2 numbers are chosen at random from set $\{1,2,3..5n\}$ (without replacement). Find the probability that $n_1^4-n_2^4$ is divisible by 5 All the numbers can be written in the form of $5k,5k+1,5k+2,5k+3,5k+4$
Also the given expression can be written as
$$(n_1-n_2)(n_1+n_2)(n_1^2 + n_2^2)$$
Case 1
Let $n_1-n_2$ be a multiple of 5
Which implies that both numbers have to be of the from $5k$, ie. $n$ possible numbers
Case 2
Let $n_1+n_2$ be a multiple of 5
So $(n_1,n_2)$ can be $(5k,5k), (5k+1, 5k+4), (5k+4, 5k+1), (5k+2,5k+3), (5k+3, 5k+2)$
With $n$ cases for each ordered pair
Case 3
Let $n_1^2 + n_2^2$ be a multiple of 5
So here cases can be 5k and 5k only (IMO)
The cases with $(5k,5k)$ are repeated 3 times, so count that as one case only
So
$$P=\frac {\binom n2 + 2\times \binom n2 +2\times \binom n2}{\binom {5n}{2}}$$
But that doesn’t match with the given answer $\frac{17n-5}{5(5n-1)}$
Where am I going wrong?

Idea:  separate into two sets $$S_1=\{5,10,15,..,5n\}\\S_2=\{1,2,3,4,6,7,8,9,11,...,5n-1\}$$now note that dividing by $5$ for each number (to the power of 4)has two remainders
$$a=5k+0\to r_1=\underbrace{a^4}_{mod 5}=0\\
a=5k+1\to r_2=\underbrace{a^4}_{mod 5}=1^4=1\\
a=5k+2\to r_3=\underbrace{a^4}_{mod 5}=2^4=16(mod \ 5)=1\\
a=5k+3\to r_4=\underbrace{a^4}_{mod 5}=3^4=81(mod \ 5)=1\\
a=5k+4\to r_5=\underbrace{a^4}_{mod 5}=2^4=256(mod \ 5)=1\\$$ so $n_1^4,n_2^4$ can choose from $S_1$ and $S_2$ such that $5|n_1^4-n_2^4$ now
$$P=\frac { \binom n2 + \binom {4n}2}{\binom {5n}{2}}$$
