Comments on: Another proof of Wilson’s Theorem

By: mixedmath

mixedmath — Tue, 03 Mar 2015 09:48:51 +0000

Hi! I'm glad you're interested. Check out the Math Community Blog FAQ at meta.math.se. In short, go to math.blogoverflow.com/wp-admin and login with your math.stackexchange information, and then let one of the administrators know you're interested in the blog chat. We're looking for content!

By: Sabyasachi Mukherjee

Sabyasachi Mukherjee — Mon, 02 Mar 2015 07:20:38 +0000

Hi, How does one contribute to this blog?

By: Thomas Andrews

Thomas Andrews — Thu, 19 Feb 2015 02:41:58 +0000

I'd love to see a purely combinatorial proof. $(p-1)!$ is the number of elements of $\Sigma_p$ of order $p$. But I'm a little unclear on where to go from there.

By: mixedmath

mixedmath — Tue, 17 Feb 2015 22:53:09 +0000

That happens to be the proof that my first number theory instructor showed me (when I was younger and still bounced when I fell). It led nicely into more general polynomial congruences, ultimately leading all the way up to understanding the AKS primality test, if I recall correctly.

By: Smiley Sam

Smiley Sam — Tue, 17 Feb 2015 22:45:21 +0000

That is brilliant! Probably worth mentioning that it uses Euler's (?) theorem that $x^{p-1} = 1$ for all $x /neq 0$, where $p$ is prime. But I love that proof!

By: anonymous

anonymous — Tue, 17 Feb 2015 15:14:16 +0000

This proof is in my old Lithuanian book on math olympiads.

It presents 3 proofs, two of which are the ones shown in this blog and the other one is: $$f(x)\equiv x^{p-1}-1\equiv (x-1)(x-2)\cdots (x-(p-1))\pmod{p}\implies f(0)\equiv -1\equiv (p-1)!\pmod{p}\ \ \ \square$$

By: Marc van Leeuwen

Marc van Leeuwen — Tue, 17 Feb 2015 13:12:00 +0000

The proof is not really that different from the standard proof. Including an initial term $0$ for symmetry, a standard argument for the identity $0+1+\cdots+n=\frac{(n+1)n}2$ would be to add $n +(n-1)+\cdots+0$ in a second row, and to observe that each of the $n+1$ columns add up to $n$. If $n$ is even, as $n=p-1$ is here, then a slight variation of the argument is: the middle term is $\frac n2$, and every pair of terms symmetric about it contributes $n$ to the sum, for a total of $\frac n2+\frac n2n$. Now passing to powers of $g$, the middle factor is $g^{n/2}=-1$, and each pair symmetric about it contributes $g^n=1$, so they are inverses: a pair also matched up in the standard argument. [I've cheated a bit: the outermost pair is is fact twice the same element $g^0=1$, but this does not contribute anything anyway.]