# Prove that $p^2-1$ is divisible by 24 for any prime number $p > 3$.

## Answer

\[N=p(p^2-1)=(p-1)p(p+1).\]

One of the three consecutive numbers $p - 1$, $p$, $p + 1$ is divisible by $3$. Also, since $p - 1$ and $p + 1$ are two consecutive even numbers one of them must be divisible by $2$ and the other should be divisible by $4$. Thus $N=(p-1)p(p+1)$ is divisible by $24=3 · 2 · 4,$ i.e.

\[(p-1)p(p+1)=24 k, k \in \mathbb{N}.\]

Since $p>3$ is prime, $2$ or $3$ do not divide $p$. Therefore $(p-1)(p+1)=p^2-1$ is also divisible by $24$.

The answer is accepted.

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