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.

(excuse my incorrect statement earlier that 4 is the only perfect square that's = 4 mod 8. The square of twice any odd number has such a property. It doesn't affect the thrust of my point, thankfully.)

The four by four grid can be approached similarly to the 2x2 and also can never yield a solution. The lower rightmost 4x4 square has a total of 328 = 41*8. Translating (as in my previous post) to the right preserves the total; translating upward adds 160 (=40*4) to the total. In this case, the total is always = 8 mod 16 since 328 = 8 mod 16 and both translation moves preserve residue mod 16.

A square of an odd number can never = 8 mod 16 (an even number). Even numbers are either of the form 4n or 4n+2 and these have squares that are = 0 and 4 mod 16 (respectively). Accordingly, no perfect square = 8 mod 16 and so no 4x4 square is a solution to this problem.

The case of 6x6 squares can also be eliminated by noting that the lower rightmost square sums to 61*18 = 61*9*2 = 2 mod 8. Translating upwards adds 60*6 to the sum which = 0 mod 8. So the 6x6 squares all equal 2 mod 8 and are therefore not solutions.

The lower rightmost 8x8 square totals 81*32, and translating it upwards adds 80*8, so all 8x8 squares = 32 mod 64. Only even numbers potentially are candidates, and these come in 4 flavors mod 8: 8n, 8n+2, 8n+4, and 8n+6. Squaring these reveals that even perfect squares must be equal to one of (0, 4, 16, 36) mod 64 and so all 8x8 squares are excluded.

There's just the one 10x10 square, totalling 5050, which isn't a perfect square.

We can be sure then, that all solutions are squares with an ODD side length.

Posted by Paul on 2008-05-21 21:20:42