{"id":"f0687764-8403-443a-aee0-5b9e36350baa","arxiv_id":"2608.09063","paper_version":1,"verdict":"ACCEPT","confidence":"HIGH","novelty_score":5.0,"correctness_risk":"low","formal_verification":"none","parameter_count":0,"one_line_summary":"For primitive binary quadratic forms, the minimal obstruction modulus is a closed formula in the discriminant, and primitive diagonal ternary forms are also handled completely.","lead":"A number theory paper finds the smallest modulus for which some remainder is impossible for a given quadratic form, for every primitive binary form and every diagonal ternary form. The answer is a complete formula in terms of the form's discriminant, replacing case-by-case checks with one arithmetic rule.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Lemma 2.6(3) uses an unproved unimodular normalization to force q∤a and y≠0; the proof is incomplete, though the lemma admits a direct two-case proof. The Δ≡1 mod8 formula in Corollary 1.3 rests on it.","rationale":"The reader's weakest assumption coincides with the main soft spot. I independently checked the surrounding arguments: Lemma 2.1, the mod-4/no-mod-4 dichotomy in Proposition 2.2, the p=2 cases in Theorem 1.5, and the worked examples all cohere; the formulas for κ_Q appear correct. The one place the proof overreaches is Lemma 2.6(3). It is load-bearing because the q_min^2 term in the Δ≡1 mod8 branch depends on ε_{Q,q}=2 for nonresidue q. However, the lemma is true and admits a two-line direct proof that bypasses the questionable normalization. Thus the paper should be accepted only after the authors replace or justify the normalization step. This is a minor but necessary revision, hence CONDITIONAL rather than ACCEPT.","tokens_in":15475,"tokens_out":33013,"duration_ms":319140,"concrete_test":"Independently derive Lemma 2.6(3) without the normalization step: start with any (x,y) not both 0 mod q satisfying Q≡0 mod q. If y≠0, show Δ≡((2ax+by)y^{-1})^2 mod q; if y=0, show q|a and Δ≡b^2 mod q. Then verify that Q≡0 mod q implies x≡y≡0 mod q when (Δ/q)=-1, hence Q cannot represent q mod q^2. If this derivation succeeds for all possible parities and divisibilities, the gap is fixed and the Δ≡1 mod8 formula stands; if any case fails, the q_min^2 term in Corollary 1.3 is unsupported.","verdict_should_be":"CONDITIONAL","load_bearing_attack":"Lemma 2.6(3) is essential for Proposition 2.5: it is the only place showing that if (Δ/q)=-1 then Q has an obstruction modulo q^2 (and no obstruction modulo q), giving the q_min^2 term in Corollary 1.3. The proof argues that from a nontrivial solution of Q≡0 mod q one may apply the bijections (x,y)->(y,x) and (x,y)->(x,x+y) and then assume both q∤a and y≠0 mod q. No proof is given that a single substitution achieves both conditions. In the case q|a,c and q∤b, with solution (1,1), one swap leaves q|a, while (x,y)->(x,x+y) makes the new leading coefficient a+b+c nonzero but sends the lifted solution to (1,0), so y'=0. The text does not justify that further transformations (even if available) preserve the obstruction and Legendre-symbol context. This is a real gap in the written argument. It is repairable: if y≠0 then Δ≡((2ax+by)y^{-1})^2 mod q; if y=0 then q|a and Δ≡b^2 mod q, so the contrapositive and the q^2 obstruction follow without normalization. But as written, the lemma's proof is incomplete, and Proposition 2.5/Corollary 1.3 lean on it.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper defines the minimal obstruction modulus κ_Q for a primitive positive definite integral quadratic form Q, i.e. the smallest modulus k for which Q fails to represent some residue class, and determines it completely for binary quadratic forms and for diagonal ternary forms. For binary forms Q=ax^2+bxy+cy^2 with discriminant Δ, Theorem 1.2 and Corollary 1.3 give explicit formulas: for odd primes p the local exponent ϵ_{Q,p} is 1 if p|Δ, 2 if (Δ/p)=-1, and infinite if (Δ/p)=1; the 2-adic contribution is 3, 2, or infinite according to the congruence class of Δ, with a further distinction for 4|Δ. The resulting κ_Q depends only on Δ. For primitive diagonal ternary forms, Theorem 1.5 and Corollary 1.6 give the analogous complete formulas via a case analysis on the number of even coefficients and on Legendre-symbol conditions. The proofs are elementary, using pigeonhole arguments, completing the square, the Chinese remainder theorem, Hensel's lemma, and explicit modular checks.","tokens_in":15757,"tokens_out":27039,"duration_ms":274926,"significance":"If the gap in Lemma 2.6(3) is repaired, the results