TITLE: Solve the equation, for $p$ prime, $x^{2p}- x^p= [6]$, in $\mathbb {Z}_p$
QUESTION [6 upvotes]: According to the title, the equation is :
$$x^{2p}- x^p=[6]$$
for $p$ prime, in $\mathbb {Z}_p$.
It's known that $a^p \equiv a \mod p$. So, if the equation was $a^p -a=[6]$ , $a=6$.
I tried to make some manipulations using as tool this theorem, but I am lost here.
I can't figure this out after lots of time thinking, anything else could I do?

REPLY [4 votes]: As you know $x^p \equiv x \pmod{p}$, the equation you want to solve is equivalent to:
$$x^2 - x \equiv 6 \pmod{p}$$
The quadratic polynomial factors: $x^2 -x -6 = (x+2)(x-3)$.  So does that get you there?