The question is related to ring polynoms. But to what else?
Given:
- $P(X) = X^k + \cdots + a_1 X + a_0$ in $\mathbb{Q}[X]$ is normed and irreducible.
- Let $w,v \in \mathbb{C}$ be roots of $P$, such that $q:= v-w$ is in $\mathbb{Q}$.
- Define $P_n(X) := P(X-nq)$, with $n \in \mathbb{N}$.
Prove that: $P_n = P$, for all $n \in \mathbb{N}$.
It is enough to show the claim for $n = 1$, because if $P = P_1$ then $P_n(X) = P(X-nq)$ $ = P(X-q-(n-1)q)$ $ = P_1(X-(n-1)q) = P(X-(n-1)q) = P_{n-1}(X)$ and by induction $P_n = P$. To Clarify $P_n = P$, means that for all $x \in \mathbb Q$, $P_n(x) = P(x)$. This implies that for all $x\in Q$ and $z \in \mathbb Z$, $P(x+zq) = P(x)$.
Now we take a look at any $P\in \mathbb Q[X]$ and any $q\in\mathbb Q$. Assume $P(X-q) = P(X)$, so $P(x-q) = P(x)$ for all $x\in\mathbb Q$. This is clearly true if $q = 0$, so assume $q \not = 0$. Then we get, like above, that $P(x-nq) = P(x)$, so define the Polynomial $Q_{x_0}(X) = P(X)-P(x_0)$ for any $x_0\not = 0 \in \mathbb Q$. This polynomial is zero for all $x_0+nq$. Those are infinitely many different values for which $Q_{x_0}$ is zero, so $Q_{x_0}$ is the zero-polynomial $0$ and thereby $P = P(x_0)$ a constant polynomial.
With this we can see that, as constant polynomials are not irreducible, $q$ has to equal zero, so the claim now translates to: "If $v,w$ roots of an irreducible polynomial $P\in\mathbb Q[X]$ and $v-w \in \mathbb Q$, then v = w"
Assume $v-w = q \not = 0$ then the polynomial $Q(X) = P(X)+P(X-q)$ has root $v$, but $deg(Q)<deg(P)$, this is a contradiction, as $P$ is normed and irreducible with root $v$, it has to be the minimal polynomial of $v$, but $Q$ also has $v$ as a root. Therefor $v-w = 0 \implies v = w$.
EDIT: The comment by Bill leads to an interesting post on the topic, where I got the last paragraph.