This is my working
$p^2-q^2=(p+q)(p-q)$
As both the integers on right hand side are even .It is divisible by 4.
Primes are of form either $3k+1$ or $3k+2$ .we get 3 possible cases for for $p$ and $ q$ After substituting them we notice that it is also divisible by 3. It is divisible by both 4 and 3. Therefore it is also divisible by 12.But how to show that it is divisible by 24
As $p$ is odd, $$(p + q) + (p - q) = 2p \equiv \pm 2 \mod 4.$$ But if none of $p + q$ or $p - q$ was divisible by $4$, then we’d have $$(p + q) + (p - q) \equiv (\pm 2) + (\pm 2) \equiv 0 \mod 4.$$ Remember that they both are even, being the sum of two odd numbers.
All in all, one of $p + q$ and $p - q$ must be divisible by $4$, the other still being even.
(With the same trick you get easily that $p^2 - q^2 \equiv 0 \mod 3$, as $1$ is the only nontrivial square modulo $3$ and $p$ and $q$ are not divisible by $3$.)
$p=6k\pm1$ and $q=6m\pm1$ so
$p^2-q^2=(6k\pm1)^2-(6m\pm1)^2=(36k^2\pm12k+1)-(36m^2\pm12m+1)$
$=12k(3k\pm1)-12m(3m\pm1)$.
Now note that $2|k$ or $2|3k\pm1$ and same for $m$.
