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 $$n(n-1)(n+1)(n^2+1)=p^4(p^4-1).(\ast )$$ 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 $(\ast )$ 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.