e1, . . . , en−1*. So if we take P (respectively, Q) to be the 2n × n matrix formed by the first (respectively last) n columns of A, then each linear map L(J), as in theorem 2, will have a matrix that consists of some rows of P. If we require all n × n minors of P to be units in R, then all the maps L(J) with |J| ≤ n will be surjective (check this!). Similarly, for the rows take )en, en+1, . . . , e2n−1*, then as requirement we take that all n × n minors of Q must be units in R. 1

or dimension, in both cases it is the number of elements in a basis. Ex}tence

CC

The diagonals are somewhat trickier since there ρ and κ can change the picture. So what are the entries of the main diagonals? First of all if we assume that ρ = κ then the sum of the entries of the main downward diagonal is the same as when we take the sum of the entries over )e0 + en, e1 + en+1, . . . , en−1 + e2n−1*. So we also require all the minors of order n from P + Q to be units of R. If we also assume ρ to be a bijection of type c then by noting that qn − 1 − ρ(a) = ρ(−a + c) we can use )e0 − en, e1 − en+1, . . . , en−1 − e2n−1* for the upward main diagonal, similar as for the downward main diagonal. So call a 2n × 2n matrix A with entries from R an n-multimagic generator matrix if all n × n minors of P, Q, P + Q, and P − Q are units in R, where P (respectively, Q) is the 2n × n matrix formed by the first (respectively last) n columns of A. Summarizing we obtain the following corollary.

Corollary 10 Notations as in diagram (2.2), page 25. Let R be a finite ring, with |R| = q. Let ρ = κ : Rn → qn be a bijection of type c and T a weighted sum bijection. If the matrix of L is an n-multimagic generator matrix, then the qn × qn square M is a n-multimagic square.

So how to construct a n-multimagic generator matrix? Since bijections of type c ∈ R exists for all R by lemma 4, and also T will be no problem to construct, we are completely free to choose R.

A class of n-multimagic generator matrices Assume that all the elements 1, 2, . . . , 2n − 2 are units in R. If n = 2 then also assume additionally that 3 is a unit in R. Then, when 0 ≤ j < i ≤ 2n − 2, the differences i − j are units in R. By appealing to Vandermonde determinants, it then follows that all n × n minors of the following 2n × n matrix P are units in R:

          

1 1 1 1 .. .

0 1 2 3 .. .

0 1 4 9 .. .

1 (2n − 2) (2n − 2)2 0 0 0

... ... ... ...

0 1 2n−1 3n−1 .

.. ... . . . (2n − 2)2n−1 ... 1



     .    

Using this matrix 5P1 6 it is easy to make an n-multimagic generator matrix A as follows: write P = P2 , where P1 (respectively, P2) is the n × n matrix formed by the first (respectively, last) n rows of P and set Q = CJ

B. Multimagic Squar@

5 2P1 6 −2P2

. Then define A = (P|Q).

Using elementary column operations one can reduce

A= to the matrix

'

'

(

P1 2P1 P2 −2P2

P1 0 P2 −4P2

(

,

which is clearly invertible because both det P1 and det(−4P2) are units in R. It follows without difficulty that A is an n-multimagic generator matrix. In case you are wondering about 3 having to be a unit in R when n = 2, then notice that in this construction the first n rows of P + Q are 3P1. Anyway, if we take R to be a finite field with p ≥ 2n − 1 elements and p prime, then we get an nmultimagic square of order pn. So in particular for n = 7 there are 7-multimagic squares of order 137, 177, …

Size matters, whatever women say… Now that we know that n-multimagic squares exist for each positive integer n, one might wonder whether for each such n, n-multimagic squares exist of order m, say if m is sufficiently large. The next result shows that this is not the case. For example, taking e = 1 and p = 2 in theorem 3 shows that 3-multimagic squares of order m such that m ≡ 2 (mod 4) cannot exist.

Theorem 3 Let M be an n-multimagic normal square of order m. If the prime number p divides m with multiplicity e(≥ 1), then n ≤ pe+1 − 2. Proof: Consider the polynomial

' ( x f(x) = n

in Q[x]. Then f has degree n. Since M is n-multimagic, it follows that the matrix f(M) obtained by applying f to each entry of M, is a magic square consisting of (nonnegative) integers. In particular, the sum f(0) + f(1) + · · · + f(m2 − 1) is m times the magic sum. The sum

which is equal to

5 m2 6 n+1

' ( ' ( ' 2 ( 0 1 m −1 + + ··· + , n n n

, is thus divisible by m. Hence the (rational) number

m(m2 − 1)(m2 − 2) . . . (m2 − n) q := (n + 1)! Size ma…ers

CF

is an integer. Since an n-multimagic square of order m cannot exist if an (n − 1)multimagic square of order m cannot exist, it suffices to show that the existence of such a square in the case n = pe+1 − 1 leads to a contradiction. Let n = pe+1 − 1. Note that q is nonzero since by hypothesis pe divides m, implying that m2 ≥ p2e > pe+1 − 1 = n (e is positive). To obtain a contradiction we compute the multiplicity of p in q. To this end, for any nonzero integer r we denote the multiplicity of p in r by vp(r) and for any nonzero rational number r/s we put vp (r/s) = vp(r) − vp(s). It is easy to verify that vp(q1q2) = vp(q1) + vp(q2) for any pair of nonzero rational numbers q1 and q2. Observe that q is a product of 2 m/(n + 1) and the factors − i)/i that vp(q) = 5 (m 6 , for i = 1, . . . , n. It follows 2 e+1 vp (m/(n + 1)), for vp (m − i)/i = 0 when 1 ≤ i ≤ n = p − 1. Recalling that n = pe+1 − 1, we obtain that

vp(q) = vp(m) − vp(pe+1) = e − (e + 1) = −1, which contradicts the fact that q is an integer.

"

The proof uses a special polynomial f ∈ Q[x] for which f(Z) ⊆ Z. It is wellknown that the only polynomials f ∈ C[x] for which 5 x6 f(Z) ⊆ Z are precisely the polynomials from the Z-algebra generated by the n where n ∈ N.

