TITLE: Prime Exponent Polynomials
QUESTION [8 upvotes]: This was a problem that my friend had on his final for a discrete math class that he mentioned he couldn't figure out. I tried, but I don't really know how to get started.
Let $f(x)\neq 0$ be a polynomial in $\mathbb{Z}$. Prove that there exists a non-zero polynomial $g(x)$ such that $f(x)g(x)$ has only prime exponents.

REPLY [3 votes]: Let $n$ be the degree of $f$, and let $V_n$ be a $\mathbb{Q}$-vector space of polynomials of degree less than $n$. The vector space dimension of $V_n$ is $n$. Consider $n + 1$ monomials$$x^{p_1}, x^{p_2}, \dots, x^{p_{n+1}},$$where $p_1, p_2, \dots, p_{n+1}$ are arbitrary $n+1$ distinct prime numbers. Consider their remainders modulo $f$. They are $n + 1$ elements of $V_n$, and thus they are linearly dependent. Hence, there are $a_1, a_2, \dots, a_{n+1} \in \mathbb{Q}$ such that$$h(x) = a_1x^{p_1} + a_2x^{p_2} + \dots + a_{n+1}x^{p_{n+1}}$$is divisible by $f$. Clearly, we can assume that $a_1, a_2, \dots, a_{n+1} \in \mathbb{Z}$. Thus, $g$ is the quotient of $h$ by $f$. Scaling $h$ by a proper integer factor, we can also assume that $g \in \mathbb{Z}[x]$ if needed.