I have a "cool" polynomial that I want to prove is irreducible. Let $$P(X) = X^7 - X - 1 \in (\Bbb Z/7\Bbb Z)[X].$$
I wish to show that if one root of $P$ is in $\Bbb Z / 7 \Bbb Z$ then all roots of $P$ are in there.
I've made a few observations that I'd like (maybe) to develop but I'm not quite sure how I might do so! Here they are;
Suppose that $\alpha$ is a root of $P$ so that $P(\alpha) = 0$. Then $X-\alpha \mid P$. If we divide out by $X-\alpha$ we get
$$P(X) = (X-\alpha)(X^6 + \alpha X^5 + \alpha^2 X^4 + \alpha^3 X^3 + \alpha^4X^2 + \alpha^5 X + \alpha^6 - 1) = \ell(X)s(X).$$
Now suppose that $\beta \neq \alpha$ is another root of $P$. Then $\beta - \alpha \neq 0$ so $\beta$ must be such that $s(\beta) = 0$. If we evaluate $s$ at $\beta$ we get
$$s(\beta) = \beta^6 + \alpha\beta^5 + \alpha^2\beta^4 + \alpha^3\beta^3 + \alpha^4 \beta^2 + \alpha^5 \beta + \alpha^6 - 1.$$
If we define $\pi : \alpha \mapsto \beta$ then $\pi(s(\beta)) = s(\beta)$. In fact it seems if we apply $\pi$ to $P$ then we get $P(X) = (X-\beta)(X^6 + \beta X^5 + \beta^2X^4 + \beta^3X^3 + \beta^4X^2 +\beta^5X +\beta^6 - 1)$ and we know that $\alpha$ is a root of this sextic factor. Denote by $S(X)$ the division of $P$ by $X - \beta$. Then $X-\alpha \mid S(X)$ so that $P(X) = (X-\alpha)(X-\beta)F(X)$ where $F$ is some degree $5$ factor.
Can I use this method to systematically "drag" linear factors out of $P$ based on the assumption that $P$ has a root in $\Bbb Z/7 \Bbb Z$, and use this to prove that $P$ is irreducible?
Hint: Show that if $\alpha$ is a zero of $P(x)$ then so is $\alpha+1$. Induction shows that $\alpha+k$ is a zero for all $k$.
See this thread for more discussion and links. There are no roots in $\Bbb{Z}/7\Bbb{Z}$. All seven of them are in the field $\Bbb{F}_{7^7}$.