## Pivotal Square Sums

### Problem 261

Published on Friday, 23rd October 2009, 05:00 pm; Solved by 425

Let us call a positive integer k a square-pivot, if there is a pair of integers m 0 and n k, such that the sum of the (m+1) consecutive squares up to k equals the sum of the m consecutive squares from (n+1) on:

(k-m)2 + ... + k2 = (n+1)2 + ... + (n+m)2.

Some small square-pivots are

- 4: 32 + 42 = 52
- 21: 202 + 212 = 292
- 24: 212 + 222 + 232 + 242 = 252 + 262 + 272
- 110: 1082 + 1092 + 1102 = 1332 + 1342

Find the sum of all distinct square-pivots 1010.