Getting high and hyper The method described earlier can be extended to construct multimagic cubes and hypercubes. We remark at the outset that there is no consensus on the definition of multimagic cubes (or hypercubes or higher dimensional analogues) in the literature. The choice I make is the following (see [Gar88] or [Wei]): an m × m × m cube of natural numbers (respectively, the consecutive numbers 0, 1, 2, . . . , m3 − 1) is called magic (respectively, normal magic) if the sum of all elements in each row, column, and pillar is the same and is equal to the sum of all elements of each of the four space diagonals. Furthermore, if n ≥ 1 such a cube is called n-multimagic if (1 ≤ d ≤ n) the cube obtained by raising each of its entries to the dth power is magic. The construction of an n-multimagic cube goes as follows: Analogues to the construction of n-multimagic squares we define a qn × qn × qn cube M by the formula

M = T ◦ L ◦ (ρ−1 × ρ−1 × ρ−1), where ρ is a bijection of type c and L is an affine map given by an invertible 3n×3n CM

B. Multimagic Squar@

matrix A over R (a finite ring of q elements) and a vector t in R3n (i.e. ,



 a L((a, b, c)) = A  b  + t c

for (a, b, c) in R3n). In order to make M n-multimagic we choose A to be a socalled n-multimagic 3-generator matrix. This means that A = (P|Q|U), where P, Q, and U are 3n × n matrices over R that exhibit the following three properties, 1. A is invertible over R (which guarantees that all the natural numbers from 0 up to including q3n − 1 appear as entries in M); 2. all n×n minors of the matrices P, Q, and U are units in R (which guarantees that when 1 ≤ d ≤ n the sum of all elements in each column, row, and pilar of M∗d is the same, hence equal to the magic sum); 3. all n × n minors of the matrices P + Q + U, −P + Q + U, P − Q + U, and P + Q − U are units in R (which guarantees that when 1 ≤ d ≤ n the sum of all elements of each of the four space diagonals of M∗d is equal to the magic sum). At this point it should be clear to the reader how to proceed in higher dimensions (i.e. , how to construct hypercubes and hypercubes with additional properties).

Universal existence It remains only to show how, for some ring R, to construct an n-multimagic 3generator matrix A or, more generally, how to construct for each d (≥ 2) an nmultimagic d-generator matrix A (i.e. , an invertible matrix A = (A1| . . . |Ad) over R such that each Ai is a dn×n matrix with the property that for arbitrary δ1, . . . , δd in {−1, 0, 1}, not all zero, all n × n minors of each combination δ1A1 + · · · + δdAd are units in R). Before we indicate the construction of such matrices recall the following well-known fact (which is easy to prove by induction on m): If f is a nonzero polynomial in m variables over the integers, then there exist integers a1, . . . , am such that f(a1, . . . , am) is nonzero.

(3.2)

Now let n ≥ 1 and d ≥ 2 be given. To avoid complicating an easy matter we demonstrate only how to find n-multimagic d-generator matrices for the case d = 2. For other d the procedure is similar. Let Au = (Aij) and Bu = (Bij) be two universal 2n × n matrices (i.e. , the entries Aij and Bij are distinct variables). Then each n × n minor of Au, Bu, Universal ex}tence

CD

Au + Bu and Au − Bu is a nonzero polynomial with integer coefficients in the 4n2 variables Aij and Bij. Let f denote the polynomial obtained by taking the product of all these minors and the polynomial det(Au|Bu). By (3.2) we can find integers aij and bij such that f(aij, bij) is a nonzero integer. Finally, let p be a prime number that does not divide this integer. It turns out that over R = Z/pZ the matrix A = (A1|A2), where A1 = (aij) and A2 = (bij), leads to an n-multimagic 2-

Ö

generator matrix when the preceding construction is carried out. For more examples and alternate constructions of n-multimagic d-generator matrices see ‘More Generator Matrices’ at page 74. The reasoning above can be used to prove the existence of squares (or cubes) with special properties which can be expressed in terms of ‘magic’ submodules in our setting. For instance, using this technique it is not difficult to see that problem 2.22 from Abe in [Abe94] has a positive answer.

Explaining Dudeney’s bimagic square Dudeney’s bimagic square of order 8 from page 9 also fits into the framework. It is however not an example that could have been made by considering theorem 2 alone. So what do you need more? First fit the square into the framework. Let R = F2, with the standard bijections for ρ = κ and T , and build L using the translation vector (0, 0, 0, 1, 1, 0) and the matrix        

0 1 1 1 1 0

1 1 0 1 0 0

0 0 1 1 0 1

0 0 0 1 1 1

1 0 1 1 1 0

1 1 0 0 1 0

       

It is left to the reader to check the result will indeed be the square from Dudeney (exercise 16 at page 26). Consider the free submodules D = )e0 + e3, e1 + e4, e2 + e5*, C = )e0, e1, e2* and R = )e3, e4, e5*. Then the rows correspond to the parallel affine submodules of R, the columns to C . The main diagonals are two parallel affine submodules of D, see figure 3.1(a). All the coordinate maps L(i) for each of R, C and D, are surjective, so theorem 1 explains why it is magic. E.g. for D you get that L(0)(r0, r1, r2) = r2 since )row0 | e0 + e3* = )row0 | e1 + e4* = 0 and )row0 | e2 + e5* = 1. If you check all the maps that are required by theorem 2 (page 55), then R and C pose no problem. Only for D you run into trouble. If we put the matrices of the linear maps L(i), as in theorem 1 (page 34) with D the free submodule, as rows in JK

B. Multimagic Squar@

45 31

3

49 42 24

4

54

32 18 14 60 39 21

9

59 2

25 43 55

5

30 44 48

20 38 58

8

19 33 61 15

51

1

29 47 52

6

26 40

62 12 16 34 57 11 23 37 7

53 41 27

0

50 46 28

10 56 36 22 13 63 35 17 (a) parallel affine submodules of D

(b) t = (1, 0, 1, 1, 0, 1)

Figure 3.1:

one matrix, you’ll see the problem easily.

       

0 1 1 0 0 1

0 1 1 0 1 0

1 1 1 1 1 1

       

L(1) = L(2) and L(0) = L(3)! So the maps L({1,2}) and L({0,3}) are certainly not surjective. All others are surjective as is readily seen. By the proof of theorem 2 we only need to correct or compensate the following sum when considering the sum of the entries over D + x:

#$ T(1)(L(1)(r) + c1 + t1) · T(2)(L(2)(r) + c2 + t2)

r∈R3

% + T(0)(L(0)(r) + c0 + t0) · T(3)(L(3)(r) + c3 + t3)

where I have written ci for )rowi | x*. Since we are only interested in the diagonals, we only need to consider the zero vector and (0, 1, 0, 1, 0, 1) for x. Note L(0)(r) = L(3)(r) = r2 and L(1)(r) = L(2)(r) = r1 + r2 + r3. For both x, all ci are 0. We could try to adapt T to the situation, in general a viable strategy, but here we consider the possible translation vectors. So all T(i) are the same map ρ. So for x either (0, 0, 0, 0, 0, 0) or (0, 1, 0, 1, 0, 1) we need to correct:

#

r∈R3

{ρ(r0 + r1 + r2 + t1) · ρ(r0 + r1 + r2 + t2) + ρ(r2 + t0) · ρ(r2 + t3)} Explaining Dudeney

JA

By considering what the sums have to be, it is not hard to see that exactly half of the possible vectors for t will result in a bimagic square. More specific, if we write t = (t0, t1, t2, t3, t4, t5) then the translation vectors t which result in a bimagic square are precisely those for which t0 + t1 + t2 + t3 ≡2 1. An example is given in figure 3.1(b).

(a) )(1, 0, 0, 1, 0, 0)*

(b) )(0, 0, 1, 0, 0, 1)*

Figure 3.2: Not just any affine submodules of F62

Exercise 48 Show why Dudeney’s square is pandiagonal using theorem 2 and )e0 + e3*. Can you see directly in Dudeney’s square that )e2 + e5* and it’s parallel affine submodules all have the same sum? What has the rank of )e2 + e5* to do with it?

