mathschallenge.net

Problem

Prove that p²1 is divisible by 24 for all primes, p > 3.

Solution

For all primes greater than 3, p = 6k1.

When p = 6k+1, p² = 36k² + 12k + 1 and so p²1 = 36k² + 12k.

Similarly, when p = 6k1, p²1 = 36k² 12k.

Therefore, p²1 = 36k² 12k = 12k(3k 1).

If k is odd, 3k 1 will be even and so we prove that p²1 will always be divisible by 12 2 = 24.

What can you say about p³1?

What about other powers of p?