--description--

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:

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

Some small square-pivots are

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

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

--hints--

pivotalSquareSums() should return 238890850232021.

assert.strictEqual(pivotalSquareSums(), 238890850232021);

--seed--

--seed-contents--

function pivotalSquareSums() {

  return true;
}

pivotalSquareSums();

--solutions--

// solution required