Exercise 49 Take R = Fq with q ≥ 5 an odd prime and the usual bijections for ρ, κ and T . Show that if you define L by using the following matrix   0 1 2 2  1 1 0 2     2 3 3 4  3 4 3 2

then the resulting normal square of order q2 will be 2-multimagic q-partitioned. Show also that every standard q × q square is not only magic but also pandiagonal. Show that the entire square is itself not pandiagonal!

Exercise 50 Let R = F5, ρ = κ and T the usual bijections. Define L by the following matrix

JI



0  2   4 4

B. Multimagic Squar@

3 0 1 2

4 3 4 0

 3 2   2  1

(a) '(1, 0, 1, 0), (0, 1, 0, 1)(

(b) '(1, 0, −1, 0), (0, 1, 0, −1)(

(d) '(1, 0, 0, 0), (0, 0, 0, 1)(

(e) '(0, 1, 0, 0), (0, 0, 0, 1)(

(c) '(0, 1, 0, 0), (0, 0, 1, 0)(

(f)

'(1, 1, 0, 1), (0, 1, 1, 0), (1, 0, 1, 1)(

Figure 3.3: Some affine submodules of F45 Use theorem 2 to show that the following submodules are ‘2-multimagic’: )e0, e1*, )e2, e3*, )e0 + e2, e1 + e3*, )e0 − e2, e1 − e3*, )e1, e2*, and )e0, e3*. Show that )e0 + e1 + e3, e1 + e2, e0 + e2 + e3* is even ‘3-multimagic’.

Explaining Dudeney

JB

❛ The essential is to excite the spectators. If that means playing Hamlet on a flying trapeze or in an aquarium, you do it. ❜ Orson Welles

JL

4

Y3Z[4 †77 Plain and simple, and the not so plain or simple animals from the magic kingdom.

Ply magic

Example 10 The normal square to the left below was made by the author in 2004 and is a 3 × 3-ply pandiagonal associative magic square with at its center a magic square of order 3. Also nice to note is that the following sums are constant (= 120) for all r and c (indices taken modulo the order): ar,c + ar+3,c+3 + ar+6,c+6 and ar,c + ar+3,c+6 + ar+6,c+3. 20 67 33 21 71 28 25 63 32

52 62 15 47 54 10 48 58 14

43

56

0

77 38

4

78 39

8

73

9

46 57 13 50 61 17 51

57 53 10 61 45 14 56 49 15

12 49 59 16 53 60 11 45 55

30 26 64 34 18 68 79 36 5 74 40 6 75 44 1 12 62 46 16 54 50 11 58 51

79

8

2

36 73

65 31 24 66 35 19 70 27 23

25 35 69 20 27 64 21 31 68

7

80 37

29 63 19 30 67 23 34 71 24

59 47 13 60

66 22 32 70 26 33 65 18 28

29 22 69

72 41

2

48 17 55 52

76 42 9

3

39 76

42 74 5

3

0

37 75

40 77

43 80

6

7

4

41

44 78

38 72

1

The normal associative 3 × 3-ply magic square (not pandiagonal) at the right is interesting since it shows that pandiagonality is not a consequence of being 3 × 3ply magic, while pandiagonality is a easy consequence of 2 × 2-ply magic. It is also special since the determinant of this square of order nine is zero, which answers the question in [Mat00, p.781], where the existence of an associative magic square of odd order with determinant zero was conjectured about. JC

Ö

Example 11 A normal pandiagonal 4 × 4-ply 6 × 6-ply magic square of order 12. 0

134 40

99

5

133 36

98

4

135 41

97

35 117 67

80

30 118 71

81

31 116 66

82

51

89

13 120 50

88

15 125 49

44 102 10 143 45 103

8

60

24 110 64

74

28 111 65

73

23 129 55

92

18 130 59

3

137 37

96

2

32 114 70

83

33 115 68

48

86

16 123 53

47 105

7

63

25 108 62

77

20 126 58

138 46 107 93

136 39 101 85

84

78

75

95

1

27 113 61 90

139

29 109 94

132 38 100

34 119 69

12 122 52

21 127 56

9

19 128 54

87

140 42 106 11 141 43 104 76

14 124

72

79

17 121 6

142

26 112

22 131 57

91

Example 12 An example of a normal square of order 15 that is magic, associative, pandiagonal, 3 × 3-ply and 5 × 5-ply:

144 36 153 148 34 159 141 33 163 139 39 156 138 43 154 92 224 22

90 222 17

104 217 15

102 212 29

97 210 27

55 106 176

53 110 175

46 116 173

50 115 166

56 113 170

204 6 123 208 4 129 201 3 133 199 9 126 198 13 124 62 194 82

60 192 77

74 187 75

72 182 89

67 180 87

145 31 161 143 35 160 136 41 158 140 40 151 146 38 155 99 216 18

103 214 24

96 213 28

94 219 21

93 223 19

47 119 172

45 117 167

59 112 165

57 107 179

52 105 177

205 1 131 203 5 130 196 11 128 200 10 121 206 8 125 69 186 78

73 184 84

66 183 88

64 189 81

63 193 79

137 44 157 135 42 152 149 37 150 147 32 164 142 30 162 100 211 26

98 215 25

91 221 23

95 220 16

101 218 20

54 111 168

58 109 174

51 108 178

49 114 171

48 118 169

197 14 127 195 12 122 209 7 120 207 2 134 202 0 132 70 181 86 JJ

68 185 85

L. Magic Zoo

61 191 83

65 190 76

71 188 80

Example 13 A normal square of order 15 that is magic, associative, pandiagonal and 3 × 5-ply.

143 69 202 50 96

135 70 198 57 91

148 62 206 49 104

118 212 41 184 14

113 219 37 185 6

105 220 33 192 1

75 130 168 27 151

88 122 176 19 164

83 129 172 20 156

98 144 67 200 51

90 145 63 207 46

103 137 71 199 59

13 107 221 34 194

8 114 217 35 186

0 115 213 42 181

150 85 123 177 16

163 77 131 169 29

158 84 127 170 21

53 99 142 65 201

45 100 138 72 196

58 92 146 64 209

193 2 116 214 44

188 9 112 215 36

180 10 108 222 31

15 160 78 132 166

28 152 86 124 179

23 159 82 125 171

203 54 97 140 66

195 55 93 147 61

208 47 101 139 74

43 182 11 109 224

38 189 7 110 216

30 190 3 117 211

165 25 153 87 121 178 17 161 79 134 173 24 157 80 126 68 204 52 95 141

60 205 48 102 136

73 197 56 94 149

223 32 191 4 119 218 39 187 5 111 210 40 183 12 106 120 175 18 162 76

133 167 26 154 89

128 174 22 155 81

Almost any ‘line’ of 15 numbers is magic. A line (a, b) of length n starting at (p, q) is the collection of entries (p + ia, q + ib) where i ∈ n and indices are taken modulo the order of the square. A line is magic when the sum of its entries is magic. For instance any knight-path (±1, ±2) is magic. The 22 exceptions of the 152 = 225 are: (1, 5), (2, 10), (3, 1), (3, 6), (3, 11), (4, 5), (5, 10), (6, 2), (6, 7), (6, 12), (7, 5) and their ‘counter-directions,’ e.g. −(1, 5) = (−1, −5) = (14, 10).

