Belarusian Mathematical Olympiad

Answer: $(n; p) = (3; 2)$.

It is clear that $p \ne n$. If $p = 2$, then $n \ge 3$ and we have $2^8 - 2^4 = 240 = 3^5 - 3$, i.e. $p = 2$, $n = 3$ is a solution. On the other hand, if $n > 3$, then $n^5 - n = n(n^4 - 1) > 3(3^4 - 1) = 240$, i.e. for $p = 2$ there are no $n$ different from $3$ satisfying the initial equality.

Now let $p > 2$. Then $p$ is an odd prime number and $n \ge 3$. We rewrite the initial equality in the form
$$n(n-1)(n+1)(n^2+1) = p^4(p^4-1). \quad (*)$$
Note that exactly one of four co-factors in the left-hand side of the equation can be divisible by $p$. Indeed, $n$ is coprime with any of numbers $n-1, n+1, n^2+1$. From the equalities $n+1 = (n-1)+2$, $n^2+1 = (n-1)^2+2(n-1)+2$, $n^2+1 = (n+1)^2 - 2(n+1) + 2$ it follows that the greatest common divisor of any two of three numbers $n-1, n+1, n^2+1$ is equal to $1$ or $2$. Therefore, any two of them have not $p$ as a common divisor.

Thus, exactly one of four co-factors in the left-hand side of $(*)$ is divisible by $p$, and so, it is divisible by $p^4$. Then this co-factor is not less than $p^4$. In any case $n^2 + 1 \ge p^4$ or $n^2 \ge p^4 - 1$. So $p^4(p^4 - 1) = n(n^2 - 1)(n^2 + 1) \ge n(p^4 - 2)p^4$, whence $p^4 - 1 \ge n(p^4 - 2) > 2(p^4 - 1)$, which is impossible. Therefore, the pair $(n; p) = (3; 2)$ is a unique solution of the given equation.