Does a set of simultaneous equations have solution in $\Bbb Q_p$

Question

Does

$2X^2+1＝Y^2$

$2X^2-1＝2Z^2$

have solution in $\Bbb Q_2$? I know Hensel's lemma, but I couldn't use that.So I tried to go back to the definition of $p$-adic field, but failed. Thank you in advance.

Answer 1

Big hint: The problem is equivalent to finding $w,x,y,z\in\Bbb{Z}_p$ such that $$2x^2+w^2=y^2\qquad\text{ and }\qquad 2x^2-w^2=2z^2,$$ with $w\neq0$. Reducing mod $8$ and some consideration then shows that $w$, $x$, $y$ and $z$ are all even.

Answer 2

While one might be able to solve this via quadratic forms, this specific one can be solved with two basic facts, namely the well-known ultrametric principle

$$v_p(a) \neq v_p(b) \implies v_p(a \pm b) = \min(v_p(a), v_p(b))$$

as well as the fact that $v_2(a^2)$ is even, and $v_2(2a^2)$ is odd, for every $a \in \mathbb Q_2^\times$. Namely, assuming there are solutions, the first line would imply $v_2(X) \ge 0$, but the second line $v_2(X) \le -1$.