<s>[INST] Prove by mathematical induction for any prime number$ p > 3, p^2 - 1$ is divisible by $3$? Prove by mathematical induction for any prime number $p > 3,  p^2 - 1$ is divisible by $3$?
Actually the above expression is divisible by $3,4,6,8,12$ and $24$.
I have proved the divisibility by $4$ like:
$$
\begin{align}
p^2 -1 &= (p+1)(p-1)\\
       &=(2n +1 +1)(2n + 1 - 1)\;\;\;\text{as $p$ is prime, it can be written as $(2n + 1)$}\\
       &= (2n + 2)(2n)\\
       &= 4(n)(n + 1)
\end{align}
$$
    Hence $p^2 - 1$ is divisible by 4.
But I cannot prove the divisibility by $3$.  
 [/INST] If $p>3$ is a prime then $3$ does not divide $p$ i.e. $3$ and $p$ are relatively prime , now $p-1$ , $p , p+1$ are three consecutive integers , so one of them must be divisible by $3$ , hence $3$ must divide their product i.e. ,  $p(p-1)(p+1)=p(p^2-1)$ , but $3$ and $p$ are relatively prime , so $3$ must divide $p^2-1$
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