This was something cool I saw during Error Correcting Codes. Let $\gamma$ be a primitive root mod $p$, and let $f,g$ be two polynomials of degree at most $p-2$ mod $p$.

Suppose that $f,g$ vanish on $\gamma,\gamma^2, \cdots, \gamma^{p-k-1}$. Show that the coefficientwise product $f\odot g$ vanishes on $\gamma,\cdots, \gamma^{p-2k}$, defined by $$(f\odot g)(x) = \sum_{i=0}^{p-1} f_ig_i x^i$$ if $f(x) = \sum_i f_ix^i, g(x) = \sum_i g_ix^i$.

Solution.

First note that $$\sum_{i=0}^{p-2} (\gamma^k)^i = \begin{cases} 0 & \text{if }(p-1) \nmid k\\ -1 & \text{if }(p-1) \mid k \end{cases}$$

Consider the space of polynomials vanishing on $\gamma,\cdots, \gamma^{p-k-1}$ (among polynomials of degree up to $p-1$). This space should be of dimension $k$, and note that the polynomials $$f_m = \sum_{i=0}^{p-1} \gamma^{mi}x^i = \sum_{i=0}^{p-1} (\gamma^m x)^i$$ form a basis for $m=0,\cdots, k-1$ (since $f_m(\gamma^{j}) = \sum_i (\gamma^{m+j})^i = 0$ as long as $m + j \le p-2$). Hence there exists a degree $\le k-1$ polynomial $\tilde f$ such that $\tilde f(\gamma^i) = f_i$.

Reversing the above argument shows that $f$ (of degree $\le p-2$) that vanish on $\gamma, \cdots, \gamma^{p-1-k}$ are precisely the $f$ whose coefficients can be interpolated by a degree $\le k-1$ polynomial $\tilde f$.

We are done, since the coefficients of $f\odot g$ can be interpolated by a degree $\le (2k-2)$ polynomial, so $f\odot g$ vanishes on $\gamma,\cdots, \gamma^{p-2k}$.