I came up with an interesting Hypothesis.
Suppose we are in the field $F, \text{char }F \neq 2$. Let's fix an arbitrary element $a \in F$. Is it true that at least one equation $x^k=n\cdot a$, ($k\in \mathbb{N} \backslash \{1\}, n \neq 0$ in $F$) has roots in $F$?
Examples.
$F = \mathbb{C}-$ obvious.
$F=\mathbb{Z}_p$. Since every nonzero element $a$ has inverse $a^{-1}$ then $x^2 = aa^{-1} = 1$ has roots.
$F = \mathbb{R}$. Either $x^2 = a$ or $x^2 = -a$ has roots.
$F = \mathbb{Q}$. $a = \frac{p}{q}$. Then $x^2 = a \cdot (pq) = p^2$ has roots.
$F = \mathbb{Q}(\sqrt{2})$. $a = \sqrt{2}$. Then $x^3 = 2\sqrt{2}$ has roots.
Counterexamples.
- $F = \mathbb{Q}(\pi)$. $a = \pi$. Since $\pi$ is not an algebraic number then the equation $x^k = n\pi$ has no solutions.
But Can it be generalized somehow?