Rational approximations of $\sqrt{2}$ in $\mathbb{R} \times \mathbb{Q}_7$ Note: this question was updated (2) after GNiklasch's answer was posted, and taking Gro-Tsen's comment into account. The initial question (1) dealt with $\mathbb{Q}_3$.

 Original post (1). Let's try to solve the equation $x^2 - 2 = 0$ with $x = \frac{a}{b} \in \mathbb{Q}$.  We can't have $x^2 \neq 2$, so the best we can do is minimize $|x^2 - 2|$.  Let's try to find an approximation that works over two different completions.  Can we have this?

$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_3 \ll \frac{1}{a}  $$
I'm trying to write an $S$-adic approximate solution over two places $S = \{ 3, \infty\}$ and $x = \frac{a}{b} \mapsto (\frac{a}{b}, \frac{a}{b}) \in \mathbb{Q} \times \mathbb{Q} \subset \mathbb{R} \times \mathbb{Q}_3 $.   What are the correct exponents?

Edit (2). Perhaps i need to find a problem statement that has a solution. gro-tsen suggests I change $p=3$ to $p=7$ so that $\sqrt{2}\in \mathbb{Q}_7$. 
$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 2 |_7 \ll \frac{1}{a}  $$
possibly I can leave the places the same and change the thing I'm approximating.
$\sqrt{7}\in \mathbb{Q}_3$ so perhaps I can find a rational number $\frac{a}{b}\in \mathbb{Q}$ such that
$$ |(\tfrac{a}{b})^2 - 2 |_\infty \ll \frac{1}{a} \text{ and } |(\tfrac{a}{b})^2 - 7 |_3 \ll \frac{1}{a}  $$
Excuse me while I try to state an instance of weak approximation that's not vacuous.
 A: Squares of elements of the $3$-adic integers $\mathbb{Z}_3$ are congruent to $0$ or $1$ modulo $3$, thus they are all at $3$-adic distance $1$ from $2$.
Squares of elements of $\mathbb{Q}_3 \setminus \mathbb{Z}_3$ are $3$-adically even further away from the target.
Thus in $\mathbb{Q}_3$, you cannot approximate a square root of $2$ at all well - just like you can't approximate a square root of $-1$ at all well in the reals.
A: Here is a full solution for the modified problem, inspired by Gro-Tsen's valuable comment.
1. There are infinitely many rational numbers $a/b\in\mathbb{Q}$ in lowest terms such that
$$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{1}{b}\qquad\text{and}\qquad
\left|\frac{a^2}{b^2}-2\right|_7\ll\frac{1}{b}.$$
To see this, we work in $\mathbb{Z}[\sqrt{2}]$, the ring of integers of $\mathbb{Q}(\sqrt{2})$. In this ring, the rational prime $(7)$ splits as $(3+\sqrt{2})(3-\sqrt{2})$, while the totally positive units are $(3+2\sqrt{2})^m$. For any positive integer $n$, we can choose the positive integer $m$ so that
\begin{align*}a+b\sqrt{2}&=(3+2\sqrt{2})^m(3+\sqrt{2})^n\asymp 7^n,\\
a-b\sqrt{2}&=(3-2\sqrt{2})^m(3-\sqrt{2})^n\asymp 1.\end{align*}
This is because $a^2-2b^2=7^n$ holds regardless of $m$. By basic arithmetic in $\mathbb{Z}[\sqrt{2}]$, the integers $a$ and $b$ are relatively prime to each other and to $7$ as well. Moreover, $a$ and $b$ are positive and of size $\asymp 7^n$. It follows that
$$\left|\frac{a^2}{b^2}-2\right|_\infty=\frac{7^n}{b^2}\asymp\frac{1}{b}
\qquad\text{and}\qquad
\left|\frac{a^2}{b^2}-2\right|_7=7^{-n}\asymp\frac{1}{b}.$$
2. By modifying the above argument, we can achieve that
$$ \left|\frac{a^2}{b^2}-2\right|_\infty\ll\frac{c}{b}\qquad\text{and}\qquad
\left|\frac{a^2}{b^2}-2\right|_7\ll\frac{c^{-1}}{b}$$
for any constant $c>0$. This is essentially best possible, because it is straightforward to see that
$$ \left|\frac{a^2}{b^2}-2\right|_\infty\left|\frac{a^2}{b^2}-2\right|_7\geq\frac{1}{|b^2|_\infty|b^2|_7}\geq\frac{1}{b^2}.$$
