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!

Printable View

- Oct 17th 2007, 03:04 PMclockinglyWilsons 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! - Oct 17th 2007, 06:10 PMThePerfectHacker
- Oct 18th 2007, 07:59 AMclockingly
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. - Oct 18th 2007, 03:45 PMThePerfectHacker
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.