Probability that number is divisible by 5 Two numbers $x$ and $y$ are chosen at random without replacement from the set $${1,2,\cdots,5n}$$.then the probability that $x^4-y^4$ is divisible by $5$ is----
I could write down $x^4-y^4$ as $(x-y)(x+y)(x^2+y^2)$ and then i could easily figure out the possoble cases for first and second brackets but i couldnot figure out for the third bracket.
Please help me in this regard.
Thanks.

Note that if $x$ is not a multiple of $5$ then $x^4-1$ is divisible by $5$ (why?).
This implies that if $x$ and $y$ are not a multiple of $5$ then $x^4-y^4=(x^4-1)-(y^4-1)$ is divisible by $5$.
Can you take it from here?
P.S. Finally you will find that $x^4-y^4$ is NOT divisible by 5 iff the set $\{x,y\}$ ($5n(5n-1)/2$ choices) contains a multiple of $5$ ($n$ choices) and a not-multiple of $5$ ($4n$ choices). Hence the probability is
$$1-\frac{(n)\cdot (4n)}{\frac{5n(5n-1)}{2}}.$$
