# Thread: Inverse product of residues of a prime

1. ## Inverse product of residues of a prime

Hi

I'm trying to prove that $\prod_{i=1}^{p-1}\Bigl(\displaystyle\frac{1}{i}\Bigr) \equiv -1 \ (mod \ p)$

The way I see it, the inverse of $1$ is itself, and from $2$ to $p-2$, the inverses are paired up, so we are left with the inverse of $p-1$ which is itself.

So, we have that $\prod_{i=1}^{p-1}\Bigl(\displaystyle\frac{1}{i}\Bigr) \equiv p-1 \equiv -1 \ (mod \ p)$

My problem (with this proof) is in showing that the inverses are all paired up from $2$ to $p-2$.
a) Is there a way to prove/show this?

or
b) Is there another way to prove $\prod_{i=1}^{p-1}\Bigl(\displaystyle\frac{1}{i}\Bigr) \equiv -1 \ (mod \ p)$?

2. That is Wilson's Theorem actually. - see here -

To show the inverses are 'paired' show that $f:\left\{ {2,...,p - 2} \right\} \to \left\{ {2,...,p - 2} \right\}$ defined by: $
f\left( x \right) = \left( {x^{ - 1} \bmod .p} \right)
$
is injective and that $f(x)=x$ can't happen. Then for each number 2 ... p-2 you will have a unique inverse there - different from itself-, which will be the inverse of that number and no other.

3. Here is one other possible proof. $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic of order $p-1$; let $\alpha$ be a generator. Then

$\prod_{j=1}^{p-1}j^{-1}=\alpha^{-(1+2+...+(p-1))}=\alpha^{-p(p-1)/2}=\alpha^{-(p-1)/2}=-1$

because $\alpha^{\pm (p-1)/2}$ is just the Legendre symbol, and a primitive root is obviously not a square (or else all elements of $\mathbb{Z}/p\mathbb{Z})$ would be squares).

4. Yes, it is indeed Wilson's theorem, written a little differently . I didn't see that before, hehehe, my bad

$\prod_{i=1}^{p-1}\displaystyle\frac{1}{i} \equiv \prod_{j=1}^{p-1}j \equiv (p-1)! \equiv -1 \ (mod \ p)$

Hehehe ... I'm learning lots of interesting things, and I'm just in awe of how these things work out , like how there's different ways of showing the same thing