I need help finding the $\sqrt[4]{2}$ in two different p-adic regimes $\mathbb{Q}_5$ and in $\mathbb{Q}_7$. The idea is to find $x \in \mathbb{Q}_p$ such that:
$$ \Big|\,x^4 - 2 \,\Big|_p < \frac{1}{p^4} $$
or show that solution does not exist in the two completions $\mathbb{Q}_5$ and in $\mathbb{Q}_7$
Square-roots can be find via Hensel's Lemma. I am asking if I can also find 4-th roots in this way.
$$ \hspace{0.4in}x^4 \equiv 2 \mod 5 $$
has no solutions since $x^4 \equiv 1 $ (unless $x = 0$). Next, what about modulo $7$
$$ 5^4 \equiv 2^4 \equiv 2 \mod 7 $$
Can these numbers be completed to solutions over 7-adic numbers?
$$ (x + 7k)^4 \equiv x^4 + 28k \, x^3 \equiv 2 \mod 7^2$$
My computer gives $k = 3$, so that $x = 5 + 7 \times 3 = 26$ is a solution. So we have:
$$ \sqrt[4]{2} \;\approx_7 \;26 $$
Even a simple PARI-GP script would suffice.
You can use the Binomial Series to get a fourth root of $2$ in $\Bbb Z_7$, the $7$-adic integers: $$ \sqrt[4]2=\sqrt[4]{16-14}=2\bigl(1-\textstyle{\frac78}\bigr)^{1/4}\,. $$ Then use the Binomial expansion of $(1+x)^{1/4}$ with $x=-7/8$, giving you a convergent series, since the binomial denominators are all prime to $7$.
You could also use the logarithmic and exponential series, but I think the Binomial Series method is quicker.