Given $p$ an odd prime, $x^2\equiv a\pmod{p^2}$ and $(a,p)=1$, how could we know that $(x,p)=1$? If I have the congruence $$x^2 \equiv a \pmod {p^2}$$ where $p$ is an odd prime and $(a,p)=1$, how could I know that $(x,p)=1$?

More generally, $\ x^n\equiv a\pmod {\!m},\ (a,m)=1\ \,\Rightarrow\,\ (x,m)=1$.    
$x^n\equiv a\pmod {\!m}\!\iff\! x^n-a=mk\ $ for some $k\in\Bbb Z$.
Assume, for contr., $\ (x,m)\neq 1\ $ - i.e. $\ \exists q\in\Bbb P\ (\ q\mid x\ \text{ and }\ q\mid m\ )$.   
$q\mid x,m\ \,\Rightarrow\,\ q\mid x^n-mk=a\ \,\Rightarrow\, \ q\mid x,m,a\ \,\Rightarrow\, \ (a,m)\ge q>1$, contr.
