I don't know how to prove this and it's really bugging me. Thanks to anybody that can help!
Let $n$ be any natural number. Prove that $n! + 1$ contains a prime factor greater than $n$ and use that to prove that there are infinitely many primes.
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Welcome to MSE! Posting your question as the title seems to have cutoff part of it. – Amzoti Feb 26 '13 at 20:50
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Thank you for notifying me of my mistake, I have corrected it and added the whole question to the body. :D – Michael Feb 26 '13 at 20:51
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My copy paste was incorrect. My mistake. :( – Michael Feb 26 '13 at 20:52
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$d\le n\Rightarrow d\mid n!.,$ so $\bmod d!:\ n!\equiv 0,$ so $,\color{#c00}{n!+1\equiv 1},,$ so $,d\mid \color{#c00}{n!+1}\iff d\mid\color{#c00} 1,,$ by divisibility mod reduction $\ \ $ – Bill Dubuque Jun 30 '25 at 15:39
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Hint:
For any $a \in \mathbb{N}$ we have that $a$ and $a+1$ are relatively prime, that is, $\gcd(a,a+1) = 1$, or in other words $d \mid a$ and $d \mid a+1$ implies $d = 1$.
Good luck!
wythagoras
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dtldarek
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@Michael If $n!+1$ itself is a prime, then it is done. If not, it has a prime factor $p$, if $p\le n$, then $p|n!+1$, but we know $p|n!$, that means $p|1$, contradiction. – Yimin Feb 26 '13 at 22:08
