# Solve the equation, for $p$ prime, $x^{2p}- x^p= [6]$, in $\mathbb {Z}_p$

According to the title, the equation is :

$$x^{2p}- x^p=[6]$$

for $p$ prime, in $\mathbb {Z}_p$.

It's known that $a^p \equiv a \text {mod}p$. So, if the equationwas $a^p -a=[6]$ , $a=6$.

I tried to mke some manipulations using as tool this theorem, but I am lost here.

I can't figure this out after lots of time thinking, anything ekse could I do?

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$x^{2p} = (x^p)^2$. So, what equation do you get? –  Daniel Fischer Jul 2 at 23:44

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?

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Thanks a lot :D ! –  Charlie Jul 2 at 23:57