mathschallenge.net

Problem

Prove that $p$²1 is divisible by 24 for all primes, $p$ > 3.

Solution

For all primes greater than 3, $p$ = 6$k$1.

When $p$ = 6$k$+1, $p$² = 36$k$² + 12$k$ + 1 and so $p$²1 = 36$k$² + 12$k$.

Similarly, when $p$ = 6$k$1, $p$²1 = 36$k$² 12$k$.

Therefore, $p$²1 = 36$k$² 12$k$ = 12$k$(3$k$ 1).

If $k$ is odd, 3$k$ 1 will be even and so we prove that $p$²1 will always be divisible by 12 2 = 24.

What can you say about $p$³1?

What about other powers of $p$?