The numbers from 1 to 100 are arranged, snaking back and forth from the bottom, in a 10x10 grid:

100 99 98 97 96 95 94 93 92 91

 81 82 83 84 85 86 87 88 89 90

 80 79 78 77 76 75 74 73 72 71

 61 62 63 64 65 66 67 68 69 70

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

Marked off are two subsquares that, if you add up the numbers within them, the total is the square of one of the numbers within. The larger square (5x5) totals 625, the square of 25, which is found within that marked square. The other one shown is the trivial "1", a 1x1 square, the single number in which of course is its own square.

Find another subsquare where the sum of its contained numbers is equal to the square of one of those numbers.

No 2x2 square can be a solution. Here's why:

Consider the lower rightmost 2x2 square:

20  19

1     2

This square totals 42, which is not a perfect square, and more importantly = 2 mod 8. Now imagine translating the square of interest either to the right or upward. (By "translating" I mean imagining the square as an outline and moving the outline without altering the position of the numbers themselves.)

Translating the square to the right results in no change to the total. That's because one of the outgoing left numbers is replaced by an incoming right number that's two greater and the other by one that's two lesser for a net change of zero.  (Which row increases and decreases depends on the position of the square, but there's always one of each.) Since the sum doesn't change, neither does the sum mod 8.

Translating the square up one results in an increase of 40 to the total. That's because each outgoing number in the bottom row is replaced by a number 20 higher in the incoming top row--2 numbers in the row means a net of +40. Since 40 = 0 mod 8, this translation also does not change the sum mod 8.

So, since the lower right most square has a sum that's = 2 mod 8, and all translations of that square preserve the remainder mod 8, then ANY 2x2 square has a sum that's = 2 mod 8. Since all perfect squares are equal to either 1, 4, or 0 mod 8 (and only 4 itself = 4 mod 8) none of these sums can be a perfect square.

Solutions with 1x1, 3x3, and 5x5 grids have already been shown, but I don't think we've established whether the list of known solutions is complete. Hopefully this is a step in that direction, by excluding one class of squares entirely.

Posted by Paul on 2008-05-21 20:50:41