For a prime p>3, prove that p^2 - 1 is divisible by 12
For any prime p>3, p^2-1 > 6.
All odd numbers > 6 are between 2 consecutive multiples of 6.
Between every 2 consecutive multiples of 6, there are 3
consecutive odd numbers, 6k+1, 6k+3, and 6k+5. I.e.,
6k < 6k+1 < 6k+3 < 6k+5 < 6k+6=6(k+1)
Thus every odd number > 6 can be written as 6k+1, 6k+3 or 6k+5.
As all primes greater than 3 are odd, they can all be written
as 6k+1 or 6k+5, (but never as 6k+3 since that is never prime).
(6k+1)²-1 = 36k²+12k+1-1 = 36k²+12k <-- a multiple of 12
(6k+5)²-1 = 36k²+60k+25-1 = 36k²+60k+24 <-- a multiple of 12
That proves it.
Edwin
