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.
Showing $p$ divides $(p-2)!-1$ when $p$ is prime
1
$\begingroup$
elementary-number-theory
prime-numbers
-
2Do 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 $(p-2)!\equiv 1\pmod{p}$. – 2012-01-25
-
5Did you mean "$p$ divides $(p-2)!-1$"? – 2012-01-25
-
0@MichaelHardy: of course you are right, careless me. :) – 2012-01-25
-
0@ArturoMagidin: Interesting theorem, thanks! – 2012-01-25
-
0Does not compute. Only 1 and p dividies p for p prime. – 2012-01-25
-
0@all: [Laylady's comment] answers an earlier version of the question (which asked why $(p-2)!-1$ divides $p$). – 2012-01-25