If $p$ is prime, then $x^p − 2 \equiv 0 \pmod{p^k}$ has a solution for $k \ge 2$.
I'm supposed to either prove or disprove the statement above using Hensel's Lemma.
So far what I have is assuming $r$ is a solution to $x^p − 2 \equiv 0 \pmod{p^{k-1}}$. I can show that $f'(r) \equiv 0 \pmod p$ since $f'(r) = p(r^{p-1})$.
I think the next step should be showing that $f(r)\not\equiv0\pmod{p^k}$ which would show that there are no solutions but I don't know how to go about doing this without any way to properly represent $r$.
The claim is true very, very rarely (at least for "small" $p$). Even for $k=2$. In fact, the challenge here is to find a single prime $p$ for which the claim holds. Indeed, only two are known.
To see this, note that the unique solution to $x^p\equiv 2\pmod p$ is $x\equiv 2 \pmod p$.
If we try to lift this to a solution $\pmod {p^2}$ we write $x\equiv 2 +tp\pmod {p^2}$ and try to solve for $t$. We get $$x^{p}\equiv (2+tp)^{p}\equiv 2^{p}\pmod {p^2}$$
Note that this does not depend on $t$.
Thus we want a prime $p$ such that $$2^{p}\equiv 2 \pmod {p^2}$$
Such primes are known as Weiferich primes and they are staggeringly rare for small numbers. Indeed, only two examples are known, namely $1093$ and $3511$. And people have searched rather a lot. Up to about $5\times 10^{17}$ according to OEIS.