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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.

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    You can use Hensel's Lemma, see https://en.wikipedia.org/wiki/Hensel's_lemma I think you have already shown that $\sqrt[4]{2}\not\in\mathbb{Q}_5$. – 2017-01-16
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    In my answer below, I should have pointed out that by far the quickest route to a root is Newton-Raphson. I got 24 places of $7$-adic accuracy after five iterations. The value was $\dots6261662515141034331550232;$ – 2017-01-18
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    @Lubin Stackexchange lets you edit your answers. looks great! I did not say but I also wondered if it was possible to get $\sqrt[4]{2}$ at two different places. I had to work to find that $\sqrt[4]{2} \in \mathbb{Q}_{23}$ so I might try to approximate this number in $ \mathbb{Q}_{7} \times \mathbb{Q}_{23}$. – 2017-01-18
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    Yes, that’s always possible. “Weak Approximation”, I think — it should be just a matter of applying Chinese Remainder Theorem. There shouldn’t be any problem in $\Bbb Q_{31}$, either… – 2017-01-18

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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.