Prime equation

we can write the equation as $p(p-1)=(n-1)(n^2+n+1)$ .

if $p|n-1$ and $n^2+n+1|p-1$ we have $n^2+n+1\leq p-1\leq n-2$ and this is impossible.

It must be $p|n^2+n+1$ and $n-1|p-1$ . We call $a=\frac{p-1}{n-1}=\frac{n^2+n+1}{p}$ . so we have $p=an-a+1$ and $ap=n^2+n+1$ . From this two equations it follows $p(a^2-n-2)=(3a-n-2)$ so it must be $a\leq 3$ and we can easily check that for a=1,2 there is no solution and for $a=3 \rightarrow p=19 , x=7$