I'm fairly new to number theory and have just started to study p-adic numbers. In a book I'm reading about the fundamentals of number theory, there is the following statement that is supposed to be easy. However I did not find a proof.
Let $p \neq 2$ be a prime number, then $1+px^2$ is a square iff $x\in\mathbb Z_p.$
One of my first thoughts was to use the unique representation of $x\in\mathbb Z_p,$ i.e. $x = u p^n,$ where $u\in\mathbb Z_p^\times,\ n\in \mathbb N_0.$
I'd be really glad if someone might hint me into the right direction here.
For $p=2$ it is not true, $3$ isn't a square $\bmod 4$.
For $p$ odd and $v_p(x)<0$ then $v_p(1+px^2)=v_p(px^2)$ is odd so $1+px^2$ can't be a square (in $\Bbb{Q}_p$). For $v_p(x)\ge 0$ then $1+px^2$ is a square due to Hensel lemma (or the binomial series).