A Determinant Congruence Conjectured by Sun

Yaoran Yang; Yutong Zhang

arxiv: 2605.19486 · v3 · pith:YKLRQPZXnew · submitted 2026-05-19 · 🧮 math.NT

A Determinant Congruence Conjectured by Sun

Yutong Zhang , Yaoran Yang This is my paper

Pith reviewed 2026-05-20 02:48 UTC · model grok-4.3

classification 🧮 math.NT

keywords determinant congruencebinary quadratic formLegendre symbolVandermonde determinantmatrix rankSun conjecturecomposite modulusfinite field involution

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The pith

The determinant of the matrix with entries (i² + c i j + d j²)^{n-2} is congruent to 0 modulo n² for composite n>3 with no restrictions on c and d, and for prime n=p when the Legendre symbol (d/p) equals -1.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

This paper establishes a strengthened form of Sun's conjecture on the divisibility properties of a determinant constructed from a binary quadratic form raised to the power n-2. For any composite integer n greater than 3, the determinant is always divisible by n squared no matter what integers c and d are chosen. When n is a prime p, the divisibility by p squared holds precisely when d is a quadratic non-residue modulo p. A reader might care because these congruences provide concrete arithmetic constraints on matrices over the integers and confirm an open conjecture through explicit algebraic and modular arguments. The proofs rely on factoring the polynomial determinant using Vandermonde products for the composite case and on showing a rank defect modulo p induced by a specific involution for the prime case.

Core claim

We prove that if n>3 and n is composite then det[(i² + c i j + d j²)^{n-2}]_{0≤i,j≤n-1} ≡ 0 mod n² with no condition on c,d; if n=p prime the same holds whenever (d/p)=-1. For composite n, a polynomial determinant is divisible by two Vandermonde factors; after specialisation, their product already yields the required square divisor. For prime n=p, we estimate the rank of the matrix modulo p. The required rank defect follows from a coefficient cancellation obtained from the involution t↦ d/t on F_p^× and the condition (d/p)=-1.

What carries the argument

The n by n matrix with entries (i² + c i j + d j²)^{n-2}, whose determinant divisibility by n² follows from Vandermonde factorization for composite n or from a rank defect modulo p induced by cancellation under the involution t ↦ d/t when (d/p) = -1.

If this is right

The strengthened form of the conjecture holds unconditionally for every composite n>3 and every pair of integers c and d.
For every prime p the congruence holds precisely when d is a quadratic nonresidue modulo p.
The polynomial version of the determinant factors through a pair of Vandermonde products whose product supplies the n² divisor after specialization to the quadratic form.
The matrix over the prime field has rank strictly less than n because of the coefficient cancellation coming from the involution when the Legendre symbol condition holds.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

The involution-based rank argument could be adapted to study similar power matrices over finite fields for other exponents or forms.
It would be natural to determine the exact p-adic valuation of the determinant rather than only the mod n² information.
The Vandermonde approach for composite moduli suggests possible generalizations to other arithmetic progressions or polynomial families.

Load-bearing premise

The involution t maps to d/t on the multiplicative group of the prime field produces enough coefficient cancellation to force the precise rank defect in the matrix modulo p when d is a quadratic nonresidue.

What would settle it

Compute the determinant explicitly for n=4 with c=0 and d=1 and check whether the value is divisible by 16, or for prime p=5 and d=2 where (2/5)=-1 and check whether the determinant is divisible by 25.

read the original abstract

We prove a strengthened form of a conjecture of Sun on a determinant attached to a binary quadratic form. Let $n>3$ and let $c,d\in\Z$. If $n$ is composite, then \[ \det\big[(i^2+cij+dj^2)^{n-2}\big]_{0\leq i,j\leq n-1}\equiv 0\pmod {n^2} \] with no condition on $c$ and $d$. If $n=p$ is prime, the same congruence holds whenever the Legendre symbol $\leg{d}{p}$ is $-1$. For composite $n$, a polynomial determinant is divisible by two Vandermonde factors; after specialisation, their product already yields the required square divisor. For prime $n=p$, we estimate the rank of the matrix modulo $p$. The required rank defect follows from a coefficient cancellation obtained from the involution $t\mapsto d/t$ on $\Fp^\times$ and the condition $\leg{d}{p}=-1$.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Desk Editor's Note private letter to a colleague

The composite case gets a solid Vandermonde proof that delivers the full n², but the prime case only clearly reaches divisibility by p and leaves the extra p unaccounted for in the sketch.

read the letter

The paper proves the composite part of Sun's conjecture by embedding the matrix entries into a polynomial determinant and pulling out two Vandermonde factors whose product supplies the square divisor after specialization. That step is direct and uses only standard algebraic identities, so it looks reliable for any n>3 composite with no restrictions on c and d. For the prime case the authors estimate the rank of the matrix over F_p using an involution t maps to d/t on the multiplicative group when the Legendre symbol is -1, which produces enough cancellation to force a rank defect and hence divisibility by p. That argument is internally consistent on its own terms and draws on basic finite-field properties rather than circular reductions. The soft spot is the jump from mod p to mod p². A single rank drop over F_p gives only one factor of p; the abstract gives no indication of a lifting argument, explicit p-adic valuation on the entries, or separate congruence check modulo p² that would supply the second factor. If the full text contains such a step it is not visible in the strategy summary, so the central claim for primes rests on an incomplete sketch at present. This is a narrow but cleanly executed piece of elementary number theory. Readers working on explicit determinant evaluations or quadratic-form congruences would get immediate use from the composite proof as a template. It deserves a serious referee to check whether the prime-case valuation is actually completed in the manuscript or whether the p² claim needs an additional argument. I would send it to review with that specific question flagged for the referees.