Example 14 Using R = F2 and the following matrix        

0 0 0 1 0 0

0 1 0 1 1 1

1 1 1 0 0 1

1 0 0 0 0 0

1 1 1 0 1 0

0 0 1 1 1 1

       

and the standard bijections you get a family of 2 × 4-ply 4 × 2-ply 4-partitioned associative magic squares. Note that exactly half of the 2 × 2-subsquares sum to Ply magic

JF

the magic sum. Below with t the zero vector:

0

15 58 53 32 47 26 21

57 54

3

12 25 22 35 44

23 24 45 34 55 56 13 46 33 20 27 14 4

1

2

52 59

11 62 49 36 43 30 17

61 50

7

8

29 18 39 40

19 28 41 38 51 60

9

42 37 16 31 10

48 63

5

6

Bimagic squares Example 15 A family of bimagic squares of odd order. Take R = Z/qZ, where q ≥ 3 and q is odd,



0  2 A=  1 2

1 0 1 1

1 0 2 2

 0 1  , 1  2

and t an arbitrary member of R4. Let ρ = κ : R → {0, 1, . . . , q−1} be the standard bijection: ρ(i) = i (0 ≤ i ≤ q−1). Then ρ is of type −1. Take T the usual bijection. Since A is a 2-multimagic generator matrix, corollary 10 gives a family of bimagic squares of odd order. If we choose q = 3 and t the zero vector, the result is the following square:

1

35 60 23 48 79 18 40 65

70 14 39 56

9

31 78 19 53

49 74 27 44 69 10 30 61 38 72 13 33 55

8

26 51 73 12 43 68 59

3

52 77 21 4

29 63

34 81 22 47 64 17 42

75 25 50 67 11 45 62 36 58

5

2

15 37 71

6

28

46 80 24 41 66 16 7

32 57 20 54 76

Furthermore, if we take q = 3 and t = (2, 1, 0, 2)T the associative bimagic square of order nine constructed by R.V. Heath before 1933 is recovered (see [BC74, p. 212]). JM

L. Magic Zoo

Example 16 A family of associative bimagic squares of order 16. Let R be the finite ring F2[x]/(x2 + x + 1) (, F4) and use for L the 2-multimagic generator matrix



x  0   1 x

0 1 1 1

1 1 1 0

 x 1  . 0  x

Let ρ = κ = N : R → {0, 1, 2, 3} be the bijection given by N(0) = 0, N(1) = 2, N(x) = 1, and N(x + 1) = 3. Then N is of type x + 1. Take T (r0, r1, r2, r3) = " 3−i 4 N(ri). By corollary 10 and proposition 3 we obtain for each t in R4 an asi sociative bimagic square of order 16. Taking t = (0, 1, 1, 0)T yields the following example from this family: 41 252 74 155 125 176 30 207 129 84 226 51

213 8 182 103

62 239 93 144 106 187 9 220 150 71 245 40

194 19 161 116

3 210 100 177

87 134 56 229 171 122 204 25

255 46 160 77

24 197 119 166

68 145 35 242 192 109 223 14

236 57 139 90

240 61 143 94

188 105 219 10

72 149 39 246

20 193 115 162

251 42 156 73

175 126 208 29

83 130 52 225

7 214 104 181

198 23 165 120 146 67 241 36

110 191 13 224

58 235 89 140

209 4 178 99

133 88 230 55

121 172 26 203

45 256 78 159

98 179 1 212

54 231 85 136 202 27 169 124 158 79 253 48

117 168 22 199

33 244 66 147 221 16 190 111 137 92 234 59

76 153 43 250

32 205 127 174 228 49 131 82

184 101 215 6

95 142 64 237

11 218 108 185 247 38 152 69

163 114 196 17

167 118 200 21

243 34 148 65

15 222 112 189

91 138 60 233

180 97 211 2

232 53 135 86

28 201 123 170

80 157 47 254

141 96 238 63

217 12 186 107

37 248 70 151 113 164 18 195

154 75 249 44

206 31 173 128

50 227 81 132 102 183 5 216

It is nice to notice that the same square can be made over F2 using the standard bijections (of type 1). By reordering the rows of the matrix above and taking a different bijection N, given by N(0) = 0, N(1) = 1, N(x) = x+1, and N(x+1) = x, it gets outside the scope of corollary 10. You still get 2-multimagic squares though not associative any more. By shifting the columns 4 to the left and sorting the rows Bimagic squar@

JD

of the ‘generator matrix’ over F2 we get the following matrix:

           

0 0 0 0 0 1 1 1

0 0 1 1 1 0 1 1

0 1 0 0 1 1 0 0

1 0 0 1 0 0 0 1

0 1 1 0 0 1 0 0

1 1 1 1 0 1 1 0

0 1 0 0 0 0 1 1

1 0 1 0 1 0 1 1

           

By using the standard bijections and choosing t the zero vector, you get the following specimen:

0 171 67 232 246 93 181 30

100 207 39 140 146 57 209 122

145 58 210 121 103 204 36 143 245 94 182 29 76 231 15 164 186 17 249 82

3 168 64 235

40 131 107 192 222 117 157 54

221 118 158 53 43 128 104 195 185 18 250 81 79 228 12 167 59 144 120 211 205 102 142 37

95 244 28 183 169 2 234 65

170 1 233 66 92 247 31 180 206 101 141 38 56 147 123 208 119 220 52 159 129 42 194 105

19 184 80 251 229 78 166 13

230 77 165 14 16 187 83 248 130 41 193 106 116 223 55 156 7 172 68 239 241 90 178 25

99 200 32 139 149 62 214 125

150 61 213 126 96 203 35 136 242 89 177 26 75 224 8 163 189 22 254 85

4 175 71 236

47 132 108 199 217 114 154 49

218 113 153 50 44 135 111 196 190 21 253 86 72 227 11 160 60 151 127 212 202 97 137 34

88 243 27 176 174 5 237 70

173 6 238 69 91 240 24 179 201 98 138 33 63 148 124 215 112 219 51 152 134 45 197 110 225 74 162 9

20 191 87 252 226 73 161 10

23 188 84 255 133 46 198 109 115 216 48 155

The interesting thing is that it is not only 2-multimagic but also the four standard submatrices of order 8 are magic squares, so it is 8-partitioned!

