Proof of Hasse's principle for quadratic equations

Question

I am currently tackling the following problem.

Problem

Consider the equation $x^2 = q, $ where $ q \in \mathbb{Q}$. Show this has a rational solution $x$ in $\mathbb{Q}$
if and only if there are real solutions and p-adic solutions in $\mathbb{Q_p}$ for all prime $p$.

Progress

Applying the p-adic norm to both sides of the equation, I deduce $\|x\|_p^2 = \|q\|_p $,
which implies $2v_p(x) = v_p(q)$, where $v_p(x)$ is the p-adic valuation of $x$.
From here I am unsure how to proceed, I would really appreciate some pointers!

Answer 1

Extended hints:

1. If the equation $x^2=q$ has a rational solution $x=m/n$, then we can view $m/n$ as a real number as well as as a $p$-adic number for all $p$. Therefore the equation has a solution in those fields as well.
2. For the other direction we assume that a rational number $q=m/n$ has a square root in the reals as well as in all the $p$-adics. Using the prime factorizations of both $m$ and $n$ we can write $$q=(-1)^\epsilon\prod_{i=1}^kp_i^{a_i}$$ for some finite set of prime numbers $p_i$ and some exponents $a_i\in\Bbb{Z}, i=1,2,\ldots,k$, and some $\epsilon\in\{0,1\}$.

• From the assumption that $q$ has a square root in the reals show that $\epsilon$ must be even.
• From the assumption that $q$ has a square root in $\Bbb{Q}_{p_i}$ show that $a_i$ must be even.
• Deduce that $q$ has a square root in the rationals.