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

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

Can this be proved with an elementary argument? 

Answer

Let 
\[N=p(p^2-1)=(p-1)p(p+1).\]
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.