The first moment of cubic Gauss sums

Source and verification

This article explains family 023 of the October 6, 2026 OpenAI collection. Research conclusions are attributed to its manuscripts. The review covers statements, introductions, and selected proof mechanisms; it is not an independent verification of the complete proofs.

🏷️ Cubic exponential sums and a small bias

For a rational prime $p$, define

$$S_p=\sum _{x\mathrm{mod}p}\exp (2\pi i{x}^3/p).$$

When $p\equiv 2(\mathrm{mod}3)$, cubing permutes the residues, so $S_p=0$. For $p\equiv 1(\mathrm{mod}3)$, the sum detects cubic-residue structure.

The source works naturally in the Eisenstein lattice $\mathbb{Z}[\omega ]$, where $\omega =e^{2\pi i/3}$. Write $N(z)=|z{|}^2$ and $\mathrm{Tr}(z)=z+\bar{z}$. A primary element is one congruent to one modulo three. Every ideal coprime to three has one primary generator in this convention.

For primary $b$, the cubic residue character ${\chi }_b$ takes cube-root-of-unity values on invertible residues and is zero elsewhere. The normalized Gauss sum is

$$\begin{array}{c}G(b)\\ =\frac{1}{\sqrt{N(b)}}\sum _{v\mathrm{mod}b}{\chi }_b(v)e(\mathrm{Tr}(v/b)),\\ e(t)=\exp (2\pi it).\end{array}$$

For prime arguments these are unit-scale oscillatory quantities. A first moment much smaller than the number of primes can coexist with equidistribution of their phases.

🏷️ The claimed first moment

Put

$$c_{\ast }=\frac{(2\pi {)}^{2/3}}{3\Gamma (2/3)}.$$

The manuscript asserts the sharp-cutoff asymptotic

$$\begin{array}{c}\sum _{\begin{array}{c}N\pi \le X\\ \pi \equiv 1(3),\pi \mathrm{prime}\end{array}}G(\pi )\\ =\frac{6}{5}{c}_{\ast }\frac{X^{5/6}}{\log X}+o\left(\frac{X^{5/6}}{\log X}\right).\end{array}$$

All primary Eisenstein primes are included. In particular, both conjugate primes above a split rational prime occur. The assertion is unconditional within the manuscript’s stated inputs. (OpenAI, 2026)

The scale $X^{5/6}/\log X$ is smaller than the prime-counting scale $X/\log X$. Consequently this nonzero first moment does not contradict equidistribution at the larger scale.

🏷️ Checking the normalization

For a split rational prime $p=\pi \bar{\pi }$, the source proves

$$G(\pi )+G(\bar{\pi })=\frac{S_p}{\sqrt{p}}.$$

Therefore the rational-prime sum with summand $S_p/(2\sqrt{p})$ has half the primary-prime coefficient:

$$\sum _{\begin{array}{c}p\le X\\ p\equiv 1(3)\end{array}}\frac{S_p}{2\sqrt{p}}\sim \frac{(2\pi {)}^{2/3}}{5\Gamma (2/3)}\frac{X^{5/6}}{\log X}.$$

Inert rational primes have norm $p^2$, so their primary-prime contribution is $O(\sqrt{X})$, negligible here.

The manuscript explicitly flags conflicting printed normalizations in a historical source. The formulas above follow its own pairing identity; identifying the rational and primary sums without the factor two would be incorrect.

One can also derive the logarithmically weighted formula. If $A(y)$ is the primary-prime sum, partial summation gives

$$\begin{array}{c}\sum _{N\pi \le X}G(\pi )\log N\pi \\ =A(X)\log X-{\int }_2^X\frac{A(y)}{y}dy.\end{array}$$

The integral is lower order, yielding leading term $(6/5)c_{\ast }{X}^{5/6}$.

