Show that if $p$ is a prime number $> 3$ then $24 \mid p^2-1$ Hi guys can someone help me with this ?(Without using Modular arithmetic)
Show that if $p$ is a prime number $>3$ then $24$ $\mid$ $p^2-1$

Without (explicitly) using modular arithmetic:
$p^2-1=(p+1)(p-1)$. Since $p$ is odd, this is a product of two even numbers that differ by two, so one of them is in fact divisible by four, and their product is divisible by 8.
Now, $p-1$, $p$ and $p+1$ are three consecutive numbers, so 3 divides one of them. $p$ is prime, so it's not divisible by three. Thus 3 certainly divides $p^2-1=(p-1)(p+1)$.
We conclude that since both 3 and 8 divide $p^2-1$, so too does 24 (using the fact that 3 and 8 are coprime).
