Let $p$ be an odd prime, $a\in \mathbb{Z}$ with $(a,p)=1$. I am trying to show that if $a$ is a square modulo $p$ then it is a square modulo $p^k$. I managed to prove this using an exponential function and finding explicitly the square root of $a$ in $\mathbb{Z}_{p^k}$, but it is rather long. Would someone have a simpler proof using the results of quadratic residues for odd primes?
Thanks a lot,
Hensel's Lemma might be useful here. Use $x^2 - a \equiv 0 \pmod p$ as the polynomial equation and lift to $p^2$, $p^3$, ... (you can use induction to finish it off).
Since $f'(x) = 2x$, you're guaranteed that you can apply the lemma because if $X$ is a solution of the original equation, then $2X$ will never be divisible by $p$ because $a$ and $p$ are coprime.