continued fraction expression for $\sqrt{2}$ in $\mathbb{Q_7}$ Hensel's lemma implies that $\sqrt{2}\in\mathbb{Q_7}$. Find a continued
fraction expression for $\sqrt{2}$ in $\mathbb{Q_7}$

There's a bit of a problem with defining continued fractions in the $p$-adics. The idea for finding continued fractions in $\Bbb Z$ is that we subtract an integer $m$ from $\sqrt{n}$ such that $\left|\sqrt{n} - m\right| < 1$. We can find such an $m$ because $\Bbb R$ has a generalized version of the division algorithm: for any $r\in\Bbb R$, we can find $k\in\Bbb Z$ such that $r = k + s$, where $\left|s\right| < 1$. This means that we can write
$$
\sqrt{n} = m + (\sqrt{n} - m) =  m + \frac{1}{1/(\sqrt{n} - m)},
$$
and since $\left|\frac{1}{\sqrt{n}-m}\right| > 1$, we can repeat the process with this number, and so on, until we have written
$$
\sqrt{n} = m_1 + \cfrac{1}{m_2 + \cfrac{1}{m_3 + \dots}},
$$
where each $m_i\in\Bbb Z$. However, in $\Bbb Q_p$ there is no such division algorithm because of the ultrametric property: the norm $\left|\,\cdot\,\right|_p$ in $\Bbb Q_p$ satisfies the property that $\left|a - b\right|_p\leq\max\{\left|a\right|_p,\left|b\right|_p\}$, with equality if $\left|a\right|_p\neq\left|b\right|_p$. If $\sigma\in\Bbb Z_p$, then $\left|\sqrt{\sigma}\right|_p\leq 1$, as $\sqrt{\sigma}$ is a $p$-adic algebraic integer. Then we may find an integer $\rho$ such that $\left|\sqrt{\sigma} - \rho\right|_p < 1$, but there's a problem: if we write $\sqrt{\sigma} = \rho + \frac{1}{1/(\sqrt{\sigma} - \rho)}$, we have $\left|\frac{1}{\sqrt{\sigma} - \rho}\right|_p > 1$. As any element of $\Bbb Z_p$ has norm less than or equal to $1$, the ultrametric property of $\left|\,\cdot\,\right|_p$ tells us that we can never find an $p$-adic integer to subtract from $\frac{1}{\sqrt{\sigma} - \rho}$ that will give the difference norm less than $1$. So we cannot write
$$
\sqrt{2} = a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \dots}}
$$
with $a_i\in\Bbb Z_7$, as we can in $\Bbb R$. If you wished, you could remedy this by taking your $a_i\in\Bbb Q_7$, but since $\sqrt{2}\in\Bbb Q_7$ already, this is silly.
