# Let $𝔽_{125}$ be the field of $125$ elements. The number of non-zero elements $𝛼 ∈ 𝔽_{125}$ such that $𝛼^5 =\alpha$?

Let $𝔽_{125}$ be the field of $125$ elements. The number of non-zero elements $𝛼 ∈ 𝔽_{125}$ such that $𝛼^5 =\alpha?$

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.

• You showed that there should be at most four non-zero elements $\alpha\in\mathbb{F}_{125}$ such that $\alpha^5=\alpha$. Can you explain why there are four such elements? – Guy Aug 5 '16 at 15:30
• No it is just my thought. As $0$ obviously satisfies the equation, so over a field of $5$ elements, the non-zero elements remains 4 – Harry Potter Aug 5 '16 at 15:47
• Have you seen the fact that a field of 125 elements necessarily has a subfield of 5 elements? – Jyrki Lahtonen Aug 5 '16 at 15:54

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.
• Why are these the only solution of $p(x)$? – Sahiba Arora May 27 '17 at 20:20
• $p$ has degree 125, and so it cannot have more roots. – Mr. Chip May 27 '17 at 23:14