Let $p$ be an odd prime and $(a,p)=1$. Show that $x^2≡a$ (mod $p)$ has solutions, then $x^2≡a$ (mod $p^n)$ always has solution , for any $n>1$.

I have solved the first part but second part need help.

my solution for first is

If $p$ is odd and $(x^2)-a=0$ (mod $p)$ has a solution $b$, then

$(x^2)-a=(x^2)-b^2=(x-b)(x+b)=0$ (mod $p$)

$⇔x=b $(mod $p$) or $x=-b ($mod $p)$

So, it has a solution.