❛ Among the Western writings on Islamic magic squares, the following are probably the most useful: …. All of the above are more useful for the examples illustrated than for the remarks about them. FK

L. Magic Zoo

❜

[Cam69a, footnote 1, p.181]

Example 17 This example took some work, since this family is 4 × 4-ply, pandi-

agonal and bimagic! Use R = F2, the standard bijections, and let the matrix of L be

       

0 0 1 1 1 0

1 0 1 1 0 1

1 1 0 1 1 0

1 1 0 0 0 1

0 0 1 0 1 1

1 0 1 1 0 0

       

As with Dudeney’s bimagic square, exactly half of the number of possible vectors t result in a 2-multimagic square, e.g. with t = (0, 1, 0, 0, 0, 0) (left) respectively t = (1, 0, 1, 1, 1, 1) (right) the result is a bimagic 4 × 4-ply pandiagonal magic square:

16 60 27 55 33 13 42 38 10 45

1

0

4

43

2

3

36

8

35 15 25 53 18 62

51 31 56 20 5

7

2

46

9

3

36

8

30 50 21 57

25 53 18 62 40

44 58 22 49 29

30 50 21 57 47 40

47

23 59 28 48

61 17 54 26 12 32 11 39

6

46

9

4

35 15

37 51 31 56 20

52 24 63 19

5

33 13 42

16 60 27 55

6

41 14 34

23 59 28 48 38 10 45

37

12 32

41 14 34 52 24 63 19

7

1

43 61 17 54 26

58 22 49 29 11 39

0

44

Example 18 A family of 8 × 8-ply 2-multimagic squares of order 16. With R = F2, ρ = κ and T standard, and let the matrix of L be            

0 0 0 0 0 1 1 1

0 0 1 1 1 0 1 1

0 1 0 0 1 1 0 0

1 0 0 1 0 0 0 1

0 1 1 0 0 0 0 0

1 0 0 1 0 1 1 0

0 1 0 0 0 1 1 1

1 0 1 0 1 0 1 1

           

Bimagic squar@

FA

Taking t = (0, 0, 0, 0)T yields the following example: 0 171 71 236 150 61 209 122 96 203 39 140 246 93 177 26 145 58 214 125

7 172 64 235 241 90 182 29

103 204 32 139

76 231 11 160 218 113 157 54 44 135 107 192 186 17 253 86 221 118 154 49

75 224 12 167 189 22 250 81

43 128 108 199

59 144 124 215 173 6 234 65 91 240 28 183 205 102 138 33 170 1 237 70

60 151 123 208 202 97 141 38

92 247 27 176

119 220 48 155 225 74 166 13 23 188 80 251 129 42 198 109 230 77 161 10

112 219 55 156 134 45 193 106

16 187 87 252

7 172 64 235 145 58 214 125 103 204 32 139 241 90 182 29 150 61 209 122

0 171 71 236 246 93 177 26

96 203 39 140

75 224 12 167 221 118 154 49 43 128 108 199 189 22 250 81 218 113 157 54

76 231 11 160 186 17 253 86

44 135 107 192

60 151 123 208 170 1 237 70 92 247 27 176 202 97 141 38 173 6 234 65

59 144 124 215 205 102 138 33

91 240 28 183

112 219 55 156 230 77 161 10 16 187 87 252 134 45 193 106 225 74 166 13

119 220 48 155 129 42 198 109

23 188 80 251

It is interesting to note that it, besides being 8 × 8-ply and 2-multimagic, it is also covered with four magic squares of order 8.

Example 19 An associative, pandiagonal, bimagic, magic square of order 25. Take

R = F5, let ρ = κ be the standard bijection (of type −1), and use the following matrix and translation vector to define L 

1  1   1 1

1 2 3 4

2 2 3 3

  2  4  , t =    4 2

 0 4  . 0  2

Then the corresponding 25 × 25 matrix is associative, pandiagonal,and bimagic, and it has the following properties: (i) each of the twenty-five standard 5 × 5 submatrices is a pandiagonal magic square (with the same magic sum); (ii) for each pair (i, j) (1 ≤ i, j ≤ 25) the 5 × 5 matrix obtained by deleting each row with row number not equivalent to i modulo 5 and each column with column number not equivalent to j modulo 5 is a pandiagonal magic square! FI

L. Magic Zoo

544 136 483 80 322 207 429 46 268 615

538 135 477 99 316 201 448 45 262 609 394 111 333 555 172

4 371 593 190 407 317 539 131 478 100

Bimagic squar@

FB

408 5 372 594 186 96 318 540 132 479 264 606 203 450 42 552 174 391 113 335 245 462 59 276 523

369 586 183 405 22 532 129 496 93 315 225 442 39 256 603 388 110 327 574 166 51 298 520 237 459

180 422 19 361 583 493 90 307 529 146 31 253 625 217 439 349 566 163 385 102 512 234 451 73 295

11 358 580 197 419 304 546 143 490 82 617 214 431 28 275 160 377 124 341 563 473 70 287 509 226

597 194 411 8 355 140 482 79 321 543 428 50 267 614 206 116 338 560 152 399 284 501 248 470 62

252 624 216 438 35 570 162 384 101 348 233 455 72 294 511 421 18 365 582 179 89 306 528 150 492

213 435 27 274 616 376 123 345 562 159 69 286 508 230 472 357 579 196 418 15 550 142 489 81 303

49 266 613 210 427 337 559 151 398 120 505 247 469 61 283 193 415 7 354 596 481 78 325 542 139

610 202 449 41 263 173 395 112 334 551 461 58 280 522 244

441 38 260 602 224 109 326 573 170 387 297 519 236 458 55 590 182 404 21 368 128 500 92 314 531

246 468 65 282 504 414 6 353 600 192 77 324 541 138 485 270 612 209 426 48 558 155 397 119 336

57 279 521 243 465 375 592 189 406 3

518 240 457 54 296 181 403 25 367 589 499 91 313 535 127 37 259 601 223 445 330 572 169 386 108

454 71 293 515 232 17 364 581 178 425 310 527 149 491 88 623 220 437 34 251 161 383 105 347 569

290 507 229 471 68 578 200 417 14 356 141 488 85 302 549 434 26 273 620 212 122 344 561 158 380

95 312 534 126 498 258 605 222 444 36 571 168 390 107 329 239 456 53 300 517 402 24 366 588 185

526 148 495 87 309 219 436 33 255 622 382 104 346 568 165 75 292 514 231 453 363 585 177 424 16

487 84 301 548 145 30 272 619 211 433 343 565 157 379 121 506 228 475 67 289 199 416 13 360 577

323 545 137 484 76 611 208 430 47 269 154 396 118 340 557 467 64 281 503 250 10 352 599 191 413

134 476 98 320 537 447 44 261 608 205 115 332 554 171 393 278 525 242 464 56 591 188 410 2 374

564 156 378 125 342 227 474 66 288 510 420 12 359 576 198 83 305 547 144 486 271 618 215 432 29

400 117 339 556 153 63 285 502 249 466 351 598 195 412 9

331 553 175 392 114 524 241 463 60 277 187 409 1 373 595 480 97 319 536 133 43 265 607 204 446

167 389 106 328 575 460 52 299 516 238 23 370 587 184 401 311 533 130 497 94 604 221 443 40 257

103 350 567 164 381 291 513 235 452 74 584 176 423 20 362 147 494 86 308 530 440 32 254 621 218

(iii) all the 189 submatrices of order 5 with the top-left corner at row r, column c with either r or c (or both) a multiple of five, have the same sum (7825), including wrap-around.

