If $p^{a} \mid\mid n$, and $n$ is not a perfect square, can $n$ be a square modulo all powers of $p$?

Question

We have an integer $n$ which is not a perfect square and $p^{a} \mid\mid n$ for some $a \geq 1$. I was wondering if $n$ can be a square modulo all powers of $p$?

My strong feeling is that the answer is NO but how do I prove it?

Answer 1

Yes, it’s possible if $a$ is even (and impossible if $a$ is odd). For example $-25$ is a square modulo all powers of $5$ (use Hensel’s lemma).

This condition is equivalent to asking whether $n$ is a square in the $p$-adic integers $\mathbb{Z}_p$.

Answer 2

From the binomial series for $(1+z)^{1/2}$ we get $$-100 = 5^2 (1-5)= (5\sum_{k=0}^l {1/2 \choose k} (-5)^k)^2 \bmod 5^l$$ where $1/2$ means an inverse of $2 \bmod 5^l$

Hensel lemma is its generalization.