Marked as Duplicate question: (Updated) Is there an expression I can derive to get this?
I know the gcd>1 but I don't know how to get an expression for it? Or if this is not possible, how do I show the gcd>1?
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1See the recently asked Explanation why $\frac{(x-1)!-(x-1)}x$ is always an integer for prime $x$. – dxiv Feb 24 '18 at 05:25
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Read that again. The $,\gcd,$ is in fact $,p,$. – dxiv Feb 24 '18 at 05:29
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1@kemb Prime numbers don't have a lot of divisors. If the gcd is not $1$, what else could it be? – Arnaud D. Feb 24 '18 at 18:08
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Hint: use Wilson's theorem to show that $\,p \mid (p-1)! - (p-1) = (p-1)\big((p-2)!-1\big)\,$, then $\,\gcd(p,p-1)=1\,$ implies that $\,p \mid (p-2)!-1\,$ by Euclid's lemma.
dxiv
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Note that by Wilson's theorem
$$(p-1)!+1\equiv 0 \pmod p\iff(p-1)(p-2)!+1\equiv 0 \pmod p \iff p(p-2)!-(p-2)!+1\equiv 0 \pmod p\iff (p-2)!-1\equiv p(p-2)!\equiv 0 \pmod p$$
thus since $$p|[(p-2)!-1]\implies \gcd((p-2)!-1,p)=p$$
user
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