On p.112-113 of volume 10-1 of Gauss's werke, which contain an unpublished fragment dated to 1805, Gauss states some results on cubic and biquadratic "Gaussian periods" in a trigonometric form. While the result on the cubic period was already published by Gauss in a footnote to article 358 of his Disquisitions Arithmeticae (1801), the result on biquadratic period has apparently not been published by him. Here is a Google translation of this fragment:

Cubic: Let $4p = a^2+27b^2$, so that: $\frac{a}{\sqrt{4p}} = cos\varphi, \frac{b\sqrt{27}}{\sqrt{4p}} = sin\varphi$. Then the period of order $\frac{1}{3}(p-1)$ is: $$-\frac{1}{3}+\frac{2}{3}cos\frac{1}{3}\varphi \sqrt{p}$$ Biquadratic: Let $p = a^2+4b^2$, so that: $\frac{a}{\sqrt{p}} = cos\varphi, \frac{2b}{\sqrt{p}} = sin\varphi$. Then the period of order $\frac{1}{4}(p-1)$ is: $$-\frac{1}{4}+\frac{1}{4}\sqrt{p}+\frac{1}{2}\sqrt{p}\cdot cos\frac{1}{2}\varphi = \prod$$, and: $${\prod}^0 + i\prod'-\prod''- i\prod''' = \sqrt{p}\cdot (\frac{a+b\sqrt{-4}}{a-b\sqrt{-4}})^{\frac{1}{4}}$$

As far as i understand, a "Gaussian period" of $\frac{1}{4}(p-1)$ terms is a complex exponential sum $\sum e^\frac{2\pi in}{p}$ when $n$ goes through all the quartic residues modulo $p$. Although Gauss doesn't define ${\prod}^0, \prod',\prod'',\prod'''$, i believe that ${\prod}^0$ represents a complex exponential sum over values of $n$ which belong to the subgroup of quartic residues modulo $p$ (subgroup of the larger group of invertible residues modulo $p$), and $\prod',\prod'',\prod'''$ are complex exponential sums over its cosets, arranged according to some order. One can see that Gauss weighted ${\prod}^0, \prod',\prod'',\prod'''$ by the four units in the Gaussian integers $+1,+i,-1,-i$, and then summed them up.

Indirectly connected with this result is Gauss's footnote to article 358 of D.A, which includes the result from his 1805 fragment as well as several other corollaries:

Corollary: Let $\epsilon$ be the root of the equation $x^3-1=0$ and we will have $(p+\epsilon p'+{\epsilon}^2p'')^3 = n(M+N\sqrt{-27})/2$. Let $M/\sqrt{4n} = cos\phi, N\sqrt{27}/\sqrt{4n} = sin\phi$ and as a result $$p = -\frac{1}{3}+\frac{2}{3}cos\frac{1}{3}\phi \sqrt{n}$$

Note that here Gauss denotes the prime number by $4n = M^2+27N^2$ (the prime number is $n$), and the Gaussian period by $p$. The result $$p+\epsilon p'+{\epsilon}^2p'' = (n(M+N\sqrt{-27})/2)^{\frac{1}{3}}$$ for the cubic case is analogous to the summation of ${\prod}^0 + i\prod'-\prod''- i\prod'''$ (for the biquadratic case), since $i$ is a root of the equation $x^4-1=0$.


I'm aware that Gauss sums and Gaussian periods are intimately connected, and according to the survey article "THE DETERMINATION OF GAUSS SUMS" by Bruce C. Berndt and Ronald J. Evans, in the D.A and in his two articles on biquadratic residues, Gauss determined cubic and quartic Gauss sums. However, i can't see how the two identities for the sums $p+\epsilon p'+{\epsilon}^2p''$ and ${\prod}^0 + i\prod'-\prod''- i\prod'''$ are connected to the cubic and quartic Gauss sums (which don't involve weighting by $\epsilon$ or $i$), nor did i find any article that references this particular result of Gauss. Therefore, i will sum up the main content of my question:

  • Are those identities an easy corollary to the determination of cubic and quartic Gauss sums? or maybe there are additional inherent difficulties to imply them from the values of Gauss sums? In any case, i'd like to get some guidelines on how to move from the values of Gauss sums to those particular identities.
  • If the answer is too long, i'd also be glad if anyone can give a link to a source that discusses those identities. For me linking to such a source can be an alternative answer, since i didn't find any article that refers to those identities.

Any useful comment will be blessed!