are correct and give the first complete determination of the minimal obstruction modulus for the two stated classes of forms. The binary result is clean and structurally interesting: κ_Q depends only on the discriminant, not on the individual coefficients. The paper is self-contained, does not use fitted parameters or numerical searches, and derives all formulas from standard lemmas; the case analyses are explicit and checkable, and the worked examples are useful. The ternary diagonal results are a substantial extension, and the paper honestly states that non-diagonal ternary forms are left open. This is a suitable contribution to the elementary and computational number-theory literature, once the load-bearing normalization issue below is fixed.","major_comments":[{"comment":"The proof of the contrapositive asserts that after applying the bijections (x,y)↦(y,x) and (x,y)↦(x,x+y) one may assume both q∤a and y not≡0 (mod q). No argument is given that a single substitution, or a specified finite sequence, achieves both conditions while preserving the form and the solution; the later phrase 'Arranging q∤a as in (3)' in Lemma 2.6(4) inherits this gap. This is load-bearing because Proposition 2.5 and Corollary 1.3 depend on Lemma 2.6(3)-(4). The gap is local and repairable: one can prove the contrapositive directly from the identity 4aQ ≡ (2ax+by)^2 − Δy^2 (mod q). If y not≡0 (mod q), this gives Δ ≡ ((2ax+by)y^{-1})^2 (mod q); if y≡0 (mod q), then q|a and Δ≡b^2 (mod q). In both cases the Legendre symbol (Δ/q) is 0 or 1, so the contrapositive follows without any normalization. I recommend replacing the normalization step with this direct two-case proof.","section":"§2.2, Lemma 2.6(3)"}],"minor_comments":[{"comment":"The set defining q_min is not declared to have the convention min∅=∞; this convention is explicitly used in Corollary 1.6 and should also be stated for the binary case for completeness.","section":"§1, Corollary 1.3 and §2.2, Proposition 2.5"},{"comment":"After completing the square, the text says 'it suffices to consider X^2−ΔY^2'; for this reduction one should explicitly note that the map (x,y)↦(X,Y)=(2ax+by,y) is bijective modulo p because 2a is invertible.","section":"§2.1, Lemma 2.1(3-i)"},{"comment":"The argument that an obstruction exists when p divides exactly two coefficients relies on the value set of cz^2 modulo p having size (p+1)/2; stating this explicitly would make the inequality (p+1)/2 < p less abrupt.","section":"§3, Proposition 3.1(2)"},{"comment":"The long 2-adic case analysis, especially the proof that the residues congruent to 2 modulo 4 modulo 16 are fewer than four in case (3), would benefit from a short table or an explicit statement of the parity patterns being used; readability is currently strained.","section":"§3.3, Proposition 3.6"}],"recommendation":"major_revision","confidential_remarks":"The manuscript is a solid elementary contribution and the main result appears correct. The only load-bearing defect is the unproved normalization in Lemma 2.6(3); it is clearly repairable by the direct two-case argument described in the major comment. I would be happy to see the revision. I do not have concerns about scope or novelty."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"The paper delivers what it promises: a closed-form formula for κ_Q, the minimal obstruction modulus, for primitive binary quadratic forms and for diagonal ternary forms. I worked through the main arguments and the central claim is correct. The binary case is solid: Lemma 2.1 handles the odd-prime cases cleanly, Proposition 2.2 exhausts the mod-4 cases for 4|Δ, and the lifting arguments for the Δ≡1 mod 8 case are valid. The formulas in Corollary 1.3 and Corollary 1.6 are new as far as I can tell from the cited literature, and they are derived from first principles (CRT, pigeonhole, Hensel) rather than quoted. No fitted constants, no circularity.\n\nThe genuine soft spot is Lemma 2.6(3). The stress-test note is right: the proof asserts that a unimodular change (x,y)→(y,x) or (x,y)→(x,x+y) lets you assume both q∤a and y≢0 mod q, but no argument shows a single substitution achieves both. As written, the contrapositive does not go through. However, this is a small, repairable gap. A direct two-case proof works: if y≠0 then Δ ≡ ((2ax+by)y^{-1})² mod q, and if y=0 then q|a and Δ ≡ b² mod q. Either way the Legendre-symbol condition follows, and the q² obstruction goes through. So the lemma is true, but the written proof is incomplete.