## Let us call a positive integer $k$ a square-pivot, if there is a pair of integers $m \gt 0$ 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 $\le 10^{10}$.

### The problem involves finding square-pivot numbers that are less than or equal to $10^{10}$. A square-pivot is defined in such a way that the sum of the squares of any $(m+1)$ consecutive integers from $k-m$ to $k$, can be equal to the sum of squares of any $m$ consecutive integers from $n+1$ to $n+m$.

The above equation can be re-written as $$\sum_{i=0}^m{(k-i)}^2 = \sum_{i=1}^m{(n+i)}^2$$

Solving the above equation further, we get:

$$k^2+m*(m+1)*(2m+1)/6+m*(m+1)*k/2 = n*(n+1)*(2n+1)/6+n*m*(n+1)/2+(n+1)*m*(m+1)/2.$$

Given $k$, $m$ and $n$ are all positive integers, and $m < k$, we need to find all distinct $k$ that are less than or equal to $10^{10}$.
To solve this problem, use a computer program that can check each possible value of $k$ and $m$ and check the above equations are satisfied. The algorithm requires using a nested loop where the outer loop runs through each possible value of $k$ and for each $k$, the inner loop runs through each possible $m$.
For optimal performance, it might be appropriate to consider some potential enhancements:
1. We can give the boundary conditions as $m

Reachable Numbers

Stone Game