🏷️ A model term that survives Poisson summation

The source compares $G(\pi )$ with the elementary model $c_{\ast }N(\pi {)}^{-1/6}$. The prime ideal theorem predicts the latter’s main term from

$$c_{\ast }{\int }_2^X\frac{t^{-1/6}}{\log t}dt\sim \frac{6}{5}{c}_{\ast }\frac{X^{5/6}}{\log X}.$$

The constant $6/5$ is the reciprocal of the integrated exponent $5/6$.

The hard step is showing that the difference from this model cancels. Cubic-theta transformation formulas handle linear, or Type I, sums. In bilinear sums, Poisson summation produces nonzero cube frequencies that contribute the model itself. Discarding all nonzero frequencies as error would lose the bias.

Corrected dispersion subtracts the retained main term before applying large-sieve estimates. Two difficult conductor-length configurations require structured moments and a stronger off-diagonal Gram estimate. These estimates apply to independent smooth convolutions of prime sequences, not arbitrary coefficients.

🏷️ Prime decomposition and sharp cutoffs

A prime-detecting decomposition must preserve that coefficient structure. The source orders norm bins and uses a stopping rule within the last bin; binomial weights separate the two coefficient sides exactly. Near a transition involving three similarly sized primes, prime sparsity supplies an additional logarithmic saving.

Removing smoothing introduces large Mellin heights. At those heights, the proof localizes both norm variables, reducing diagonal contributions near the product boundary; oscillation handles the remaining region. A low-height theorem alone would not establish the sharp upper cutoff in the main statement.

The imported level transformation formula is unconditional. The earlier GRH-dependent first-moment cancellation theorem is not used as an input. The source does not claim a general improvement of the cubic large sieve.

🏷️ Angular modes and limits of the conclusion

For each fixed nonzero integer $\ell$, the manuscript also claims

$$\begin{array}{c}\sum _{N\pi \le X}G(\pi ){\left(\frac{\pi }{|\pi |}\right)}^{\ell }\\ =o_{\ell }(X^{5/6}/\log X).\end{array}$$

There is no uniformity for growing $\ell$. Its fixed-power consequences cover powers $G(\pi {)}^k$ with $3\nmid k$ and $|k|>1$; nonzero multiples of three require stronger angular prime information.

🏷️ Key steps and proof outlines

Corrected dispersion estimates the difference from the bias

For ideal norms $Na\asymp A$, $Nb\asymp B$, with $AB\asymp X$, define

$$\begin{array}{rl}{F}_u(a)& =\sum _b{\beta }_b(Nb{)}^{iu}G(b)\overline{{\chi }_b(a)},\\ S(u)& =\sum _b{\beta }_b(Nb{)}^{iu-1/6},\\ H& =B^2/A,\mathcal{Q}=AB{H}^{1/3}.\end{array}$$

The coefficients ${\beta }_b$ belong to the source’s specified smooth prime-convolution class; at low heights they also satisfy roughness and uniform nonprincipal-character cancellation hypotheses. The corrected estimate controls the mean square of

$$F_u(a)-c_{\ast }\overline{G(a)}(Na{)}^{-1/6}S(u),$$

with upper bound $\mathcal{Q}(\log X{)}^{-K}$ after suitable parameter choices.

Proof outline. Expand the square and apply Poisson summation to the lattice variable. Nonzero cube frequencies have a principal character contribution, so they survive the prime summation and generate the displayed model. Subtract this term before estimating the remaining frequencies. Ordinary and cubic large sieves control most conductor-length ranges; the two remaining ranges use the source’s structured moment and off-diagonal Gram estimates. This is why unrestricted coefficients cannot be inserted: the moment bounds use their independent prime factors. (OpenAI, 2026)

Cauchy–Schwarz produces the precise first-moment scale

Proof outline. Gauss-sum multiplicativity writes the bilinear expression as a sum of ${\alpha }_{a}G(a)(Na{)}^{iu}{F}_u(a)$. Insert the model from the previous estimate and apply Cauchy–Schwarz to the remainder. A coefficient bound ${\sum }_a|{\alpha }_a{|}^2\ll A(\log X{)}^{O(1)}$ gives a square-root scale $(A\mathcal{Q}{)}^{1/2}$. The scale definitions imply the exact relation

$$A\mathcal{Q}=A^{2}B(B^2/A{)}^{1/3}=(AB{)}^{5/3},$$

so the remainder is $X^{5/6}$ times a controllable negative logarithmic power. The model term becomes the bilinear sum of $c_{\ast }{\mu }^2(ab)N(ab{)}^{-1/6}$ after reinstating coprimality. Pairs sharing a prime are sparse because the coefficients are rough; summing reciprocal prime squares and applying their moment bounds makes this correction negligible.

