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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