Equidistribution of powers of primitive roots modulo $p$

Let me start with a nice experimental observation. Fix a large prime, say $$p = 5003$$. It turns out that $$g = 2$$ is a primitive root mod $$p$$. If we plot the powers of $$g \in \Bbb F_p^{\times}$$ (considering representatives between $$0$$ and $$p-1$$), we obtain this$$^{[1]}$$ :

$$\hspace{2cm}$$

We see that it looks pretty random! So we might wonder how powers of a given primitive root modulo $$p$$ distribute in $$\{1, ..., p-1\}$$, as $$p$$ goes to infinity.

To make it more precise, for every prime $$p$$, fix a primitive root $$g_p$$ (i.e. it generates $$\Bbb F_p^{\times}$$), and define $$\phi_p : \Bbb Z / (p-1) \Bbb Z \to \Bbb F_p^{\times}, k \mapsto g_p^k.$$

let $$e_p(x) = \exp(2 \pi i x / p)$$ and consider the finite subset $$E_p = \{ ( \,e_p(k) \; , \; e_p(\phi_p(k)) \,) \mid 1 \leq k \leq p-1 \} \subset S^1 \times S^1$$ of the complex torus (we might think simply of the fundamental domain $$[0,1[^2$$ as well, as in the above picture).

The question is :

When $$p \to +\infty$$, do the sets $$E_p$$ equidistribute in $$S^1 \times S^1$$ with respect to Lebesgue (Haar) measure ?

By Weyl theorem, it can be shown that it boils down to prove that for every $$(m, n) \in \Bbb Z^2 \setminus \{ (0,0) \},$$ one has $$\Sigma_{n, m}(f ; p) := \sum_{1 \leq x \leq p-1} e_p(mx + n g_p^x) = o(p), \quad p \to \infty$$

I did not find any reference dealing with such exponential sums. Any help would be appreciated.

$$^{[1]}$$ This is done using the following code in SAGE (https://cocalc.com/ or https://sagecell.sagemath.org/)

powers = [[0, 1]];

N = 2;
p = next_prime(5000);
for j in range(1, p):
powers = powers + [[j, mod(N, p)^j]];

Plot1 = list_plot(powers, size=5, color='blue')
show(Plot1, figsize=[8,8])
save(Plot1, "powers_of_2_modulo_5003.pdf", figsize=[8,8])

• Just to make sure. In your plot the exponent is on one axis and the power on the other? So for example there is exactly one dot on each row and column? – Jyrki Lahtonen Jun 26 '20 at 16:28
• Also, in my opinion it would be more natural to use $e_{p-1}(mx)$ instead of $e_p(mx)$, because the arithmetic of the exponents really is modulo $p-1$. Furthermore, unless I misinterprested something then you would be looking at Gauss sums, known to have absolute value $\sqrt{p}$ in the interesting cases. – Jyrki Lahtonen Jun 26 '20 at 16:35
• Admittedly I haven't completely wrapped my head around your question yet. My first reaction was to use Vinogradov-Polya method to get bounds like $O(\sqrt p\log p)$, but that applies to different sums, I think.... – Jyrki Lahtonen Jun 26 '20 at 16:39
• @JyrkiLahtonen : thanks for your comments. So my plot is $(k, g^k)$, i.e. it is the graph of $\phi$. So there is exactly one dot on each row and column. – Watson Jun 26 '20 at 19:24
• I agree for your comment about $e_{p-1}$, maybe looking at $(e_{p-1}(k), e_p(g^k))$ would be better (I still use $e_p$ in the second variable though, because $g^k$ really belongs to $\Bbb F_p$). In any case, I don't know if this would help. In particular, I am not sure how to relate my sum to a Gauss sum... – Watson Jun 26 '20 at 19:26

Taking a look at a little bit different exponential sums that I find more natural in this context. The exponent $$x$$ more naturally takes values in the additive group $$\Bbb{Z}/(p-1)\Bbb{Z}$$. So I want to look at the picture as a subset of a $$(p-1)\times p$$ grid as opposed to a $$p\times p$$ grid. All this because that is a point of view, where a well-known answer can be given. My guess is that this change would not disturb the equidistribution property. If I'm wrong then we can scratch this.
Let $$\xi=e^{2\pi i/(p-1)}$$ and $$\zeta=e^{2\pi i/p}$$. For all intagers $$m$$ the function $$\psi_m:\Bbb{F}_p^*\to\Bbb{C}^*, g^x\mapsto \xi^{mx}$$ is a multiplicative character of $$\Bbb{F}_p$$. In other words, $$\psi_m(ab)=\psi_m(a)\psi_m(b)$$ for all $$a,b\in\Bbb{F}_p^*$$. This is because $$\psi_m(g)=\xi^m$$ is a root of unity of order $$d\mid p-1$$ and in $$\Bbb{F}_p^*$$ we have $$g^x\cdot g^y=g^{x+y}$$ with the addition of the exponents done modulo $$p-1$$.
Similarly, for all integers $$n$$ the function $$\chi_n:\Bbb{F}_p\to\Bbb{C}^*, a\mapsto \zeta^{na}$$ is an additive character of $$\Bbb{F}_p.$$ In other words, $$\chi_n(a+b)=\chi_n(a)\chi_n(b)$$ for all $$a,b\in\Bbb{F}_p$$.
Then the substitution $$a=g^x$$ converts the sums $$\Sigma_{m,n}(p)=\sum_{x\in[0,p-2]}\xi^{mx}\zeta^{ng^x}$$ to the character sums $$\Sigma_{m,n}(p)=\sum_{a\in\Bbb{F}_p^*}\psi_m(a)\chi_n(a).$$ These sums are one of the most studied hybrid character sums also known as Gauss sums. It is easy to show that $$\left|\Sigma_{m,n}(p)\right|= \begin{cases} p-1,&\ \text{if p-1\mid m and p\mid n},\\ 0,&\ \text{if p-1\nmid m and p\mid n},\\ 1,&\ \text{if p-1\mid m and p\nmid n, and}\\ \sqrt{p},&\ \text{when neither character is trivial.} \end{cases}$$ Only the last case is somewhat non-trivial. The divisibility constraints come from the facts that when $$p-1\mid m$$ we have $$\psi_m(a)=1$$ for all $$a$$, and similarly when $$p\mid n$$ we have $$\chi_n(a)=1$$ for all $$a$$.
• Ok, I see. The trick was to consider a sum over $g^x$ and not over $x$. Very nice and elementary, thank you! – Watson Jun 27 '20 at 9:13