Prime numbers and divisibility 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 substracting odd numbers, we get a even number, $p^4-p^4$ is divisible by $2^3$. How about another $2$ and $3$?

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.
