Find the minimum $k >2$, such that there are $k$ consecutive numbers such that their sum of squares is a square.
Solution
Let $k=2m+1$ and let the $k$ numbers be $(n-m),(n-(m-1)), \dots, n, \dots, (n+m)$. Then after adding the squares, we have
$(2m+1)\left(n^2+\frac{m(m+1)}{3}\right)$
has to be a square, which working case by case, has the smallest solution for $m=5$ and $n=23$ (or $n=1$ for any integers).
Letting $k=2m$, then adding the squares gives
$m\left(2n^2+2n+\frac{2m^2+1}{3}\right)$
which analyzing the first cases gives no solution for $m<6$ and hence $k=11$ is the minimum.
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