Take any polynomial $f \in \mathbb{R}[x]$ with degree $n \ge 1$. Prove that there are real numbers $a_{0}, a_{1}, ... a_{n}$ not all zero such that $f$ divides $\sum_{i=0}^n a_{i}x^{2^i}$
How I want to tackle this problem is that I want to divide each of $x^{2^i}$ by $f$ and I would probably want to show that there is a linear dependence among the remainders of each quotient. However, I am not sure how to actually demonstrate this. How can I do this? Thank you.
On
You are right on track: The remainders are polynomials of degree at most $n-1$, depending on $n$ coefficients, hence they are in a linear space of dimension $n$ (with $\{ 1, X, X^2, \ldots, X^{n-1}\}$ as a basis). However, there are $n+1$ values from $0$ to $n$, hence you have $n+1$ remainders - they can't be linearly independent.
You had the right idea as to how to approach the problem . . .
For convenience, define the degree of the zero poynomial as $-\infty$.
By the division algorithm, for $0\le i\le n$, we can write $$x^{2^i}=q_if+r_i$$ where $q_i,r_i\in\mathbb{R}[x]$, and $\deg(r_i) < n$.
Let $P_n=\{p\in\mathbb{R}[x]:\deg(p) < n\}$.
Then $P_n$ is an $n$-dimensional vector space over $\mathbb{R}$, hence $r_0,...,r_n$ are linearly dependent over $\mathbb{R}$.
Choose $a_0,...,a_n\in\mathbb{R}$, not all zero, such that $$\sum_{i=0}^n a_ir_i=0$$ Then we get \begin{align*} \sum_{i=0}^n a_ix^{2^i}&=\sum_{i=0}^n a_i(q_if+r_i)\\[4pt] &=\sum_{i=0}^n a_iq_if+\sum_{i=0}^n a_ir_i\\[4pt] &=\sum_{i=0}^n a_iq_if\\[4pt] \implies\;\sum_{i=0}^n a_iq_if&=\sum_{i=0}^n a_ix^{2^i}\\[4pt] \implies\;f\sum_{i=0}^n a_iq_i&=\sum_{i=0}^n a_ix^{2^i}\\[4pt] \implies\;f&{\,\Large{\mid}\!}\left(\sum_{i=0}^n a_ix^{2^i}\right)\\[4pt] \end{align*} as was to be shown.