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This is a paragraph in David M. Burton, "elementary number theory, seventh edition:

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But I do not understand:

1- How Wilson's theorem implies the existence of an infinitude of composite numbers of the form $n! + 1$?

This is the statement of Wilson's theorem that I know:

$P$ is a prime iff $(p-1)! \equiv -1 (mod p)$

Could anyone explain this for me please?

2- I do not understand the second statement in the paragraph, especially in comparison to the first statement, does they mean that the form $n! +1$ can give us prime and composite numbers ?

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  • $\begingroup$ if $p>3$ is prime, then $(p-1)!+1$ is composite $\endgroup$ – Angina Seng May 24 '19 at 3:11
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    $\begingroup$ Because by Wilson: $(p-1)!\equiv -1\pmod{p}$, and thus, for every prime $p$, $p\mid (p-1)!+1$. Now, since $(p-1)! \geqslant 2^{p-2}$, it is not hard to see that for sufficiently large $p$, $(p-1)!+1>p$, thus, it is a positive integer, being larger than $p$ and divisible by $p$, has to be composite. $\endgroup$ – TBTD May 24 '19 at 14:43
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1) this follows if you also use Euclid's result that there are infinitely many primes, as $(p-1)!+1$ will be composite for each such prime $p$.

2) Indeed, this implies that while we are sure there are infinitely many $n$ such that $n!+1$ is composite, we don’t know if there are infinitely many $n$ for which it’s prime. In other words if $n-1$ is not prime, this doesn’t necessarily imply that $n!+1$ is prime.

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  • $\begingroup$ why you are chosoing $n-1$ exactly that should be not prime? $\endgroup$ – Secretly May 24 '19 at 8:33
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Wilson's Theorem says that if $p$ is prime, then $p$ divides the number $m:=(p-1)!+1$. Since clearly $p < m$ if $p>3$, it follows that $m$ is composite.

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