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How can we prove that if : $n+1|n!+1$ then $n$ is a prime number.
My try:
I try to do contrapositive: If $n$ is composite then $n+1 \nmid n!+1$ and then to get that if $n$ is composite so $n+1=ab$ when $2 \le a \le n $ so that $a|n+1$ but here I don't have idea how to continue, is this the good way?

Bill Dubuque
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  • I want to understand something: if $n+1|n!+1$ so that $n+1 \equiv n!+1 (mod p)$ ? –  Jun 09 '21 at 22:34
  • @Xavi: No, $n+1\mid n!+1$ is the same as $n!+1\equiv 0\pmod{n+1}$, or equivalently $n!\equiv -1\pmod{n+1}$. – Troposphere Jun 09 '21 at 22:36
  • @Troposphere And then how you can know that $n$ is a prime number? –  Jun 09 '21 at 22:37

1 Answers1

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If $n+1=ab$ then what you now want to disprove is that $ab \mid n!+1$.

However, $a$ is among the factors of $n!$. How could $n!+1$ then be a multiple of $ab$?

This is one of the directions of Wilson's theorem, which is usually phrased:

$n\in\mathbb N$ is a prime number if and only if $(n-1)!\equiv -1 \pmod n$.

(Note that the $n$ in statement is what your statement calls $n+1$).

Troposphere
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  • Yes, Wilson's Theorem does have 2 directions. The easier direction is the exercise given here. While the other direction, showing that $n+1$ a prime implies $n+1$ does divide $n!+1$ is much harder. – Mike Jun 09 '21 at 22:49
  • Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jun 09 '21 at 23:48