Let $p$ be an odd prime number, and let $h\in\{1,\ldots,p-1\}$ be a primitive element of the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^{\times}$, i.e., $\{h^{k}\,(\mathrm{mod}\,p):1\leq k\leq p-1\}=\{1,\ldots,p-1\}.$ Also, let $\omega$ be a primitive $(p-1)$st root of unity, and consider the sums $$\lambda_{\ell}^{(m,p)}:=\sum_{k=1}^{p-1}(h^{k}\,(\mathrm{mod}\,p))^{m}\omega^{k\ell},$$ where $m\geq 1$ is an integer, and $\ell\in\{1,\ldots,p-1\}.$ The important thing here is that in the formula above, we're computing $h^{k}\,(\mathrm{mod}\,p)$ to be some number in $\{1,\ldots, p-1\},$ and then just treating it as a regular integer from that point on, i.e., the $m$th power, multiplication with $\omega^{k\ell},$ and summation all occur as usual for complex numbers. For example, with $h=2,$ $\lambda_{\ell}^{(1,5,2)}=2i-4-3i+1,\;-2+4-3+1,\;-2i-4+3i+1,\;2+4+3+1$ as $\ell$ goes from $1,\ldots, 4$, which are respectively $-3-i,0,-3+i,10.$
My question is this: when is $\lambda_{\ell}^{(m,p)}=0?$ I can prove that when $p>3,$ there are always some values $\ell$ for which $\lambda_{\ell}^{(1,p)}=0,$ and I can prove that when $m\geq p-1$, $\lambda_{\ell}^{(m,p)}\neq 0$ for all values of $\ell.$
My conjecture is that for any prime $p>3,$ whenever $m\geq 2,$ $\lambda_{\ell}^{(m,p)}\neq 0$ for all values of $\ell,$ and I checked this numerically for the first 100 primes greater than 3. I've been at a loss for about three years about how to prove this, though.