More Generator Matrices Example 20 A 1-multimagic 3-generator matrix for R = Fq with q prime and q ≥ 11 :



 1 2 4  1 2 −4  1 −2 −4

So this will result in (3-dimensional) magic cubes.

The Vandermonde-like matrix P as given in ‘A class of n-multimagic generator matrices’ is special because it tends to give ‘small’ primes q that are not in the occurring determinants, and hence P as seen in Fq is a multimagic generator matrix. And small prime q is nice because then the order q2n will be ‘small.’ It is not hard to show that the same construction from the section ‘A class of nmultimagic generator matrices’ generalizes to higher dimensions. Define P0, P1, . . . , Pd−1 by dividing the dn × n matrix P in n × n matrices like before.



     P=    

1 1 1 1

0 1 2 3

.. .

.. .

0 1 4 9 .. .

··· ··· ··· ··· ..

0 1 2n−1 3n−1

.

.. .

1 (dn − 2) (dn − 2)2 · · · (dn − 2)2n−1 0 0 0 ··· 1

Stack these Pi’s in one n-multimagic d-generator matrix as follows:

          

5 6 V(n, d) := ∆i,j 2j Pi i,j=0...d−1

with ∆i,j = 1 if i + j ≤ d − 1 and −1 otherwise.

Example 21 V(2, 3) is a 2-multimagic 3-generator matrix for R = Fq with q prime and q ≥ 11 :

       

1 1 1 1 1 0

 0 2 0 4 0 1 2 2 4 4   2 2 4 −4 −8   3 2 6 −4 −12   4 −2 −8 −4 −16  1 0 −2 0 −4

So there are bimagic perfect cubes of order 112, 132, 172 etc. . FL

L. Magic Zoo

A small table of the required units in R for V(n, d) for small n and d: 2 3 4 5 d \n 1 2, 3 2, 3 2, 3 2, 3, 5 2, 3, 5, 7 2 2, 3, 5, 7 2, 3, 5, 7 2, 3, 5, 7 2, 3, 5, 7 2, 3, 5, 7, 11, 13 3 2, 3, 5, 7, 11, 13 2, 3, 5, 7, 11, 13 2, 3, 5, 7, 11, 13 2, 3, 5, 7, 11, 13 2, 3, 5, 7, 11, 13, 17 4 5

2, 3, 5, 7, 11, 13, 17,

2, 3, 5, 7, 11, 13, 17,

2, 3, 5, 7, 11, 13, 17,

2, 3, 5, 7, 11, 13, 17,

19, 23, 29, 31

19, 23, 29, 31

19, 23, 29, 31

19, 23, 29, 31

As was noted before in ‘A class of n-multimagic generator matrices’ the primes occurring in the minors of V(n, 2) are precisely the primes less than 2n − 2 (if n ≤ 2 then also 3 has to be a unit). It is rather cumbersome to check this. It is somewhat worse to do this for V(n, d). It is also possible to construct an n-multimagic 2-generator matrix by stacking the identity n × n matrix and a matrix with no vanishing minors, e.g. a Cauchymatrix, and then use the same method as for the V(n, 2). Vice versa, any n-multimagic 2-generator matrix can, using elementary column operations, be reduced to a block matrix where the top left and bottom right, n × n matrices are identity matrices. The two remaining n × n block matrices are than automatically matrices with no vanishing minors. Generalizations to higher dimensions are straightforward. Note that any permutation of the rows will not disturb the n-multimagic dgenerator property. Likewise any permutation that will permute the columns of each n-column-blocks in the same way will not disturb it.

Example 22 Let x ∈ R! then the following matrix is a 2-multimagic 2-generator matrix if there will be certain restrictions on R:



1  0   x x

 0 x −1 1 −x x   1 1 0  x 0 1

Following the reasoning at the end of the section ‘Getting high and hyper’ you get the restriction 5 65 65 6 2 x (x − 1) (x + 1) −1 − x + x25 1 − x + x26 5 1 + x + x2 6· 1 + 3 x + x2 1 − 3 x + x2 ∈ R!

So x ∈ / {−1, 0, 1}. It is easy to deduce that in all cases {2, 3, 7} ⊆ R! . If x = 2 then the factors of the unit above are {2, 3, 7, 11}. If x = 3 then the factors of the unit above are {2, 3, 5, 7, 13, 19}. If x = 2 and R = F5 then the resulting 2-multimagic square is also pandiagonal. More Generator Matƒc@

FC

Example 23 A 2-multimagic 2-generator matrix with R = Fq with q prime and q ≥ 11, q #= 17 :

       

1 0 0 1 1 2

0 1 0 2 1 1

 0 −1 −2 −1 0 −1 −1 −2   1 −2 −1 −1   1 1 0 0   2 0 1 0  1 0 0 1

Miscellaneous Example 24 It is not too hard to think of variants for cards to send at Christmas:

Fijne Feestdagen en een

Fantastisch 2007 Breng alles in Balans evenveel

sterren als bomen gekleurd als niet groot als klein omrand als niet

in elke rij

horizontaal diagonaal en verticaal

FJ

L. Magic Zoo

!

‡8:ˆ10: 97 :1•1491‰ 1Š104}@ Chapter 0 3 1,0,81 .

5 i) At most n2 − 2n − 1 if the order is n (cf. [Coa83], [WI80], [Rat59])2

6 A connection figure is made by connecting two numbers if they are complementary.

7 Check (ai,j) of order n with ai,j = i + n · j. There are no associative or pandiagonal normal magic squares of order 4k + 2 as is proven in lemma 6 at page 29.

8 iv) Since ai,j = a0,j + (ai,0 − a0,0), the sequences are: a0,0, a0,1, . . . , a0,n−1 and 0, a1,0 − a0,0, a2,0 − a0,0, . . . , an−1,0 − a0,0. 56 34

8

57 18 47

33 20 54 48

7

5

4

49

35 30 53 12 46 60

15 25 63 6

31

29 59 10

26 43 13 23 64 38 19

9

2

41 24 50 40

55 17 11 36 58 32 45

61 16 42 52 27

1

39 22

44 62 28 37 14 51 21

3

Although not an exercise, this is the solution to the bimagic square puzzle of G. Pfefferman from 1891 (page 9) 1

all rotations and reflections of the Lo Shu. if the −1 surprised you, the requirement was for normal squares, so the magic sum is known beforehand. 2

FF

Chapter 2 13 Step (1, 2), cross-step (2, 1) and startposition (0, 0). No, in the sense of the previous chapter the maps ρ, κ and T were fixed. The step from 1 to 2 is not the same step as from 0 to 1, so this is impossible.

