WHAT'S REALLY MAGIC ABOUT MAGIC SQUARES.

We propose several new concepts about magic squares suitable for all sorts of audiences--even for television. Our first example is an adaptation of an old trick, which is an ideal presentation for classrooms of all ages.

Presentation: "I am going to fill this 4x4 grid with integers of your choice. Roll a die for our starting number and tell me what it is. If you don't have any dice you may do this mentally."

Suppose they roll a 5. You write 5 on a tablet and secretly write 50 on the tablet's reverse as it is placed on the chalk tray of the blackboard. The class is then asked to choose an order of the grid's rows and you write the numbers in order in their choice of rows. Perhaps this is the result.
9    10   11   12
17   18   19   20
5    6    7    8
13   14   15   16

"Now someone choose any number on the grid." Suppose they select 10. "I circle 10 and cross out the rest of the numbers in the 10's row and column. Now someone else choose a number not already crossed off." This is repeated until four numbers are selected. Suppose the four are 10, 16, 5 and 19.

"Now someone please add these four." No matter what four they choose the sum will always be 50 for a 5 start and your unexpected reveal of 50 on the back of the tablet will usually astonish the audience.

This is our version of an effect described by Martin Gardner in Chapter 2 of his book Mathematical Puzzles and Diversions (1962 Simon & Schuster), a collection of his "Mathematical Games" columns from Scientific American. It also appears in one form as the "Egyptian Fortune Telling Tablet" by Subir Kr. Dhar in issue 24, Dec. 1973 of Sam Dalal's Swami.

Any n X n grid with the integers consecutively placed in the rows (or in the columns) will always have the same sum in our selection process. For any start integer 111, positive, zero or negative, the final sum for an n x 11 grid will be m*n + n([n[sup.2]-1)/2.

The most popular n's are n=3, 4, or 5 and nvn will thus be added respectively to 12, 30, or 60. In the above example m = 5 so we add 5(4) = 20 to 30 to obtain the final 50. Negatives are OK too. If a 2 is rolled and the roller chooses to make it-2, we subtract 8 from 30 to obtain a new constant of 22 and count in order-2,-1, 0, 1, and so on.

It is of some value to give the class the problem of explaining fully why this technique always works. Perhaps they can notice that the rows and columns each differ by the same numbers so that our choice will always choose the same 2n numbers.

Other methods can expand our magic to more common squares. For example in the 4x4 case we start by using the numbers in this grid whose construction is straight forward.
4   5   6   7
3   4   5   6
2   3   4   5
1   2   3   4

These numbers sum to 4[sup.3] and can be rearranged into a remarkable magic square.

The bold 4x4 outlined grid represents one solution. The border cells mark the start of a tiling of the plane by the solution grid. Alternately it can be regarded as a folding of the 4x4 into a torus. There are 36 sets of four squares that sum to the magic constant 16.

Before we comment further about this square we perform some magic using it. First we hide the number 16 under a slip of paper. Then ask a member of our audience to choose any number. Suppose he selects and circles a 5. Other members are asked to form a square with the circled 5 as a corner and then add the four numbers. The four always total 16!

The square is also magic with constant 16, not only on the rows and columns but also on all eight diagonals including the broken ones. Note also the four corners of any 2x4 rectangle for instance.

It is a fact that for all n equaling 3 or greater our method will always produce an n x n magic square with constant n[sup.2] at least on rows, columns, and the two main diagonals.

For n=3 this grid is used.
3   4   5
2   3   4
1   2   3

We leave to the reader the problem of using these nine numbers in a 3 x 3 magic square with constant 3[sup.2] = 9.

For n = 5 we like the "Greek Cross" square constructed from the following grid.
5   6   7   8   9
4   5   6   7   8
3   4   5   6   7
2   3   4   5   6
1   2   3   4   5

The bold-face 5x5 grid is a solution. The outer border describes the folding of the grid into a torus so as to identify 25 small Greek crosses that each sum to the magic constant 25. The 25 are each centered at one of the 25 entries in the 5x5 grid.

A Greek Cross has orientation of one of two types:

For example the two crosses centered at 1 are:

Before pointing out for the audience that this square is magic on all rows, columns and diagonals we ask volunteers to select a Greek Cross and add the five numbers. The sum will always be our previously predicted 25, the magic constant of the square.
13    3    4   10
6     8    15  1
11    5    2   12
0     14   9   7

The 4x4 square above is our modification of one using 16 consecutive numbers.

Our solution requirements make this puzzle one of what is called by Dame Kathleen Ollernshaw a most-perfect magic square. This remarkable lady published when she was 87 years old (Most-Perfect Pandiagonal Magic Squares, The Institute of Mathematics and its Applications, 1998, Great Britain, University Press, Cambridge) a complete solution set for all 4k=n such squares. With her co-author David Bree, this was the first time a complete enumeration of an infinite subset of magic squares was completed.

Some mathematical background. There are (16!)/(4!)(3!)(3!)(2!)(2!) = 6054048000 ways of filling the grid with the 16 tokens that will look different to the eye If we don't count reflections and rotations as different this reduces to 756756000 different placements.

The generic term "tokens" usually refers to the numbers that are used in our squares but need not always to be so. We later will employ playing cards and even colors and shapes as tokens.

In his November 1999 Scientific American column "Mathematical Recreations" Ian Stewart notes that Ollerenshaw's "most perfect" requirement is for any magic square of size 4n x 4n with the property that any two-by-two block of adjacent entries have the same sum. He adds "The discovery of the (complex) formula and its proof, leads deeper into combinatorics, so I'll stop here, except to say that for the doubly even orders of 4, 8, 12 and 16, the numbers of different most-perfect magic squares are 48; 368, 640; 2.22953 x 10[sup.10] and 9.322433 x 10[sup.14]."

Martin Gardner in vol. 395, 17 Sept. 1998 ofNature also reports on the achievement. " Dame Kathleen Ollerenshaw, one of England's national treasures, has solved a long standing, extremely difficult problem ... ".

For the 4x4 Gardner points out that every broken diagonal sums to 30 as well as every row and column. Also, every 2x2 square sums to 30 and also any two cells a diagonal hop apart add up to 15.

We can add more magic to the 4x4 most perfect magic square. Our handout dedicated to Martin Gardner at the "Gathering for Gardner V", April 2002, Atlanta, Georgia was entitled "Five Card Study" and follows.

Caption: Dame Kathleen Ollerenshaw (1912-2014)