\n\nA minor issue: the paper cites [LO18] on binary quadratic forms modulo n without stating that paper's results. That leaves the novelty claim slightly unsharpened, since I cannot tell from the text whether [LO18] already characterizes universalities modulo n. This is a presentation issue, not a mathematical flaw.\n\nThe ternary section is long but consistent. The coefficient-labeling conventions are heavy but needed, and the case distinctions in Propositions 3.3, 3.5, and 3.6 are exhaustive. The examples check out.\n\nWho is this for? Number theorists working on congruence representation by quadratic forms, and anyone teaching local obstructions. A serious referee should engage; the paper is correct and complete enough that the one gap is repairable in revision. I recommend sending to peer review, with the request that the author patch Lemma 2.6(3) and add a sentence summarizing the relevant part of [LO18].","headline":"A complete, elementary determination of the minimal obstruction modulus for binary and diagonal ternary forms; the main theorem holds up, with one repairable gap in Lemma 2.6(3).","tokens_in":16304,"tokens_out":1209,"would_cite":true,"duration_ms":13705,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11E16"],"pacs":[],"model":"deepseek-v4-flash","headline":"For a primitive positive definite integral binary quadratic form, the minimal obstruction modulus $\\kappa_Q$ is a function of the discriminant $\\Delta$ alone, and the paper gives complete explicit formulas for it and for the analogous…","keywords":["minimal obstruction modulus","binary quadratic forms","local obstructions","Legendre symbol","ternary diagonal forms","congruence representation","discriminant"],"falsifier":"Compute $\\kappa_Q$ for the form $Q=x^2+3xy+14y^2$ (with $\\Delta=-47$); the formula predicts $\\kappa_Q=25$ and specifically that the residue $5$ is not represented modulo $25$. If an exhaustive check of all pairs $(x,y)\\in(\\mathbb{Z}/25\\mathbb{Z})^2$ produces a solution to $Q(x,y)\\equiv5\\pmod{25}$, the central claim is false.","tokens_in":15231,"feed_emoji":"🧮","tokens_out":8414,"duration_ms":72693,"temperature":0.7,"pith_summary":"This paper establishes that for a primitive positive definite integral binary quadratic form $Q=ax^2+bxy+cy^2$, the minimal modulus $\\kappa_Q$ at which $Q$ fails to represent at least one residue class is determined entirely by the discriminant $\\Delta=b^2-4ac$. The author works out complete explicit formulas for $\\kappa_Q$ and the prime-power obstruction exponents $\\epsilon_{Q,p}$, covering the two parity families $\\Delta\\equiv 0 \\pmod 4$ and $\\Delta\\equiv 1 \\pmod 4$. The same method yields analogous complete formulas for primitive diagonal ternary quadratic forms. These results turn a previously studied existence question into a short, finite list of cases that can be checked by inspection of $\\Delta$ and the coefficient parity.","feed_headline":"Smallest modulus a quadratic form cannot cover: discriminant decides","feed_subtitle":"Explicit formulas give the exact obstruction modulus for every primitive binary form and every diagonal ternary form.","key_machinery":"The load-bearing object is the local obstruction exponent $\\epsilon_{Q,p}$, the smallest $e$ such that $Q$ misses a residue class modulo $p^e$, together with the prime-by-prime factorization $\\kappa_Q=\\min_p p^{\\epsilon_{Q,p}}$ that follows from the Chinese remainder theorem. The proofs combine completing the square with conditions on the Legendre symbol $(\\Delta/p)$: when $(\\Delta/p)=1$ the form represents every class modulo $p^k$ by an explicit Hensel-type construction, when $(\\Delta/p)=-1$ a pigeonhole argument forces a first obstruction exactly at $p^2$, and when $p\\mid\\Delta$ an obstruction occurs already modulo $p$. The 2-adic analysis is separate and case-based, depending only on $\\Delta$ modulo powers of 8 and 4.","core_discovery":"The central discovery is that the minimal obstruction modulus $\\kappa_Q$ is an invariant of the discriminant. In the case $4\\mid\\Delta$, writing $\\Delta=-2^n m$ with $n\\ge2$ and $m$ odd, the paper proves that $\\kappa_Q=8$ when $m=1,n=3$; $\\kappa_Q=4$ when $m=1,n\\neq3$; $\\kappa_Q=\\min\\{8,p_{\\min}\\}$ when $m>1,n=3$; and $\\kappa_Q=\\min\\{4,p_{\\min}\\}$ when $m>1,n\\neq3$, where $p_{\\min}$ is the least odd prime divisor of $\\Delta$. In the case $\\Delta\\equiv1\\pmod4$, it proves that $\\kappa_Q=3$ if $3\\mid\\Delta$, $\\kappa_Q=4$ if $3\\nmid\\Delta$ and $\\Delta\\equiv5\\pmod8$, and $\\kappa_Q=\\min\\{p_{\\min},q_{\\min}^2\\}$ if $3\\nmid\\Delta$ and $\\Delta\\equiv1\\pmod8$, where $q_{\\min}$ is the least odd prime $q$ with $(\\Delta/q)=-1$. For primitive diagonal ternary forms, Corollary 1.6 gives the corresponding complete list of formulas.","pith_inferences":["A natural extension the author leaves implicit is to non-diagonal ternary forms; the paper's closing remark suggests that cross terms may make the 2-adic analysis substantially harder, but the odd-prime part may still obey a Legendre-symbol rule.","The formula $\\min\\{p_{\\min},q_{\\min}^2\\}$ in the $\\Delta\\equiv1\\pmod8$ case hints at a class-group flavor: one might interpret $q_{\\min}^2$ as the smallest square of a prime whose Legendre symbol is $-1$, suggesting a genus-theoretic description of the obstruction.","One could test the theory mechanically: for a fixed $\\Delta$, enumerate all reduced primitive forms of that discriminant and check the predicted obstruction residues modulo $\\kappa_Q$; the formulas predict a specific missing residue class, which is verifiable by brute-force computation.","The methods may apply to the analogous $n$-ary diagonal forms when $n\\ge4$; the paper does not address whether the minimal obstruction modulus for such forms remains finite or follows a similar prime-power pattern."],"forward_implications":["The value of $\\kappa_Q$ can be read off from $\\Delta$ alone, without ever inspecting the coefficients $a,b,c$; in particular two forms with the same discriminant share the same minimal obstruction modulus.","For binary forms the only possible obstruction moduli are powers of 2, 3, a prime $p\\mid\\Delta$, or $q^2$ for a prime $q$ with $(\\Delta/q)=-1$; no other integers occur.","The complete list for primitive diagonal ternary forms gives $\\kappa_Q$ in terms of coefficient parity, the least prime dividing exactly two coefficients, and the least prime $q$ with $(-r_q s_q/q)=-1$.","The examples $x^2+y^2+8z^2$ and $x^2+2y^2+4z^2$ show that the discriminant-only dependence fails for ternary forms, so the binary case is genuinely special."],"supporting_citations":[{"why":"Supplies the classical theorem that $4|\\Delta$ forces an obstruction modulo 8 (Lemma 2.1) and the local representation results used for odd primes.","marker":"[Cas08]"},{"why":"Provides the background on binary quadratic forms, class groups, and the local obstruction phenomenon that motivates $\\kappa_Q$.","marker":"[Cox13]"},{"why":"Gives the local-global framework showing every positive definite form in at most three variables fails locally at some finite prime.","marker":"[Jon50]"},{"why":"The multivariable Hensel lifting lemma that underlies the no-obstruction arguments for ternary diagonal forms.","marker":"[Con20]"}],"fun_headline_variants":["Discriminant decides smallest modulus a quadratic form misses","Exact obstruction modulus formulas for binary and ternary forms","Minimal modulus a form cannot cover: complete solution","Quadratic forms: exact minimal obstruction modulus from discriminant"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof that a prime $q$ with $(\\Delta/q)=-1$ forces an obstruction modulo $q^2$ assumes that the unimodular coordinate changes $(x,y)\\mapsto(y,x)$ and $(x,y)\\mapsto(x,x+y)$ can be applied to a nontrivial solution of $Q\\equiv0\\pmod q$ without changing the obstruction behavior, so that one may assume $q\\nmid a$ and $y\\not\\equiv0\\pmod q$; if this step fails, the contrapositive argument that forces $q\\mid x,y$ collapses.","fun_headline_variants_meta":{"raw":{"variants":["Discriminant decides smallest modulus a quadratic form misses","Exact obstruction modulus formulas for binary and ternary forms","Minimal modulus a form cannot cover: complete solution","Quadratic forms: exact minimal obstruction modulus from discriminant"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000252,"raw_usage":{"total_tokens":1564,"prompt_tokens":953,"completion_tokens":611,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":569,"completion_tokens_details":{"reasoning_tokens":548}},"tokens_in":569,"tokens_out":611,"duration_ms":7776,"temperature":1.0,"reasoning_tokens":548,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-12T00:25:58.790279+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Compute $\\kappa_Q$ for the form $Q=x^2+3xy+14y^2$ (with $\\Delta=-47$); the formula predicts $\\kappa_Q=25$ and specifically that the residue $5$ is not represented modulo $25$. If an exhaustive check of all pairs $(x,y)\\in(\\mathbb{Z}/25\\mathbb{Z})^2$ produces a solution to $Q(x,y)\\equiv5\\pmod{25}$, the central claim is false.","supporting_citations":[],"review_version":1}