16

i) With the standard bijections for ρ, κ and T and the ring Z/3Z, take L by using the following matrix and translation-vector:

'

2 1 2 2

(

'

,

1 0

(

.

ii) Use F22 and the ρ, κ and T as in ‘Other flavours of T’. Take L by using the following matrix and translation-vector:



1  0   0 1

0 1 1 1

0 1 1 0

 1 1  , 0  1



 0  1   .  1  0

iii) See ‘Explaining Dudeney’ at page 60. iv) All of Benjamin Franklin’s squares fit into the framework with R = F2 and the usual bijections.

17 First note that for any weighted sum bijection T the coefficients ti are a permutation of 1,q,q2,…,qn−1, by the same reasoning as the uniqueness of the base q representation. Secondly just write out T (−r) as follows # # # tiρ(i)(−ri) = ti(q − 1) − tiρ(i)(ri + ci) = qn−1 − 1 − T (r + c) i

i

i

where c = (c0, c1, . . . , cn−1) ∈ Rn.

26 Since bjrj ∈ S there are rjk ∈ R such that bjrj = Im(Bi) 1

d−1 # j=0

)rowi | bj*rj = =

d−1 # j=0

)rowi |

"

d−1 # k=0

k skrjk.

So

skrjk*

  d−1 # )rowi | sk*  rjk ∈ Im(L(i))

d−1 # k=0

j=0

27 (a) )(1, 1, 1, 0)*, (b) )(0, 0, 1, 1), (1, 1, 0, 0)*, (c) )(0, 1, 0, 1)*, (d) )(1, 1, 1, 0), (0, 0, 0, 1)*, (e) )(0, 0, 0, 1), (1, 0, 0, 0), (0, 1, 0, 0)*.

28 When mρ(0),κ(0) = 0. Since S ⊕ S⊥ = R2n and S ∩ S⊥ = {0}. FM

A. Answers

29 Yes: i) ii) iv), No: iii) 30 iii) hint: consider the first three primes. 32 The matrix used to define L is

'

3 1 1 2

(

.

39 Remember that the columns of a pandiagonal square can be cyclically permuted without disturbing it’s pandiagonality. So if it would be true, any cyclically permuted pandiagonal sub-square would not disturb the diagonal sums. This is not true. Take for instance the downward broken diagonal starting at the top right corner of the first (pandiagonal) sub-square. Since this sub-square only attributes one corner to the sum of the broken diagonal, and so precisely one number to that broken diagonal, any cyclical permutation of columns of this sub-square will affect the sum directly if it is to have all distinct entries. Also see exercise 49 at page 62.

42 E.g. 0 2 3 1 1 2

1 0 1 3 1 3

1 0 2 0 3 3

2 3 1 3 0 0

2 2 2 0 3 0

3 2 0 2 1 1

44 Let X be the square of order 4 below. Denote Xσ = (σ(xi,j)) where σ is a permutation of 4. The sought after detrivializer can be build by putting X ⊗ (0)s in the top-left quadrant, X(02)(13) ⊗ (0)s top-right quadrant, X(01)(23) ⊗ (0)s bottom-left quadrant, and X(03)(12) ⊗ (0)s in the bottom-right quadrant. In other words for s = 1 you’ll get the following detrivializer, with next to it the result.

6 0 2 1 3 1 3 0 2

1 3 0 2 0 2 1 3

2 0 3 1 3 1 2 0

3 1 2 0 2 0 3 1

2 0 3 1 3 1 2 0

3 1 2 0 2 0 3 1

0 2 1 3 1 3 0 2

1 3 0 2 0 2 1 3

27 32 61 38 59 7

31

57 36 63 34 25

56 37 30

3

1

0

33 60 2

26

0 2 1 3

1 3 0 2

2 0 3 1

3 1 2 0

29

28 39 58

24

5

4

62 35

22 11 48 45 54 43 16 13 49 44 23 10 17 12 55 42 15 18 41 52 47 50 40 53 14 19

8

9

20

21 46 51

It is left to the reader to check the properties for each s ≥ 1. A. Answers

FD

Chapter 3 49 Use corollary 10 (page 10) to prove the matrix of L is a 2-multimagic generator matrix as long as q #∈ {2, 3}. To prove the pandiagonal magic subsquares use theorem 1 in conjunction with the following submodules: )e1*, )e3*, )e1 + e3* and )e1 − e3*. That the resulting bimagic square is itself not pandiagonal is left to the reader.

❛ I hear and I forget. I see and I remember. I do and I understand. MK

A. Answers

❜

Confucius

‹[Œ•[7Z03S• [Abe94]

Gakuho Abe. Unsolved problems on magic squares. Discrete Mathematics 127 pp. 3–13 (1994) (Cited on pages 6, 9, 29, and 60.)

[Adl97]

Allan Adler. Magic n-cubes form a free monoid. the electronic journal of combinatorics 4 (R15) pp. 1–14 (1997) (Cited on page 44.)

[And17]

W. S. Andrews (ed.). Magic Squares and Cubes. Dover Publications Inc., second edition, 1960 reprint ed. (1917). ISBN 0-486-20658-0 (Cited on pages 6, 9, 23, 28, and 51.)

[And60]

W. R. Andress. Basic properties of pandiagonal magic squares. The American Mathematical Monthly 67 (2) pp. 143–152 (February 1960) (Cited on page 8.)

[ASY77]

Allan Adler and Robert Li Shuo-Yen. Magic cubes and prouhet sequences. American Mathematical Monthly 84 pp. 618–627 (October 1977) (Cited on page 44.)

[Bar88]

F. A. P. Barnard. Theory of magic squares and of magic cubes. In The National Academy of Sciences Memoirs, vol. IV, part 1, pp. 207–270 (1888) (Cited on page 10.)

[BC74]

W.W. Rouse Ball and H.S.M. Coxeter. Mathematical Recreations and Essays. University of Toronto Press, 12 ed. (1974) (Cited on page 68.)

[BC87]

W.W. Rouse Ball and H.S.M. Coxeter. Mathematical Recreations and Essays. Dover Publications Inc., 13th ed. (1987). ISBN 0-486-25357 (Cited on pages 23 and 50.)

[Boy]

Christian Boyer. Multimagic squares. URL http://www.multimagie. com/ (Cited on page 53.)

[Boy01]

Christian Boyer. Les premiers carrés tétra et pentamagiques. Pour la science pp. 98–102 (August 2001) (Cited on page 53.)

[Cam69a] Schuyler Cammann. Islamic and indian magic squares, part i. History of Religions 8 (3) pp. 181–209 (February 1969) (Cited on page 70.) MA

[Cam69b] Schuyler Cammann. Islamic and indian magic squares, part ii. History of Religions 8 (4) pp. 271–300 (May 1969) (Cited on page 10.) [Can34]

A. L. Candy. To construct a magic square of order 2n from a given square of order n. National Mathematics Magazine 8 (7) pp. 147–160 (April 1934) (Cited on pages 29 and 50.)

[Can35]

A. L. Candy. To construct a magic square of order 2n from a given square of order n. National Mathematics Magazine 9 (4) pp. 99–105 (Januari 1935) (Cited on pages 29 and 50.)

