Solve the eqation $p^8-p^4=n^5-n$ Solve the equation for primes $p$ and natural numbers $n$ $p^8-p^4=n^5-n$.
For $p=2$ we get $n=3$, bjt for the next 5 prime numbers we get irational numbers. I cant prove (if its true) that there are no more solutions.

if $p\ge 3$ be prime number,then $n\ge 3$,and
$$n(n^2-1)(n^2+1)=p^4(p^4-1)$$
Note
$$\gcd{(n,n^2-1)}=\gcd{(n,n^2+1)}=1,\gcd{(n^2-1,n^2+1)}=1\textbf{or}2$$
then have
$$n^2+1\ge p^4$$
so
$$p^4(p^4-1)=n(n^2-1)(n^2+1)\ge n(p^4-2)p^4$$
$$\Longrightarrow p^4-1\ge n(p^4-2)>2(p^4-1)$$
a contradiction
On the other hand,if $p=2$,then we have
$$2^8-2^4=240=3^5-3\Longrightarrow (n,p)=(3,2)$$
