Quote by mikeyflex
Hi Dick,
The problem stated that p is a prime number that is greater than 3. When you work out p^2(mod 12) you get a remainder of 1 for every prime number square that is greater than 3. This is not the case with 2 or 3. I was wondering if there was some proof behind this.

A direct way to see this is that p(mod 12) can only be 1,5,7 or 11. Otherwise it is divisible by 2 or 3. But 1^2, 5^2, 7^2 and 11^2 are all equal to 1 mod 12. This is what matt was alluding to when he said all primes are equal to +1 or 1 mod 6.