I'm seeking to prove or disprove the following statement: If $a \in \mathbb{Q}^*$ has a square root in $\mathbb{Q}_p$ for all primes $p$, then it has a square root in $\mathbb{Q}$. (It is not assumed that $a$ has a root in $\mathbb{R}$).
Now from much trial and error I think this statement is true.
If $a$ is positive and there is some $x \in \mathbb{Q}_p$ such that $x^2 = a$ for all primes $p$ then by considering norms it follows that each power of $p$ in $a$ occurs with even exponent, and so we can clearly find a square root of $a$ in $\mathbb{Q}$.
However this doesn't work when $a$ is negative - how can we get a root of $a$ in $\mathbb{Q}$ in this case?