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Pivotal Square Sums
Problem 261
Published on 23 October 2009 at 05:00 pm [Server Time]
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.
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