Is it true that every polynomial $f(x) $ ( not identically $0$ ) has a multiple $g(x)=f(x)h(x)\ne0$ in which every exponent is prime , that is $g(x)$ is of the form $\sum_{p \text{ is prime}} a_p x^p$ ?
Asked
Active
Viewed 97 times
1 Answers
3
It can be proven that, for any infinite set $S$ of natural numbers and any polynomial $f$ (assumed to be over $\mathbb{C}$, which easily implies it over $\mathbb{R}$ as well), there must be some multiple $g$ of $f$ such that $$g(x)=\sum_{s\in S}a_sx^s.$$
To prove this, simply note that the condition $f$ divides $g$ can be rewritten as:
If $f$ has a root of degree $d$ at $x$, then $g$ has a root of at least degree $d$ at $x$ as well
which is a consequence of the fact that any polynomial can be written as a product of linear factors. Moreover, this means that, since "has a root of degree $d$ at $x$" is equivalent to saying that $f^{(i)}(x)=0$ for any $0\leq i < d$ where $f^{(i)}$ is the $i^{th}$ derivative of $f$. Thus, we can expand the condition that $f$ divides $g$ into a series of $\deg(f)$ equalities of the form (for sequence $n_i$ and $x_i$): $$g^{(n_i)}(x_i)=0$$ or, equivalently, $$\sum_{\substack{s\in S\\s\geq n_i}}a_s\frac{s!}{(s-n_i)!}x_i^{s-n_i}=0.$$ This means we have to solve a system of $\deg(f)$ equations to find an appropriate selection of $a_s$ - but this is easy; in particular, let $I\subset S$ be a set such that $|I|=\deg(f)+1$ and consider the vector space of polynomials expressible as $\sum_{i\in I}a_ix^i$. This has dimension $\deg(f)+1$. Notice that the set of solutions to each equation $$\sum_{\substack{s\in I\\s\geq n_i}}a_s\frac{s!}{(s-n_i)!}x_i^{s-n_i}=0$$ is a linear subspace of dimension $\deg(f)$. The intersection of two such subspaces has dimension at least $\deg(f)-1$ - and the intersection of the subspaces associated to each of the $\deg(f)$ equations has dimension at least $1$, implying there is some non-zero solution to every equation and that there is, thus, some multiple with exponents only in $I\subset S$.
Milo Brandt
- 60,888