It is possible to find rational numbers that approximate square roots in $\mathbb{R}$. For example,
$$ \frac{7}{5} < \sqrt{2} < \frac{3}{2}$$
but we can also solve the equation $x^2 \equiv 2$ in $7$-adic numbers since
$$ 3^2 \equiv 2 \mod 7$$
so at least we know the first digit $\sqrt{2} = \dots 3$. Can we find rational numbers that approximate $\sqrt{2}$ in both spaces at once:
$$ x \approx \sqrt{2} \text{ in } \mathbb{R} \times \mathbb{Q}_7$$
To clarify there are two different norms here:
the norm over $\mathbb{R}$ - what we call "absolute value" - is an non-archimedian norm . It's just plain old: absolute value.
the norm over $\mathbb{Q}_7$ is such that $|| 7^k|| = \frac{1}{7^k}$. This leads to p-adic numbers
So I would like a solution that is close in both norms: $$ || x^2 - 2 ||_\mathbb{R} < \frac{1}{7^2} \quad\text{ and }\quad || x^2 - 2 ||_{\mathbb{Q}_7} < \frac{1}{7^2}$$
This looks good. Start with $p/q = 3/1.$ at each stage, multiply the column vector by $$ \left( \begin{array}{cc} 75 & 106 \\ 53 & 75 \end{array} \right) $$ The factored row is $p^2 - 2 q^2.$