Referee Report

1 major / 1 minor

Summary. The paper proves a strengthened version of Sun's conjecture: for n>3 and c,d integers, the determinant of the n×n matrix with entries (i² + c i j + d j²)^{n-2} (indices 0 to n-1) is congruent to 0 modulo n². When n is composite this holds unconditionally; when n=p is prime it holds provided the Legendre symbol (d/p) equals -1. The composite case is proved by factoring a polynomial determinant into two Vandermonde factors whose product supplies the square divisor after specialization. The prime case proceeds by showing a rank defect of the matrix over F_p arising from coefficient cancellation under the involution t ↦ d/t on F_p^× when (d/p)=-1.

Significance. If the central claims hold, the result supplies a clean arithmetic statement linking determinants of quadratic forms to n²-divisibility, with the composite case furnishing an explicit algebraic factorization that is a clear strength. The approach via Vandermonde divisibility for composites and involution-induced cancellation for primes is conceptually economical and could extend to related matrix congruences in combinatorial number theory.

major comments (1)

[Prime-case argument] § Prime-case argument (the paragraph beginning 'For prime n=p, we estimate the rank...'): the rank defect established modulo p via the involution t ↦ d/t and the condition (d/p)=-1 implies only that the determinant is divisible by p. The target statement requires divisibility by p², yet the text contains no auxiliary step (p-adic lifting, explicit p-adic valuation of entries, congruence of the matrix modulo p², or lifting of the kernel) that would supply the second factor of p. This is load-bearing for the prime-case claim.

minor comments (1)

[Abstract] The abstract and introduction could state more explicitly that the composite case yields exactly the square divisor from the product of two Vandermonde determinants, while the prime case requires a separate argument for the extra p.

Simulated Author's Rebuttal

1 responses · 0 unresolved

We thank the referee for the careful reading and for identifying a potential gap in the prime-case argument. We address the major comment below and will revise the manuscript to strengthen the proof of divisibility by p².

read point-by-point responses

Referee: [Prime-case argument] § Prime-case argument (the paragraph beginning 'For prime n=p, we estimate the rank...'): the rank defect established modulo p via the involution t ↦ d/t and the condition (d/p)=-1 implies only that the determinant is divisible by p. The target statement requires divisibility by p², yet the text contains no auxiliary step (p-adic lifting, explicit p-adic valuation of entries, congruence of the matrix modulo p², or lifting of the kernel) that would supply the second factor of p. This is load-bearing for the prime-case claim.

Authors: We agree that establishing a rank defect of 1 over F_p via the involution t ↦ d/t (under (d/p)=-1) yields only that the determinant vanishes modulo p. To reach the required p²-divisibility, an additional step is needed. In the revision we will insert an explicit computation of the p-adic valuation: we lift the kernel vector obtained from the involution to a vector over the p-adics whose inner product with the rows gives a first-order zero, then show that the second-order term also vanishes by direct expansion of the quadratic form entries modulo p². This supplies the extra factor of p without altering the overall strategy. revision: yes

Circularity Check

0 steps flagged

No significant circularity; proof uses standard algebraic identities and direct rank estimation.

full rationale

The derivation for composite n invokes the known factorization of a polynomial determinant into two Vandermonde factors whose product supplies the n² divisor after specialization. For prime p the argument estimates matrix rank over F_p via explicit coefficient cancellation under the involution t ↦ d/t when (d/p) = -1. Both steps rely on field-automorphism properties and polynomial identities that are independent of the target congruence and do not reduce to a fitted parameter, self-definition, or load-bearing self-citation. The provided description contains no self-citation chain or renaming of a known result as a new derivation. The paper is therefore self-contained against external algebraic benchmarks.

Axiom & Free-Parameter Ledger

0 free parameters · 2 axioms · 0 invented entities

The proof draws on standard facts about Vandermonde determinants and finite-field involutions without introducing new free parameters, ad-hoc axioms, or postulated entities.

axioms (2)

standard math Vandermonde determinant factorization over polynomial rings
Invoked for the composite-n case to obtain the square divisor after specialization.
domain assumption Existence of the involution t ↦ d/t on F_p^× and its action on coefficients when (d/p)=-1
Used to produce the rank defect modulo p in the prime case.

pith-pipeline@v0.9.0 · 5707 in / 1416 out tokens · 37712 ms · 2026-05-20T02:48:43.607357+00:00 · methodology

discussion (0)

Lean theorems connected to this paper

Citations machine-checked in the Pith Canon. Every link opens the source theorem in the public Lean library.

IndisputableMonolith/Cost/FunctionalEquation.lean washburn_uniqueness_aczel unclear

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unclear
Relation between the paper passage and the cited Recognition theorem.

the required rank defect follows from a coefficient cancellation obtained from the involution t↦d/t on F_p^× and the condition (d/p)=-1
IndisputableMonolith/Foundation/AbsoluteFloorClosure.lean absolute_floor_iff_bare_distinguishability unclear

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unclear
Relation between the paper passage and the cited Recognition theorem.

a polynomial determinant is divisible by two Vandermonde factors

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