Let $_{125}$ be the field of $125$ elements. The number of non-zero elements $ ∈ _{125}$ such that $^5 =\alpha?$
I don't know how to approach in a correct way, but $\alpha^5=\alpha\implies \alpha^5-\alpha=0$. So we need to consider a sub-field containing $5$ elements such that every element is a root of above equation. Right? Since we need non-zero elements so the answer should be $4$. Is my reasoning correct? Any help would be great. Thanks.
A typical construction of $\mathbb{F}_{125}$ is as the splitting field of $p(x) = x^{125} - x$ over $\mathbb{F}_5$. Since $x^5 - x$ is a factor of $p(x)$, it follows that $\mathbb{F}_{125}$ contains all the roots of $x^5 - x$ over $\mathbb{F}_5$. We know there are 5 of these, namely the elements of $\mathbb{F}_5$ themselves, so your conclusion is correct.