Let us call a positive integer $k$ a square-pivot, if there is a pair of integers $m>0$ and $n\ge k$, 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:

$$(k-m{)}^2+\cdots +k^2=(n+1{)}^2+\cdots +(n+m{)}^2.$$

Some small square-pivots are

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

Find the sum of all distinct square-pivots $\le {10}^{10}$.