Proving this modulo question Let a,b be integers and p is prime

$a^pb^{p^2} \equiv 0\pmod p \Rightarrow a \equiv 0\pmod p$ or $b \equiv 0\pmod p$

and using elemination method

$a^pb^{p^2} \equiv 0\pmod p$ and $a \not\equiv 0\pmod p \Rightarrow b \equiv 0\pmod p$

from here
since $a \not\equiv 0\pmod p \Rightarrow p \nmid a \Rightarrow gcd(a,p) = 1$
by Fermat's little theorem we get

$a^p \equiv a\pmod p$

so, 

$ab^{p^2} \equiv 0\pmod p$

again since $ p \nmid a$ 
so 

$p \mid b^{p^2}$

I'm pretty sure $p \mid b^{p^2} \Rightarrow p \mid b$ is wrong
I can't seem to find a way to make it so that $p \mid b$ to get 
$b \equiv 0\pmod p$

$Z/p$ is a field so $a^pb^{p^2}=0$ mod $p$ implies that $a^p=0$ mod $p$ or $b^{p^2}=0$ mod $p$.
If $a^p=0$ $mod$ $p$, Little fermat implies that $a^p=a$ mod $p$ done.
If $b^{p^2}=0$ mod $p$, you have $b^{p^2}=(b^p)^p=b$ mod $p$ by little fermat.
