Problem 34 *

Prove that for all primes p > 3, 24 \mid (p^2 - 1).

[originally posted by Star-girl on TSR]

Write:

p^2 - 1 = (p - 1)(p + 1)

Note that one of p - 1, p, p +1 is divisible by 3. Since p is prime and not 3 itself, this must be either p - 1 or p +1. Since p is odd, (p > 3) both p - 1 and p + 1 are even. Since p - 1 and p + 1 are “consecutive” even numbers, one of them must also be divisible by 4. (since every other even number is divisible by 4, if one is 0 \pmod 4, the next is 2 \pmod 4, then 0, etc.)

So (p - 1)(p + 1) is divisible by 3 \times 2 \times 4 = 24.