Showing that $1$ is the only root in $f=x^5 -1 \in \mathbb{F}_p[x]$ if $p-1$ is not divisible by $5$

Question

I am trying to show that if $p-1$ is not divisible by $5$ then $1$ is the only root in $f=x^5 -1 \in \mathbb{F}_p[x]$, where $p$ is a prime.

I can see that $f = (x-1)(x^4 + x^3 + x^2 + x + 1)$ so it suffices to show that if $p-1$ is not divisible by $5$ then $x^4 + x^3 + x^2 + x + 1 \in \mathbb{F}_p[x]$ has no roots.

I wanted to show that $x^4 + x^3 + x^2 + x + 1$ is irreducible when $p-1$ is not divisible by $5$ (if $p-1$ is divisible by $5$ then $x^4 + x^3 + x^2 + x + 1$ is not always irreducible, e.g. for $p=11$ then $x=3$ is a root), but got stuck. How can I proceed? Any hints are much appreciated.

Answer 1

Suppose $a\in \mathbb F_p$ is a root, then $a^5=1$. You also have $a^{p-1}=1$ in $\mathbb F_p$ and if $5$ and $p-1$ are coprime this implies $a=1$.

Answer 2

By Fermat Little Theorem, if $x \in (\mathbb{Z}/p\mathbb{Z})^*$ you have

$$x^{p-1}-1=0 \mod p$$

and the order of every non zero element of $(\mathbb{Z}/p\mathbb{Z})^*$ divides $p-1$

Now, if $a$ is a root of $x^5-1$, the order of $a$ must divide both $5$ and $p-1$ and then $a$ must have order equal to $GCD(p-1,5)=1$ and then $a=1$.

Answer 3

Generalizing: let p and q be primes, such that q does not divide p-1.Then we have the following:

The map from $$F_p*$$ to itself defined by $$f(x)=x^q$$ is an isomorphism.

Proof:it obviously conserves multiplication, so we only have to prove that $$f(x)\neq f(y)$$ for distinct x and y.Assuming we have $$x^q=y^q,$$ then the element $$a=xy^{-1}$$ would have order q, but $$a^{p-1}=1$$ and (p-1,q)=1, a contradiction.Therefore f is an isomorphism and exactly one element maps to 1, namely 1.In fact, we have showed that every member of our group is a q-th power.