TITLE: Elementary proof for: If x is a quadratic residue mod p, then it is a quadratic residue mod p^k
QUESTION [7 upvotes]: In article solving quadratic congruences, it is shown how to use Hensel's lemma to iteratively construct solutions to to $x^2 \equiv a \pmod{p^k}$ from the solutions to $x^2 \equiv a \pmod{p}$. The case where $p=2$ is treated separately.
While the construction is elegant, it is a tad lengthy and its reliance on Hensel's lemma makes it a bit far from elementary number theory.
If we are only concerned with an existential proof (rather than a constructive one), can we simplify the proof? That is, is it possible to succinctly prove the following theorem without resort to Hensel's lemma?

For any prime $p$ and any $k \in \mathbb{N}$, if $x$ is a quadratic residue mod $p$, then it is a quadratic residue mod $p^k$.

REPLY [2 votes]: I have long appreciated this result in LeVeque's Fundamentals of Number Theory, Theorem 4.4:
Suppose $p$ is a prime relatively prime to $a$.  Let $t_n$ be the order of $a$ modulo $p^n$ and assume that $p^z$ exactly divides $a^{t_1} - 1$.  Then if $p > 2$ or $z > 1$,
$$ t_n = \begin{cases} t_1, \quad &\text{for $n \leq z$}\\\\ t_1 p^{n-z}, \quad &\text{for $n > z$.}\end{cases}$$
This result can be used to prove several congruences mod a prime power that are otherwise messy and annoying to prove (for instance, see Other ways to deduce Cyclicity of the Units of certain groups?).
In your question, if $a$ is a quadratic residue modulo $p$, then $t_1$ divides $\frac{p-1}{2}$.  Thus in both of the above cases, $t_n$ is a divisor of $p^{n-1} \cdot\frac{p-1}{2} = \phi(p^n)/2$, so $a$ is a quadratic residue modulo $p^n$.