In $Q_p$: $f(x)=0$ iff $p(f(x))^2$ is a square

Question

I am reading the paper by Jan Denef: the rationality of the Poincaré series associated to the p-adic points on a variety. In the proof of his first lemma he makes the claim that

For $f\in\mathbb{Z}_p[x_1,\ldots,x_m]$ a polynomial in $m$ variables over the $p$-adic integers and $x=(x_1,\ldots,x_m)\in Q_p^m$ we get $f(x)=0$ if and only if $p(f(x))^2$ is a square, for arbitrary prime $p$.

This seems a triviality but I am not sure how to show the more difficult direction.

Answer 1

For $0\neq y\in \mathbb Q_p$, we have $py^2=z^2$ for some $z$ iff $p=(z/y)^2$ is a square, but $p$ is not a square in $\mathbb Q_p$. Apply to $y=f(x)$.