Thread: Wilsons Proof help please

The proof is:

Prove that the value of (p-1)! mod p is always p-1.

Not really sure how to approach this so any help would be greatly appreciated!

Originally Posted by clockingly

The proof is:

Prove that the value of (p-1)! mod p is always p-1.

Not really sure how to approach this so any help would be greatly appreciated!

This is Wilson's theorem.

The idea is to pair elements say and where . Given we can find so that because this is the solution to congruence. And and so on ... until we only have and . So have .

Thanks! So basically the proof is this?:

In order to compute (p-1)!, you have to find the product of the integers 1 through p-1. In this set of integers, there are groups of two elements (excluding the group 1 and p-1) that can be multiplied together to get a number (p-1)!/(p-1), which equals p-1, which is congruent to 1 mod p.

Originally Posted by clockingly

Thanks! So basically the proof is this?:

In order to compute (p-1)!, you have to find the product of the integers 1 through p-1. In this set of integers, there are groups of two elements (excluding the group 1 and p-1) that can be multiplied together to get a number (p-1)!/(p-1), which equals p-1, which is congruent to 1 mod p.

Exactly. It is like the trick Gauss did when he was a little boy when he was asked to compute he paired them like where appears times. Now here the idea is similar except that the pairing is not as straightforward as in Gauss' trick. But however such a pairing exists. And that is what we do.