1 Answer 1


Let $p \equiv 1 \bmod 3$ be a prime number, let $g$ be a be a primitive root modulo $p$, and $\zeta$ a primitive $p$-th root of unity. The three cubic periods are \begin{align*} \eta_0 & = \zeta + \zeta^{g^3} + \zeta^{g^6} + \ldots + \zeta^{g^{p-4}}, \\ \eta_1 & = \zeta^g + \zeta^{g^4} + \zeta^{g^7} + \ldots + \zeta^{g^{p-3}}, \\ \eta_2 & = \zeta^{g^2} + \zeta^{g^5} + \zeta^{g^8} + \ldots + \zeta^{g^{p-2}}. \end{align*} We have $$ \eta_0 + \eta_1 + \eta_2 = -1, $$ and $$ \tau = \eta_0 + \rho \eta_1 + \rho^2 \eta_2 \quad \text{and} \quad \overline{\tau} = \eta_0 + \rho^2 \eta_1 + \rho \eta_2 $$ are the cubic Gauss sum $$ \tau = \sum_{a=1}^{p-1} \chi_3(a) \zeta^a $$ and its complex conjugate $\overline{\tau}$; here $\chi_3$ is a suitably chosen nontrivial cubic character on $({\mathbb Z} /p{\mathbb Z})^\times$. It is known that $\tau^3 = \pi^2 \overline{\pi}$, where $\pi$ is a prime factor of $p$ in ${\mathbb Z}[\rho]$ ($\rho$ is a primitive cube root of unity; Gauss used $\epsilon$).

We find \begin{align*} \tau + \overline{\tau} & = 2\eta_0 - \eta_1 - \eta_2, \\ \rho \tau + \rho^2 \overline{\tau} & = - \eta_0 - \eta_1 + 2\eta_2,\\ \rho^2 \tau + \rho \overline{\tau} & = - \eta_0 + 2\eta_1 - \eta_2. \end{align*} Adding $\eta_0 + \eta_1 + \eta_2 = -1$ to these equations we get $$ \eta_0 = \frac{-1 + \tau + \overline{\tau}}3, \quad \eta_1 = \frac{-1 + \rho^2 \tau + \rho \overline{\tau}}3, \quad \eta_2 = \frac{-1 + \rho \tau + \rho^2 \overline{\tau}}3. $$ Since $\tau^2 = \pi^2 \overline{\pi} = p\pi$ and $\tau \overline{\tau} = p$ we have $$ (\tau + \overline{\tau})^3 = p\pi +p \overline{\pi} + 3p(\tau + \overline{\tau}). $$ If we write $\pi = \frac{L + 3M\sqrt{-3}}2$, then $\tau + \overline{\tau} = L$, hence $\tau + \overline{\tau}$ is a root of the cubic polynomial $$ f_p(X) = X^3 - 3pX - pL. $$

Now if $\xi$ is a root of a polynomial $f(x)$, then $\frac{\xi-1}3$ is a root of $f(3x+1)$. Thus $\eta_0$ is a root of the polynomial $$ P_3(x) = \frac1{27} f_p(3x+1) = x^3 + x^2 - \frac{p-1}3 x - \frac{(L+3)p-1}{27}. $$

Since $4p = L^2 + 27M^2$ we know that $|\frac{L}{2\sqrt{p}}| < 1$ and $|\frac{3M\sqrt{3}}{2\sqrt{p}}| < 1$; thus there exists a unique angle $\phi$ with $$ \cos \phi = \frac{L}{2\sqrt{p}} \quad \text{and} \quad \sin \phi = \frac{3M\sqrt{3}}{2\sqrt{p}}. $$ With $T = \tau + \overline{\tau}$ we have $|T| \le 2 \sqrt{p}$, hence there is an angle $\alpha$ such that $\cos \alpha = \frac{T}{2\sqrt{p}}$. Now $T^3 = 3pT + pL$ implies $$ 4\Big(\frac{T}{2\sqrt{p}}\Big)^3 - 3 \cdot \frac{T}{2\sqrt{p}} = \frac{L}{2\sqrt{p}}, $$ which can be written as $$ 4 (\cos \alpha)^3 - 3 \cos \alpha = \frac{L}{2\sqrt{p}}. $$ Since $4 \cos(x)^3 - 3 \cos (x) = \cos(3x)$, this implies $\alpha = 3\phi$ (for some suitable choice of $\alpha$).

The biquadratic case is similar; $\Pi^0$, $\Pi'$ etc. are the four quartic periods, the linear combination on the left side of your equation is the usual quartic Gauss sum, and the equation is the prime factorization of the Gauss sum (up to a fourth root of unity).

  • $\begingroup$ thanks for your comprehensive answer! I guess i was mistaken when i thought that establishing the identity for $p+\epsilon p' + {\epsilon}^2 p''$ is different from the determination of the cubic Gauss sum, so i have to sharpen my uderstanding before being able to make a good reading of your answer. Anyway, thanks! $\endgroup$
    – user2554
    Aug 20, 2021 at 16:45

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