Recall that $$\sum_{j=1}^nj^2=\frac{n(n+1)(2n+1)}{6}.$$ When is this quantity a perfect square? It appears that the only solutions are $n=0,1,24.$ By setting $x=12n+6$, the problem reduces to finding rational points on the elliptic curve $$ y^2=x(x+6)(x-6), $$ but I do not know where to go from here. How can I show that these are the only solutions?
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user236182
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3This is the Cannonball Problem. The only positive integer solutions to $\sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6}=m^2$ are $(n,m)=(1,1),(24,70)$. One elementary solution can be found here (by W. S. Anglin). – user236182 Jan 01 '16 at 12:37
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@user236182, please add your comment as an answer. – lhf Jan 01 '16 at 12:48
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http://math.hashcode.ru/questions/71079/диофантовы-уравнения-суммы-последовательных-квадратов?страница=1&focusedAnswerId=71097#71097 – individ Jan 01 '16 at 13:36
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Here is JSTOR link to the Anglin's paper. – Martin Sleziak Jan 08 '16 at 13:18