Probability that $5 \mid x^4 - y^4$ for random $x, y$ Two numbers $x$ and $y$ are chosen at random without replacement from the set $\{1,2,3,\cdots,100\}$. Find the probability that $x^4 - y^4$ is divisible by $5$.
I don't know how to proceed with this problem. So any help would be appreciated.

Here are the fourth powers modulo $5$:
$$
0^4 \equiv 0, 1^4 \equiv 1, 2^4 \equiv 1, 3^4 \equiv 1, 4^4 \equiv 1
$$
(You can calculate these by hand, or conclude using Fermat's little theorem.)
Anyway, $x^4 - y^4$ is a multiple of $5$ if
and only if $x^4 \equiv y^4 \pmod 5$.
There are $20$ multiples of $5$ and $80$ non-multiples, 
so the number of unordered pairs of numbers where $x^4 \equiv y^4 \pmod 5$ is
$$
{20 \choose 2} + {80 \choose 2}
$$
You should then divide this by the total number of ways of choosing $2$ numbers out of $100$.
