Cubic Polynomials and Sums of Two Squares
Pith reviewed 2026-05-21 19:39 UTC · model grok-4.3
The pith
An irreducible monic cubic with negative discriminant is a sum of two squares for ≫ x^{1/3-o(1)} many n up to x when the constant term satisfies h ≡ 2 mod 4.
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
We establish a lower bound for the frequency with which an irreducible monic cubic polynomial with negative discriminant can be expressed as a sum of two squares. This provides a quantitative answer to a question posed by Grechuk concerning the infinitude of such values. Our proof relies on a two-dimensional unit argument and the arithmetic of degree six number fields. For example, we show that if h ≡ 2 mod 4, then the number of n from 1 to x such that n^3 + h is a sum of two squares is at least on the order of x to the power 1/3 minus little o of one.
What carries the argument
a two-dimensional unit argument together with the arithmetic of degree six number fields generated by the cubic polynomial
Load-bearing premise
The two-dimensional units and the arithmetic in the degree-six fields generated by the cubic must yield a sufficiently high density of representations for the polynomials in the chosen family.
What would settle it
A numerical check for a fixed h congruent to 2 modulo 4 showing that the count of n up to a large X where n cubed plus h is sum of two squares grows slower than X to the power 0.3 would challenge the lower bound.
read the original abstract
We establish a lower bound for the frequency with which an irreducible monic cubic polynomial with negative discriminant can be expressed as a sum of two squares ($\square_{2}$). This provides a quantitative answer to a question posed by Grechuk (2021) concerning the infinitude of such values. Our proof relies on a two-dimensional unit argument and the arithmetic of degree six number fields. For example, we show that if $h \equiv 2 \pmod{4}$, then \begin{align*} \# \{n : n^3+h \in \square_{2}, \ 1 \leq n \leq x \} \gg x^{1/3-o(1)}. \end{align*} These arguments may be generalised to study the representation of irreducible monic cubic polynomials by the quadratic form $x^2+ny^2$, where $n \in \mathbb{N}$.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The manuscript establishes a lower bound on the number of positive integers n ≤ x such that an irreducible monic cubic polynomial f(n) = n³ + h with negative discriminant takes a value that is a sum of two squares. For the family with h ≡ 2 (mod 4) the count is shown to be ≫ x^{1/3-o(1)}. The argument proceeds by embedding the problem into the arithmetic of the sextic field obtained as the compositum of the cubic field Q(α) with Q(i) and applying a two-dimensional unit argument to produce sufficiently many principal ideals of the required norm.
Significance. If the central estimate holds, the result supplies the first quantitative lower bound answering Grechuk’s 2021 question on the infinitude of such representations. The method, which combines unit rank in degree-six fields with norm equations, is potentially generalizable to representations by other binary quadratic forms x² + n y² and therefore contributes a new arithmetic-statistics tool for values of cubic polynomials.
major comments (1)
- [§3] §3 (two-dimensional unit argument): the claimed density ≫ x^{1/3-o(1)} requires that the regulator of the sextic field remains at most polynomial in the height of h and that the class group does not systematically obstruct the existence of principal ideals of norm f(n) below x^{1/3}. The manuscript must supply explicit bounds or effective versions of the unit theorem that rule out superpolynomial loss; without them the lower bound does not follow from the lattice-point count in the logarithmic embedding.
minor comments (2)
- [Abstract] The notation □₂ for the set of sums of two squares is introduced without an explicit definition; a parenthetical reminder would aid readability.
- [Introduction] A short table or list of the precise families of h for which the argument applies would clarify the scope of the generalization mentioned in the final sentence.
Simulated Author's Rebuttal
We thank the referee for their thorough review and positive evaluation of the manuscript. We appreciate the suggestion to clarify the dependence on the regulator and class group in the two-dimensional unit argument. Below we respond to the major comment.
read point-by-point responses
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Referee: §3 (two-dimensional unit argument): the claimed density ≫ x^{1/3-o(1)} requires that the regulator of the sextic field remains at most polynomial in the height of h and that the class group does not systematically obstruct the existence of principal ideals of norm f(n) below x^{1/3}. The manuscript must supply explicit bounds or effective versions of the unit theorem that rule out superpolynomial loss; without them the lower bound does not follow from the lattice-point count in the logarithmic embedding.
Authors: The sextic field is fixed once h is fixed, as it is the compositum of the cubic field Q(α) with α³ = −h and Q(i). Its regulator and class number are therefore constants independent of x. In the lattice-point count within the logarithmic embedding, these quantities contribute only constant factors to the main term. The notation ≫ x^{1/3−o(1)} is understood to mean that for every ε > 0 the count is at least c_ε x^{1/3−ε} for a constant c_ε > 0 and all sufficiently large x. Any fixed multiplicative constant is absorbed into c_ε, and any polylogarithmic factors arising from effective versions of the unit theorem are likewise absorbed since (log x)^C ≪ x^ε. Consequently, no superpolynomial loss occurs, and explicit bounds on the regulator in terms of the height of h are not required for the asymptotic statement with h fixed. We have inserted a short paragraph in §3 explaining this point to address the referee’s concern. revision: partial
Circularity Check
No circularity: lower bound derived from independent arithmetic of sextic fields and units
full rationale
The derivation relies on the existence of units in the compositum of the cubic field with Q(i) and the arithmetic of the resulting degree-6 fields to produce a positive density of representations as sums of two squares. No parameter is fitted to the target count, no self-citation supplies a uniqueness theorem that forces the result, and the x^{1/3-o(1)} lower bound is not obtained by renaming or by construction from the input data. The argument is self-contained against external number-theoretic benchmarks.
Axiom & Free-Parameter Ledger
axioms (1)
- standard math Properties of units in two-dimensional lattices arising from the ring of integers of a degree-six number field generated by the cubic.
Lean theorems connected to this paper
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IndisputableMonolith/Cost/FunctionalEquation.leanwashburn_uniqueness_aczel (J uniqueness) unclear?
unclearRelation between the paper passage and the cited Recognition theorem.
Lemma 3.5: rank of units of O_{Q(θ,i)} is two (Dirichlet, all roots complex). Lemma 3.6: two-dimensional unit argument with |l1|≠1, |l2|=1 to bound #R_θ,1(X,X) ≫ X^{1/3-o(1)}
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IndisputableMonolith/Foundation/AlexanderDuality.leanalexander_duality_circle_linking (D=3 forcing) unclear?
unclearRelation between the paper passage and the cited Recognition theorem.
Lemma 3.4: θ+i generates degree-6 field; minimal polynomial Q(x) product over roots ±i
What do these tags mean?
- matches
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- extends
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- uses
- The paper appears to rely on the theorem as machinery.
- contradicts
- The paper's claim conflicts with a theorem or certificate in the canon.
- unclear
- Pith found a possible connection, but the passage is too broad, indirect, or ambiguous to say the theorem truly supports the claim.
discussion (0)
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