IMC 2007 · Day 1 · P5

Let us define a subset $I$ of the polynomial ring $\mathbb{R}[X]$ as follows: $$I = \left\{ P(X) = \sum_{j=0}^{m} b_j X^j : \sum_{j=0}^{m} b_j f(k + j\ell) = 0 \text{ for all } k, \ell \in \mathbb{Z},\ \ell \ne 0 \right\}.$$ This is a subspace of the real vector space $\mathbb{R}[X]$. Furthermore, $P(X) \in I$ implies $X \cdot P(X) \in I$. Hence, $I$ is an ideal, and it is non-zero, because the polynomial $R(X) = \sum_{i=1}^{n} X^{a_i}$ belongs to $I$. Thus, $I$ is generated (as an ideal) by some non-zero polynomial $Q$.

If $Q$ is constant then the definition of $I$ implies $f = 0$, so we can assume that $Q$ has a complex zero $c$. Again, by the definition of $I$, the polynomial $Q(X^m)$ belongs to $I$ for every natural number $m \ge 1$; hence $Q(X)$ divides $Q(X^m)$. This shows that all the complex numbers $$c, c^2, c^3, c^4, \dots$$ are roots of $Q$. Since $Q$ can have only finitely many roots, we must have $c^N = 1$ for some $N \ge 1$; in particular, $Q(1) = 0$, which implies $P(1) = 0$ for all $P \in I$. This contradicts the fact that $R(X) = \sum_{i=1}^{n} X^{a_i} \in I$, and we are done.