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How to prove that the number $1!+2!+3!+\dots+n!$ is never square?

Show that the sum $$\sum_{k=1}^nk!\neq m^2$$for any integer $m$, for $n\geq4$.

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marked as duplicate by Chris Eagle, Jonas Meyer, martini, Davide Giraudo, draks ... Dec 7 '12 at 10:23

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Using $m^2 \equiv 0,1,4 (\mathrm{mod}\;5)$. –  cjackal Dec 7 '12 at 7:08

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At $4$ the sum is $33$. Beyond $4$, every new term is divisible by $5$. So at $4$ or beyond, the sum is $\equiv 3\pmod{5}$. Nothing congruent to $3$ modulo $5$ can be a perfect square.

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