[Coa83]

R. B. Coates. 67.46 the dimension of the space of magic squares. The Mathematical Gazette 67 (422) pp. 303–304 (December 1983) (Cited on page 77.)

[DEEa]

Harm Derksen, Christian Eggermont, and Arno van den Essen. Multimagic squares. URL http://arxiv.org/abs/math/0504083 (Cited on pages ii and 53.)

[DEEb]

Harm Derksen, Christian Eggermont, and Arno van den Essen. Multimagic squares, cubes and hypercubes (report 04_12). URL http://www.math.ru. nl/onderzoek/reports/rep2004/rep04_12.ps.gz and 53.)

[Dud17]

(Cited on pages ii

Henry E. Dudeney. Amusements in Mathematics. Dover, 1970 reprint ed. (1917). ISBN 0-486-20473-1. URL http://www.gutenberg.org/etext/ 16713 (Cited on pages 6 and 8.)

[Fen]

Pan Fengchu. Announcement of 6-multimagic square. URL http://www. zhghf.net/China/ (Cited on page 53.)

[Fro92]

M. Michel Frolov. Égalités à deux et à trois degrés. Bulletin de la Société Mathématique de France 20 pp. 69–84 (1892). URL http://www. numdam.org/item?id=BSMF_1892__20__69_1 (Cited on page 9.)

[Ful74]

John Lee Fults. Magic Squares. Open Court (1974). ISBN 0-87548-197-3 (Cited on pages 9, 28, and 44.)

[Gar88]

Martin Gardner. Time Travel and Other Mathematical Bewilderments, chap. Magic squares and cubes, pp. 213–225. W.H. Freeman, New York (1988) (Cited on page 58.)

[Kar94]

Vladimı́r Karpenko. Two thousand years of numerical magic squares. Endeavour, New Series 18 (4) pp. 147–153 (1994) (Cited on page 10.)

[Mat00]

R. Bruce Mattingly. Even order regular magic squares are singular. The American Mathematical Monthly 107 (9) pp. 777–782 (November 2000) (Cited on page 65.)

MI

[McC97]

Emory McClintock. On the most perfect forms of magic squares, with methods for their production. American Journal of Mathematics 19 (3) pp. 99–120 (April 1897) (Cited on page 43.)

[OB98]

K. M. Ollerenshaw and D. S. Brée. Most-Perfect Pandiagonal Magic Squares: Their Construction and Enumeration. Institute for Mathematics and Its Applications, Catherine Richards House, 16 Nelson Street, Southend-on-Sea, Essex SS1 1EF, UK (1998). URL http://www.ima.org. uk/mathematics/publications.htm (Cited on page 7.)

[Oll86]

Kathleen Ollerenshaw. On ’most perfect’ or ’complete’ 8×8 pandiagonal magic squares. Proceedings of the Royal Society of London Series A, Mathematical and Physical Sciences (1934-1990) 407 (1833) pp. 259–281 (October 1986). URL http://www.journals.royalsoc.ac.uk (Cited on page 7.)

[Pfe91]

G. Pfeffermann. Les Tablettes du Chercheur, Journal des jeux d’esprit et de combinaisons, Paris (1891) (Cited on page 9.)

[Rat59]

Jr. Ratliff, L. J. The dimension of the magic square vector space. The American Mathematical Monthly 66 (9) pp. 793–795 (November 1959) (Cited on page 77.)

[RW38]

B. Rosser and R. J. Walker. On the transformation group for diabolic magic squares of order 4. Bulletin of the American Mathematical Society 44 pp. 416–420 (1938) (Cited on page 7.)

[RW39]

Barkley Rosser and R.J. Walker. The algebraic theory of diabolic magic squares. Duke Mathematical Journal 5 (4) pp. 705–728 (1939). URL http: //projecteuclid.org/euclid.dmj/1077491563 (Cited on page 7.)

[Wei]

Eric W. Weisstein. Magic cube, from mathworld-a wolfram web resource. URL http://mathworld.wolfram.com/MagicCube.html (Cited on page 58.)

[WI80]

James E. Ward III. Vector spaces of magic squares. Mathematics Magazine 53 (2) pp. 108–111 (March 1980) (Cited on page 77.)

[WR89]

Daniel B. Widdis and R. Bruce Richter. It’s magic! multiplication theorems for magic squares. The College Mathematics Journal 20 (4) pp. 301–306 (September 1989) (Cited on page 44.)

[Yok91]

Ho Peng Yoke. Chinese science: The traditional chinese view. Bulletin of the School of Oriental and African Studies, University of London 54 (3) pp. 506–519 (1991) (Cited on page 10.)

❛ The secret to creativity is knowing how to hide your sources.

❜

Albert Einstein

MB

❛ ‘When I use a word’, Humpty Dumpty said, in a rather scornful tone, ‘it means just what I choose it to mean — neither more nor less’. ❜

ML

Ž8‰1Š The first page number is usually, but not always, the primary reference to the indexed topic.

M!d, 47 )u | v*, 33 )e0, e1*, 32 (0)s, 45 s, 22 A ⊗ B, 44

addition table, 7 affine affine submodule, 33 affine subspaces, 32 affine map, 19, 22 Agrippæ, Henrici Cornelii, 1 associative, 6 bijection of type c, 26 usual, 23 weighted sum, 26, 33 bimagic, 8 broken diagonal, 7 Cardano, Girolamo, 1 classical, 4 complement, 6, 28, 29 complementary numbers, 6 compounding,⊗, 44 connection figures, 5, 6 cross-step, 16 cube

magic, 58 n-multimagic, 58 detrivializer, 48 diagonal bent, 39 broken, 6 downward, 7 upward, 7 Dudeney, Henry Ernest, 6, 8 Frolov, M., 9 Khajuraho, 1, 49, 50 Lo Shu, 1 magic constant, 29 magic square, 4 associative, 6, 65 bimagic, 8 p-multimagic, 47 pandiagonal, 7, 65, 79 r × c-ply, 8, 65 magic sum, 4 mixed radix representation, 42 module free, 34 parallel to S, 33 modulo, 17, 18 MC

multimagic d-generator matrix, 59

Yang Hui, 10

normal, 4 numbers complementary, 6, 29

zero matrix, (0)s, 45

order, 4 p-multimagic, 47 pandiagonal, 7 pandiagonal square, 6 Pfeffermann, G., 9 Picatrix, 1 property additionable, 46 can be blown up, 46 compoundable, 46 tilable, 46 semi-magic square, 4 square associative, 6, 27 bimagic, 47 classical, 4 most-perfect, 7 normal, 4 order of, 4 pandiagonal, 6, 7, 15, 29, 79 n-partitioned, 10 primitive, 7 r × c-ply, 8 semi-magic, 4, 6 trivial, 4 uniform step square, 18 subspace affine, 32 linear, 32 Sudoku, 42 n × m blocks, 42 trick, calendar , 14 trivial, 4 MJ