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.

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    $\begingroup$ The value of $\lambda_{\ell}^{(m,p)}$ depends on the choice of primitive element $h$, but this is not shown in the notation. $\endgroup$ – Servaes Jun 27 '17 at 7:14
  • $\begingroup$ Changing $h$ only permutes $\lambda_{\ell}^{(m,p)},$ so this isn't really important for my question. However, for definiteness, when we want to be explicit about the choice, let's use $\lambda_{\ell}^{(m,p,h)}.$ $\endgroup$ – RideTheWavelet Jun 27 '17 at 7:19
  • $\begingroup$ What do you mean by 'permutes $\lambda_{\ell}^{(m,p)}$'? In your example $\lambda_1^{(1,5,2)}=-3-i$ whereas $\lambda_1^{(1,5,3)}=-3+i$. Perhaps you want to show that if $\lambda_1^{(1,5,h)}=0$ for some primitive root $h$, then $\lambda_1^{(1,5,h)}=0$ for every primitive root $h$? $\endgroup$ – Servaes Jun 27 '17 at 9:38
  • $\begingroup$ It might be worth noting that $\lambda_{\ell}^{(m,p,h)}$ is contained in a prime ideal of $\Bbb{Z}[\omega]$ lying over $p$. $\endgroup$ – Servaes Jun 27 '17 at 10:45
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    $\begingroup$ By "permutes," I mean that $\lambda_{\ell}^{(m,p,h)}=\lambda_{\sigma(\ell)}^{(m,p,h')},$ for some permutation $\sigma$ (and all $\ell$). It is certainly true that if you want to say for fixed $m$ and $p$ for what $\ell$ $\lambda_{\ell}^{(m,p,h)}=0,$ you need to specify $h,$ but my conjecture is about $\lambda_{\ell}^{(m,p,h)}\neq0$ for all $\ell,$ for specific $p$ and $m,$ so the primitive becomes less important. $\endgroup$ – RideTheWavelet Jun 27 '17 at 11:22

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