$48|p^4-q^4$ for all prime numbers $p,q>3$

Question

Prove that for all prime numbers $p,q>3$ holds $$48|p^4-q^4$$

My solution:

$$48=2^43$$ $$p^4-q^4=(p-q)(p+q)(p^2+q^2)$$ Since adding or subtracting odd numbers, we get an even number, $p^4-q^4$ is divisible by $2^3$. How about another $2$ and $3$?

Answer 1

Since $p,q\equiv 1,3\pmod 4$, one of $p-q,p+q$ is $\equiv 0\pmod 4$.

Since $p,q\equiv 1,2\pmod 3$, one of $p-q,p+q$ is $\equiv 0\pmod 3$.

Since each of $p-q,p+q,p^2+q^2$ is even, $p^4-q^4$ is divisible by $48$.

Answer 2

For $3$, note that $2^4 \equiv 1^4 \equiv 1 \pmod 3$, so $p^4 - q^4 \equiv 1 - 1 \equiv 0 \pmod 3$.

For $2^4$, you have $(p + q) \equiv (p - q) \equiv 0 \pmod 2$. Since $p, q$ are odd, they are equivalent to $1$ or $3 \pmod 4$. $3^2 \equiv 1^2 \equiv 1 \pmod 4$ so $p^2 - q^2 \equiv 1 - 1 \equiv 0 \pmod 4$.

Putting these together gives the result.