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For $n > 5$ and $1 < k < n+1$ prove that $n! + k$ has a prime divisor that is greater than $n$.

This question appears as an exercise in the book "Not always buried deep" by Paul Pollack. There is a reference, but the paper is in German! It seems to be enough to prove the existence of a prime divisor of $n!+k$ that is greater than $k$ for if all prime divisors of $n!+k$ were less than $n+1$, then they would divide n! and hence would have to divide $k$. The case $k = 1$ is easy to handle, but am stuck for $k > 1$.

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This isn't a complete solution, but maybe it leads to one:

Suppose $1 \le k \le n$, and all prime divisors of $n!+k$ are $\le n$.

Suppose a prime $p$ divides the integer $\frac{n!}{k}+1$. Then $p \mid n!+k$, so $p \le n$, $p \mid n!$, and $p \mid k$. If we had $p < k$, then $p$ divides $(k-1)!$ which in turn divides $\frac{n!}{k}$. But this would imply that $p \mid 1$, which is impossible. Therefore, $k = p$ is prime, and it is the only prime factor of $\frac{n!}{k}+1$; that is, $\frac{n!}{k}+1 = k^r$ for some integer $r > 0$; so $n! = k(k^r-1)$. Also note $k > n/2$ (because otherwise $k^2 \mid n!$ which is impossible).

So it remains to show that this is impossible when $n > 5$. Unfortunately I'm not sure how to do this.


edit: The cited book references Chowdhury, which gives the same argument as me. Then it cites a result by Grundhöfer that says that $n!+k \;(2 \le k \le n)$ can't be a prime power when $n > 5$.

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    $\begingroup$ If $r$ is even, you get $n!=k(k^{r/2}-1)(k^{r/2}+1)$. This looks very unlikely to be possible. $\endgroup$ – xavierm02 Oct 18 '16 at 21:56
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    $\begingroup$ Added some references. I can't read German, but typing into Google Translate does a decent job. $\endgroup$ – arkeet Oct 19 '16 at 1:22
  • $\begingroup$ Thanks you folks! Very informative and useful. $\endgroup$ – student Oct 20 '16 at 10:48

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