Two numbers and are chosen at random (with out replacement) from the set {1, 2, 3, …., 5n}. The probability that $n_1^4-n_2^4$ is divisible by 5, is equal to 

$\frac{17 n-5}{5(5 n-1)}$

$n_1^4-n_2^4=\left(n_1^2+n_2^2\right)\left(n_1+n_2\right)\left(n_1-n_2\right)$

We can segregate the numbers as multiple of five

$(5 \lambda), 5 \lambda+1,5 \lambda+2,5 \lambda+3,5 \lambda+4$, as follows:

$R_1 \rightarrow 1 \quad 6 .....5 n-4 \longrightarrow 5 \lambda+1$

$R_2 \rightarrow 2 \quad 7 .....5 n-3 \longrightarrow 5 \lambda+2$

$R_3 \rightarrow 3 \quad 8 .....5 n-2 \longrightarrow 5 \lambda+3$

$R_4 \rightarrow 4 \quad 9 .....5 n-1 \longrightarrow 5 \lambda+4$

$R_5 \rightarrow 5 \quad 10 .....5 n-\longrightarrow 5 \lambda$

We can select either both the number from R₅, or any two numbers from the first four rows.

(As square of any number that is not a multiple of three will be in the form of 5λ' + 1 or 5λ' – 1)

Thus, required probability

$=\frac{{ }^n C_2+{ }^{4 n} C_2}{{ }^{5 n} C_2}$

$=\frac{n(n-1)+4 n(4 n-1)}{5 n(5 n-1)}$

$=\frac{17 n-5}{5(5 n-1)}$