I want to prove the statement "Any prime factor of $x!+1$ is larger than $x$."

Any slight hint will be ok.


Assume $p \leq x$ divides $x! + 1$. Then, $p$ also divides $x!$ (as any $y \leq x$ divides $x!$). Thus, $p$ divides both $x!$ and $x! + 1$. Do you see what this forces $p$ to be?

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  • $\begingroup$ now I see what I missed. thanks! $\endgroup$ – fbg Feb 26 '16 at 9:46

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