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.
 A: 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 $100$.
A: HINT:
If $\displaystyle 5|(x^4-y^4)$ 
Case $1:$
$\displaystyle 5|x^4$
as $5$ is prime, it must divide $x$ and subsequently $5|y^4\iff 5|y$
Case $2:$
As $5$ is prime, if $5\nmid x,(5,x)=1$
Now, $\displaystyle x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2-4+5)=(x-1)(x+1)(x-2)(x+2)+5(x^2-1)$
If $5\nmid x,5$ must divide exactly one of $(x-1),(x+1),(x-2),(x+2)$
$\displaystyle\implies 5\mid(x^4-1)$
Similarly, $5\nmid y,(5,y)=1$
Now there are $\displaystyle\frac{100}5=20$ multiples of $5$ in the given set
We need $$P\{xy, (xy,5)=1 \text{ or } (5|x\text{ and } 5|y)\}$$
Can you take it from here?
