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Problem 261

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Pivotal Square Sums

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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平方枢轴数

我们称满足如下性质的正整数k为平方枢轴数：存在整数对m > 0和n ≥ k，使得从k开始往前数(m+1)个连续的数的平方和等于从(n+1)开始往后数m个连续数的平方和。

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

一些较小的平方枢轴数为

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

求≤ 10¹⁰的范围内，所有不同的平方枢轴数之和。.

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