How might one show that $p$ divides $(p-2)!-1$ where $p$ is a prime number? I am not even sure if it is true but I have been randomly trying it on some primes and it seems to be true.
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3Do you know Wilson's Theorem? $(p-1)!\equiv -1\pmod{p}$. Hence $(p-2)!(p-1) \equiv (p-1)\pmod{p}$, and cancelling $p-1$ (which is relatively prime to $p$) we get $(p-2)!\equiv 1\pmod{p}$. – Arturo Magidin Jan 25 '12 at 17:46
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5Did you mean "$p$ divides $(p-2)!-1$"? – Michael Hardy Jan 25 '12 at 17:46
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@MichaelHardy: of course you are right, careless me. :) – Yves Jan 25 '12 at 18:41
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@ArturoMagidin: Interesting theorem, thanks! – Yves Jan 25 '12 at 18:43
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Does not compute. Only 1 and p dividies p for p prime. – Laylady Jan 25 '12 at 17:45
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@all: [Laylady's comment] answers an earlier version of the question (which asked why $(p-2)!-1$ divides $p$). – Srivatsan Jan 25 '12 at 17:52
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Indeed it is, this is a simple corollary of Wilson's theorem: If $p$ is prime, then $(p-1)! \equiv -1$ mod p (the converse is also true, but we don't need that).
So, assuming Wilson's theorem, $-1 \equiv (p-1)! \equiv (p-1)(p-2)!\equiv -(p-2)!$ from which we immediately get $(p-2)! \equiv 1$ mod p.
To prove Wilson's theorem, see here: http://en.wikipedia.org/wiki/Wilson%27s_theorem
Edit: Whoops, I see I was beaten to this answer in the comment thread, sorry.
Hunter Spink
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Thank you, I don't know why the accept this answer button would't work... – Yves Jan 25 '12 at 18:42