For $p$ prime, every prime divisor of $p!+1$ is greater than $p$

Question

I am not sure how to start this queston. Could you give me some hint?

Show that for any prime number $p$, every prime divisor of $p!+1$ is greater than $p$.

Thanks

The assumption that $p$ is a prime is unnecessary. See this earlier Question Every prime divisor of $n! + 1$ is an odd integer greater than $n$ — hardmath, Jan 28 '22 at 17:46

score 1 · Answer 1 · answered Sep 18 '16 at 20:38

Notice that all primes less than or equal to $p$ divide $p!$. Therefore, they cannot also divide $p!+1$. Therefore, $p!+1$ can be represented as a product of primes which are all greater than $p$. This is actually one of the nice proofs showing that there are infinitely many primes.

Mitchell Spector · Answer 2 · 2016-09-18T20:46:02.480

1

Here's a hint: If $q$ is any prime (in fact, any positive number) less than or equal to $p,$ and you divide $p!+1$ by $q,$ what is the remainder?

edited Sep 18 '16 at 20:46

answered Sep 18 '16 at 20:38

Mitchell Spector

10,681

score 1 · Answer 3 · answered Sep 18 '16 at 20:43

the key idea here is that any two consecutive integers are coprimes: for any positive integer $n$ we have: $(n,n+1)=1$.

Thus, $p!+1$ and $p!$ have no common divisor (apart from $1$). But all primes less than $p$ are (by the definition of the factorial) divisors of $p!$ thus none of them can be a divisor of $p!+1$.

Thus: the least prime divisor of $p!+1$ must necessarily be greater than $p$.

score 0 · Accepted Answer · answered Sep 18 '16 at 20:39

0

Why don't you start by assuming that there is a prime divisor which is less than or equal to $p$ and try to show contradiction?

answered Sep 18 '16 at 20:39

Daniel

127

score -1 · Answer 5 · answered Sep 18 '16 at 20:38

-1

Hint. If $q\leq p$ then $q\mid p!$

answered Sep 18 '16 at 20:38

Bart Michels

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For $p$ prime, every prime divisor of $p!+1$ is greater than $p$

5 Answers5