Originally Posted by

**Defunkt** Well, if $\displaystyle \left(\frac{(2m-1)!}{m},m\right)=1 \Rightarrow $ $\displaystyle \exists x,y \in \mathbb{Z} ~ s.t. ~ x\frac{(2m-1)!}{m} + ym = 1 \overbrace{\Rightarrow}^{t = -y} x\frac{(2m-1)!}{m} = tm + 1 \equiv 1 \pmod{m}$

So these results are actually equivalent,

Not at all! For example, $\displaystyle (2,5)=1$ , but of course this doesn't mean $\displaystyle 2=1\!\!\!\!\pmod 5$ ...The same can be said, of course, for non-primes.

The problem is precisely to prove that $\displaystyle \frac{(2m-1)!}{m}=1\!\!\!\!\pmod m$ whenever $\displaystyle m$ is a prime. Of course this fraction is an integer: that's trivial, as you note. What isn't trivial is that it equals precisely 1 in the (multiplicative) group $\displaystyle \left(\mathbb{Z}\slash m\mathbb{Z}\right)^{*}$, and this is what I attempted, and suceeded (unless proven wrong), to prove in my prior post.

Tonio

however I don't agree that what you showed proves the claim... It is fairly trivial, even without using Wilson's theorem, that $\displaystyle (2m-1)! \equiv m \pmod{m}$; if we let $\displaystyle t = (m-1)!, ~ k = (m+1)(m+2)...(2m-1)$ then $\displaystyle (2m-1)! = ktm \equiv m \pmod{m}$