## Pivotal Square Sums

### Problem 261

Published on Friday, 23rd October 2009, 05:00 pm; Solved by 457Let 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}+ ... +`k`^{2}= (`n`+1)^{2}+ ... + (`n`+`m`)^{2}.Some small square-pivots are

**4**: 3^{2}+**4**^{2}= 5^{2}**21**: 20^{2}+**21**^{2}= 29^{2}**24**: 21^{2}+ 22^{2}+ 23^{2}+**24**^{2}= 25^{2}+ 26^{2}+ 27^{2}**110**: 108^{2}+ 109^{2}+**110**^{2}= 133^{2}+ 134^{2}

Find the sum of all **distinct** square-pivots ≤ 10^{10}.