# Franklin's magic squares

The page Franklin's Magic Squares describes certain magic squares constructed by Benjamin Franklin.

I then confessed to him that in my younger days, having once some leisure which I still think I might have employed more usefully, I had amused myself in making these kind of magic squares...

A magic square (of order n) is an nxn array such that each number between 1 and n2 occurs precisely once, and all row sums and column sums are equal.

Magic squares are easy to construct, and one usually tries to impose additional restrictions. For example, it is common to ask that the sum of the numbers along a diagonal also equals the row sum.

Franklin claims other restrictions for his squares, namely that the sum of the numbers along the first or second half of each row or column is half the row sum. Also that the sum of the numbers along a "bent row" (half of the main diagonal together with half of the back diagonal) equals the row sum.

Example quoted from the above web site:

```                 52 61  4 13 20 29 36 45
14  3 62 51 46 35 30 19
53 60  5 12 21 28 37 44
11  6 59 54 43 38 27 22
55 58  7 10 23 26 39 42
9  8 57 56 41 40 25 24
50 63  2 15 18 31 34 47
16  1 64 49 48 33 32 17

```
with "bent row" patterns
```      # - - - - - - -   # - - - - - - #
- # - - - - - -   - # - - - - # -
- - # - - - - -   - - # - - # - -
- - - # - - - -   - - - # # - - -
- - - # - - - -   - - - - - - - -
- - # - - - - -   - - - - - - - -
- # - - - - - -   - - - - - - - -
# - - - - - - -   - - - - - - - -

- - - - - - - -   - - - - - - - #
- - - - - - - -   - - - - - - # -
- - - - - - - -   - - - - - # - -
- - - - - - - -   - - - - # - - -
- - - # # - - -   - - - - # - - -
- - # - - # - -   - - - - - # - -
- # - - - - # -   - - - - - - # -
# - - - - - - #   - - - - - - - #
```

## Subsquares

A further conspicuous feature of Franklin's squares is that each 2x2 subsquare has the same sum.

From this 2x2 subsquare property many other properties follow immediately. For example, one sees that a horizontally adjacent pair of cells in row i and columns j and j+1 gives a sum that depends only on j and on the parity of i, so that two such pairs in the same columns and on rows with an even distance have the same sum.

And from that it follows again that the bent rows need not have their halves along the diagonals: translations will work just as well.

## Counting

A magic square of order n has entries 1, ..., n2 summing up to n2(n2+1)/2, so each row has row sum n(n2+1)/2, and if all 2x2 subsquares have the same sum, then this sum is 2(n2+1).

## Construction

In order to construct a magic square with the 2x2 subsquare property, it suffices to give the top row and the first column. Then all other entries follow immediately by completing 2x2 subsquares.

Note that a magic square with the 2x2 subsquare property necessarily has even order, so we may assume that n is even.

Since cyclically permuting rows and columns does not change the requirements for a magic square of even order with 2x2 subsquare property, we may assume that the number 1 is in the top left corner.

(Concerning permutations: all permutations of rows and columns can be used provided that they respect the parity of row or column number.)

Let Sij be the number in row i and column j. Put aj = S1j and bi = Si1 so that a1 = b1 = 1. Now all Sij can be expressed in terms of the aj and bi.

Let M=n2+1. In order to get slightly simpler expressions (avoiding factors (−1)i) put cj = aj if j is odd, and cj = M−aj if j is even, and similarly di = bi if i is odd, and di = M−bi if i is even. (So that c1 = d1 = 1.) Now the expression for Sij becomes Sij = M+1−cj−di if i+j is odd, and Sij = cj+di−1 if i+j is even. (Check that this correctly gives the first row and the first column, and has 2x2 subsquares summing to 2M.)

The numbers cj and di cannot be chosen arbitrarily: we must make sure that the Sij are all distinct numbers in 1, ..., n2.

Assume the additional restriction di = dn+1−i. We have to make sure that the n2/2 numbers cj+di−1 (j=1,...,n and i=1,...,n/2) pick one of the two numbers x and M−x for x=1,...,n2/2.

Assume that all numbers di are equal mod n. (Then they are all 1 mod n.) Assume that the n numbers cj are distinct mod n, and that the n/2 numbers di are distinct. Then the value of cj+di−1 mod n determines j and we see that these n2/2 numbers are all distinct. Clearly, they are all larger than or equal to 1. In order to make sure that cj+di−1 is never larger than n2, assume that the cj are the numbers 1,...,n. Finally, in order to make sure that no two numbers add up to M it suffices that no two (di−1)/n add up to n−1.

These are very light restrictions, so we easily find squares of order n with the 2x2 subsquare property for all even n. Note however that there is no guarantee yet that all row sums will be equal. (But the column sums will, due to the symmetry in the di.)

Example For n=2 the above yields the square

```     1 3
4 2
```
which does not have equal row sums. (And, clearly, there is no 2x2 magic square.)

Example For n=4 the above yields the square

```      1 15  4 14
8 10  5 11
9  7 12  6
16  2 13  3
```
that achieves constant row and column sums, but no uniform half row and half column sums. (Here we had c1 = 1, c2 = 2, c3 = 4, c4 = 3, d1 = 1, d2 = 9. The condition to fulfil for constant row sums was c1+c3 = c2+c4.)

Example For n=12 one can write down the magic square

```   1 142   6 138   8 140   2 141  10 136  12 134
132  15 127  19 125  17 131  16 123  21 121  23
97  46 102  42 104  44  98  45 106  40 108  38
36 111  31 115  29 113  35 112  27 117  25 119
73  70  78  66  80  68  74  69  82  64  84  62
96  51  91  55  89  53  95  52  87  57  85  59
49  94  54  90  56  92  50  93  58  88  60  86
72  75  67  79  65  77  71  76  63  81  61  83
109  34 114  30 116  32 110  33 118  28 120  26
48  99  43 103  41 101  47 100  39 105  37 107
13 130  18 126  20 128  14 129  22 124  24 122
144   3 139   7 137   5 143   4 135   9 133  11
```
with constant row and column sums 870, and constant half row and half column sums 435, but not all "bent rows" have the same sum 870.

# HSA

The reason to write down the above were Dutch newspaper articles today (2007-03-23) that mentioned that three secondary school pupils Petra Altena (15), Jesse Hoekstra (17), Willem Schilte (17) attempted to construct a Benjamin Franklin magic square of order 12 and made the below.
```   1 142  11 136   8 138   5 139  12 135   2 141
120  27 110  33 113  31 116  30 109  34 119  28
121  22 131  16 128  18 125  19 132  15 122  21
48  99  38 105  41 103  44 102  37 106  47 100
73  70  83  64  80  66  77  67  84  63  74  69
60  87  50  93  53  91  56  90  49  94  59  88
85  58  95  52  92  54  89  55  96  51  86  57
72  75  62  81  65  79  68  78  61  82  71  76
97  46 107  40 104  42 101  43 108  39  98  45
24 123  14 129  17 127  20 126  13 130  23 124
25 118  35 112  32 114  29 115  36 111  26 117
144   3 134   9 137   7 140   6 133  10 143   4
```
with constant row and column sums 870, no constant half row and half column sums, but with constant third row and third column sums, and good "bent rows" and good diagonals.

A magic square of order 12 with constant half row and half column sums and also constant "bent row" sums seems missing so far.