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Prove that $\displaystyle \forall n\geq 2, \sum_{k=1}^n (k!)^2$ is never a perfect square.

I'm far from well-read in number theory and I can't make any significant progress with this problem.

I tried to look at the sum $\text{mod}$ some small numbers, to no avail.

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    $\begingroup$ This is not the case: $(5!+1)^2 = 14641 < 15017 = \sum_{k=1}^5 (k!)^2$. $\endgroup$ – user133281 Sep 5 '16 at 7:35
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    $\begingroup$ Did you try modulo $5$? $\endgroup$ – Arthur Sep 5 '16 at 7:37
  • $\begingroup$ @Arthur yeah, that's right, $2$ is not a square mod $5$... $\endgroup$ – Gabriel Romon Sep 5 '16 at 7:40
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Hint: In base $10$, a square number can end only with digits $1, 4, 6, 9, 0$, or $5$. Now note that from $\left(5!\right)^{2} $ all factorials are multiple of $10$ and $$\sum_{k=1}^{4}\left(k!\right)^{2}=617.$$

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