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So, can $\sum_{i=1}^n i^2$ be written as a sum of $n-1$ different perfect squares? Surely if we are looking at this problem with small numbers, the answer is both yes and no. If we take $n$ to be 3, there can not be two different perfect squares numbers which satisfy the statement. If we take $n$ to be 4, we can see that $3^{2}+4^{2}=5^{2}$ which means that the statement is satisfied.

So, my truly question is: how can we find if a certain $n$ number is solution or not for this statement? (for example: Can $\sum_{i=1}^{2015} i^2$ be written as a sum of $2014$ different perfect squares?)

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  • $\begingroup$ I was pointing out the main idea of that sum. $1^{2}+2^{2}+3^{2}+4^{2} = 1^{2}+2^{2}+5^{2}$ $\endgroup$ – KroTeK Dec 11 '18 at 22:30
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Let us replace 2015 by $N$. What you found, is that: if there exists $1 \leq u<v \leq N$ such that $u^2+v^2$ is the square of some integer that is greater than $N$, then your statement holds. Now, let $w=45^2+4^2=2041, u=2*45*4=360, v=45^2-4^2=2009.$ Then $w^2=u^2+v^2$.

To generalize, let $m$ be the smallest integer such that $m^2>N$. Let $k$ be the smallest integet such that $v=m^2-k^2 \leq N$. Then, if $0<u=2*m*k<v$, $u^2+v^2=(m^2+k^2)^2$ and the statement holds.

Now, if $v=0$, then $N < m^2-(m-1)^2=2m-1 < 2\sqrt{N}+1$, thus $N < 7$.

Thus, if $N > 6$, $k<m$ and $2mk+k^2 \leq 3mk$.

Now, $k < 1+\sqrt{m^2-N} \leq 1+\sqrt{2\sqrt{N}+1} < 1+\sqrt{2m+1} \leq m/3$ if $m \geq 24$ (corresponding to $N \geq 23^2=529$) so $0<u<v$ holds.

As a conclusion, if $N$ is large enough (at least $529$) your result holds.

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