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When is $n!+10$ is a perfect square ? I have tried and found that only for $n=3$ is $n!+10$ a perfect square. Is there any other solution to this?

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In base $10$, the digits of a perfect square must end in $1,  4,  6,  9, 00$, or $25$. However, for $n\geq 10$, $n!$ is divisible by $100$ and so the last two digits of $n!+10$ are $10$. Therefore, $n!+10$ cannot be a perfect square for $n\geq 10$. And you can just check directly that for $1\leq n\leq 10$, only $n=3$ gives a perfect square.

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    $\begingroup$ You can get even a sharper bound since if $n\geq 4$ then we have that $n!+10$ is divisible by $2$ but not by $4$. $\endgroup$
    – kingW3
    Feb 24, 2015 at 13:10
  • $\begingroup$ @kingW3 Oh nice observation, thanks. $\endgroup$
    – Casteels
    Feb 24, 2015 at 13:14
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A perfect square is congruent to $0$ or $1 \bmod 4$.

If $n\ge 4$ then $n! \equiv 0 \bmod 4$ and so $n!+10 \equiv 2 \bmod 4$ is not a square.

The case $n \le 3$ is dealt by hand.

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