What is the probability that $x^4-y^4$ is divisible by 5? Two numbers $x$ and $y$ are chosen at random without replacement from the set $\{1,2,3...,5n\}$.
What is the probability that $x^4-y^4$ is divisible by $5$?
I divided the numbers into groups of 5 $(1,2,3,4,5),(6,7,8,9,10),...$.The probability in the first group would itself be the answer.But how to find that?

$x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y)$.
If $x = 5k + j; 0 \le j < 5$ then $5|x^4 - y^4$ if $y = 5l + m; 0 \le m < 5$ where $m = j$; $m = 5 - j$; $m^2 + $j^2 = 5V$.  
If $j = 0$ $5|x^4 - y^4 \iff m = 0$.  
As $1 = 1 = 5 -4; 1^2 + 2^2 = 5; 1^2 + 3^2 = 10; 2 = 2 = 5 -3; 2^2 + 4^2 = 20; 3^2 + 4^2 = 25$.  If $j \ne 0$ $5|x^2 - x^4 \iff m \ne 0$.  
So... the ways it can be divisible by 5:
The first element is of form $a = 5k$ and the second is of form $b = 5m$.  The probability of this is: 1/5*(n-1)/(5n - 1).
The first element is not of the form $a = 5k$ and the second is not of the form $b = 5m$.  The probability of this is: 4/5*(4n -1)/(5n - 1).  
These two possibilities are mutually exclusive so Probability is: [(n-1) + 4(4n - 1)]/5(5n - 1) = [17n -5]/[25n - 5]
