Show that over $\mathbb{Z}_7$, $x^5-1$ has no roots other than 1.

Question

Show that over $\mathbb{Z}_7$, $x^5-1$ has no roots other than 1.

I know I can iterate through each member of $\mathbb{Z}_7$ and show that each one, when raised to the 5th power, is not equal to 1 mod 7, but is there an easier way, or a more general way?

Answer 1

Any nontrivial root of $x^5-1$ will be an element of order $5$ in $\mathbb{Z}_7^\times$, which has order $6$. This cannot happen because of Lagrange's theorem.

Alternatively, if $x^5=1$ then $x\ne0$ and so $x^6=1$ by Fermat's theorem. But then $1=x^6=xx^5=x$.

Answer 2

You may show that the map $\psi:x\to x^5$ is bijective over $\mathbb{F}_7^*$, since it is an involution: $(x^5)^5 = x^{25}\equiv x^1 = x\pmod{7}$ by the Fermat's little theorem. That gives $\left|\psi^{-1}(1)\right|=1$.

Answer 3

Lil' Fermat here says for all $x\neq 0$, $\;x^5=x^{-1}$. Hence $\;x^5=1\iff x^{-1}=1\iff x=1$.

Answer 4

You can use this fact for more general:

$x^4+x^3+x^2+x+1$ is the $5$-th cyclotomic polynomial. Since the minimal pozitive integer $k$ such that $$7^k \equiv 1 \mod 5$$ is $4$, this polynomial is irreducible over $\mathbb F_7$.

Hence no more root in $\mathbb F_7$.