Near the transition of three prime factors of size $X^{1/3}$, the general logarithmic saving is insufficient. Here sparsity supplies $\sum |{\alpha }_a{|}^2\ll A/(\log X{)}^2$ and $\sum |{\beta }_b{|}^2\ll B/\log X$. Substituting these into the short-dispersion estimate gives $O(X^{5/6}(\log X{)}^{-3/2})$. This stronger coefficient information is used only in that special configuration.

The stopping-bin identity preserves independent coefficients

A prime detector decomposes the sum into products whose factors must be divided between two sides. A cutoff depending on the complete product can couple those sides and invalidate the structured bilinear input.

Proof outline. Order the norm bins and stop using a surrogate that gives every prime in the final bin the same endpoint value. Fix that bin and the numbers $k_{+},k_{-}$ of factors sent to the selected and remaining sides. There are exactly $\left(\genfrac{}{}{0ex}{}{k_{+}+k_{-}}{k_{+}}\right)$ choices of the selected factors, and either every choice or none satisfies the surrogate crossing test. Weight each by the reciprocal binomial coefficient. Their sum restores the original coefficient exactly. Once the bin and counts are fixed, the crossing condition belongs entirely to one side; all other prime weights split multiplicatively. The result is a sum of genuinely independent coefficient products, with only logarithmically many outer indices. The squarefree-supported kernel kills tuples introduced by allowing repeated cross-side primes. This exact identity, rather than a rough estimate for the detector, makes the moment hypotheses available.

Removing smoothing requires a separate large-height branch

Mellin inversion of a finely smoothed interval introduces twists $N(n{)}^{it}$. At the first-moment scale, a boundary strip of width about $X/T$ must be smaller than $X^{5/6}/\log X$, motivating heights through $T=X^{1/6+\rho }$ with a small fixed $\rho >0$.

Proof outline. At low heights use the corrected Gauss-sum/model comparison. At large heights treat the Gauss-sum term itself by an averaged dispersion bound. Partition both norm variables into short cells. Near the product boundary this reduces the size of the diagonal, which oscillatory estimates cannot cancel. Away from it, integration in height exploits the nonzero logarithmic phase difference. The source chooses the cell width, height thresholds, and decomposition margins so every resulting error is below the first-moment scale. Merely extending a low-height estimate by a trivial bound would lose the sharp cutoff.

After summing the dyadic prime intervals, the remaining undecomposed model is evaluated by the prime ideal theorem and partial summation. Integrating $t^{-1/6}/\log t$ supplies $(6/5)X^{5/6}/\log X$. Thus the main coefficient comes from the cube-frequency residue, while the exponent and factor $6/5$ come from its prime summation. The split-prime pairing then gives the separate rational-prime normalization stated earlier.

🏷️ Connections

Zero-free half-planes uses related cubic-theta identities but a different reciprocal-$L$ signal. The first-moment theorem here is not an input to that family’s $7/8$ argument.

🏷️ Manuscripts and verification scope

• An unconditional first moment for cubic Gauss sums — pinned manuscript and companion source.

The family scope page describes selected formalization coverage. The page selects the all-primary-prime asymptotic and fixed angular-mode conclusions. No Lean build, axiom audit, or manuscript-to-formal-statement equivalence check was performed for these notes.

OpenAI collection provides prerequisites and the reading sequence; coverage register records completed and pending families.

🐻  OpenAI 2026. An unconditional first moment for cubic Gauss sums.