Efficiently computing $\sum_k x^{k^2}$ modulo $p$

Question

Let $p$ be prime. There is a whole host of "large" degree polynomials that can be computed efficiently modulo $p$. I was wondering if:

$$q(x) = \sum_{k=0}^{p-1} x^{k^2}$$

is a polynomial that can be evaluated efficiently modulo $p$.

A lot of polynomials like $1 + x + ... + x^l$ can be evaluated efficiently for even large $l$ by taking modular inverses, etc. If we have a rational generating function for the polynomial with a "few" terms it can be evaluated quickly.

So I was wondering if there are any nice expressions for $q(x)$ that make this easy to compute.

Somethings I looked at: https://en.wikipedia.org/wiki/Pentagonal_number_theorem https://mathworld.wolfram.com/JacobiTripleProduct.html

When $x$ is a rational root of unity whose denominator $p \equiv 2 \mod 4$ then this always 0. In general for any rational root of unity $x$ there is a simple formula based on Cesaro summation. In general because square numbers are sums of odd numbers you have the factorization $1+x(1+x^3(1+…$ which now just requires you to evaluate $x,x^3,x^5….$ this can be done efficiently via knowing $x$, and $x^2$ to generate the rest. Now can any of this become faster mod $p$? I’ll have to chew on that. — Sidharth Ghoshal
– Sidharth Ghoshal, Commented Nov 18, 2023 at 19:08

Oleksandr Kulkov · Accepted Answer · 2023-11-20 12:45:42Z

Not exactly an answer, but should provide a strong hint on where to look.

This is closely related to Gauss quadratic sums, expressions of the form $$ g(a;p) = \sum\limits_{k=0}^{p-1} \omega_p^{ak^2}, $$ where $\omega_p = e^{\frac{2\pi i}{p}}$ is the $p$-th primitive root of unity. Noteworthy, in complex numbers, $$ g(1;p) =\sum_{k=0}^{p-1}\omega_p^{k^2}= \begin{cases} (1+i)\sqrt{p} & \text{if}\ p\equiv 0 \pmod 4, \\ \sqrt{p} & \text{if}\ p\equiv 1\pmod 4, \\ 0 & \text{if}\ p \equiv 2 \pmod 4, \\ i\sqrt{p} & \text{if}\ p\equiv 3\pmod 4. \end{cases} $$

Now, getting back to $q(x)$, for any $x \neq 0$ it is kind of similar to $q(x)=1+\frac{p-1}{a}g(1;a)$, where $a = \operatorname{ord} x$. If my understanding is correct here, for any $a$ there would only be at most $4$ possible values of $q(x)$, depending on which square roots of $-1$ and $a$ within $GF(p)$ or $GF(p^2)$ they correspond to.

Unfortunately, I'm not exactly sure whether it translates to remainders modulo $p$ as easily as I hope it does, and if it does, how to understand which of the four possible values is actually $q(x)$ for any given $x$. I can only hope that a better understanding of Gauss quadratic sums would shed more light on this.

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