{"id":"ff8b4eaa-21fb-4bea-b982-afa5dd350185","arxiv_id":"2412.04253","paper_version":2,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":4.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"For five-sixths of primes p, Hilbert's 10th problem is shown unsolvable over Q(ζ3, ∛p), with a companion result for degree-12 extensions Q(ζ3, √D, ∛p).","lead":"The paper proves that for 5/6 of all primes p, Hilbert's 10th problem is unsolvable over the ring of integers of Q(ζ3, ∛p), and an infinite family of degree-12 extensions Q(ζ3, √D, ∛p). The proof uses rank properties of Mordell curves and cube-sum results, but the main theorem is already subsumed by the announced Koymans-Pagano result for all number fields.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The 5/6-density proof of Theorem 3.3 rests on the exact wording of [MS23, Thm. 1.1(b)(ii)]; if that theorem carries any extra hypothesis or a different nonresidue direction, Lemmas 3.1(b) and 3.2 lose the p≡4,7,8 mod9 classes and the 5/6 bound is not established.","rationale":"I agree with the reader that Theorem 3.3 is most threatened by the quotation from [MS23]. My read of Lemma 3.2 is that the role swap is valid: the condition q∉F_{ℓ_i}^3 is exactly the hypothesis of Theorem 1.1(b)(ii) with the 8-residue prime called q and the 4/7-residue prime called ℓ_i, and the products needed are contained in the theorem's conclusion. So the concern is not the swap itself but whether the source theorem has exactly those hypotheses. The alternative candidate concern, the unproved assertion that E_{−432·9²} has rank 1 over Q, is less load-bearing: the argument only needs positive rank, and 9=1³+2³ is a rational cube-sum, which by the equivalence in §1.1 gives positive rank for that curve. Thus the proof's support for the p≡2,5 classes is not really in question. Since the reader already assigned CONDITIONAL on essentially this point, my stress-test does not move the verdict; it only sharpens the concrete verification required. I would keep the paper as CONDITIONAL, not reject: the method is internally coherent and the main statement is already unconditional in the cited work of Koymans–Pagano.","tokens_in":9608,"tokens_out":36825,"duration_ms":351512,"concrete_test":"Obtain [MS23] (Proc. Math. Sci. 133 (2023), art. 43) and locate the theorem quoted as Theorem 1.1(b)(ii). Verify word-for-word that the hypotheses are exactly p≡4,7 mod9, ℓ≡8 mod9, and ℓ∉F_p^3, with no further condition involving 3 or the reverse residue direction. Re-run Lemma 3.2's induction using the source's actual hypotheses and recompute the conditional density of q≡8 mod9 covered; the 5/6 claim in Theorem 3.3 requires this density to be 1. Independently, compute in Magma/Sage the Q-ranks of y²=x³−432N² for N∈{7·17, 7·17², 13·53, 13·53², 31·89, 31·89²}; a positive rank for any of these would contradict the quoted theorem or reveal a product mismatch in Lemma 3.2.","verdict_should_be":"UNCHANGED","load_bearing_attack":"Section 3 reduces Theorem 3.3 to three residue-class groups. The p≡2,5 group uses Lemma 3.1(a); the p≡4,7 group uses Lemma 3.1(b); the p≡8 group uses Lemma 3.2. Lemmas 3.1(b) and 3.2 both invoke Theorem 1.1(b)(ii) from [MS23]: if p≡4,7 mod9 and ℓ≡8 mod9 with ℓ∉F_p^3, then ℓp, ℓp² and pℓ² are not cube-sums. In Lemma 3.2 the labels are interchanged: for a fixed ℓ_i∈S and a prime q≡8 mod9 with q∉F_{ℓ_i}^3, the products needed for Lemma 2.1 are ℓ_i q and ℓ_i q², which are the first and third products of Theorem 1.1(b)(ii) with ℓ=q and p=ℓ_i. This relabelling is internally coherent. The load-bearing assumption is that the published [MS23] theorem has exactly this statement and no additional condition. The density argument then gives success proportion 2/3 at each independent CRT stage, hence coverage 1−3^{−(k+1)} and finally 100% of primes q≡8 mod9. If the source contains an extra hypothesis (for instance on 3, or on the opposite cubic-residue direction), the admissible q-set changes and the 100% conclusion is not justified by the quoted theorem. Because the p≡4,7,8 classes together account for half of all primes, a failure here drops Theorem 3.3 from 5/6 to at most 1/3 within the paper's own proof, even though the announced Koymans–Pagano theorem would keep the statement true.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"This paper proves that Hilbert's 10th problem is unsolvable for the ring of integers of Q(ζ3, ∛p) for 5/6 of all primes p (Theorem 3.3), and for a family of degree-12 fields Q(ζ3, √D, ∛p) with D ranging over an infinite set S and p ≡ 2,5 (mod 9) (Proposition 3.4). The method combines Shlapentokh's rank-stabilization criterion with known results on the cube-sum problem: the rank of the relevant Mordell curves E_a is controlled by whether certain products of primes are sums of two rational cubes. The main theorem is reduced to three residue-class cases; the p ≡ 8 (mod 9) case is handled by a density argument using cube-sum theorems for products of two primes.","tokens_in":10034,"tokens_out":24961,"duration_ms":212620,"significance":"The paper's internal argument from rank conditions to unsolvability is clean and correctly applies Shlapentokh's theorem. The approach via cubic twists of Mordell curves is distinct from the quadratic-twist methods in [KP24] and [ABHS25], and the paper provides explicit computational data and raises natural open questions. The main theorem, however, is superseded by the recent announced proof of Hilbert's 10th problem for all number fields by Koymans and Pagano; the value of this note therefore lies in the method and the explicit families rather than in the statement of Theorem 3.3. The dependence on several external cube-sum theorems (some coauthored by the present authors) makes precise quoting and verification of hypotheses essential.","major_comments":[{"comment":"The assertion that the elliptic curve E_{−432·9^2} has rank 1 over Q is unsupported; since Lemma 2.1 only requires positive rank, the authors should either exhibit and prove the non-torsion of an explicit rational point (such as (36,108)) or give a citation. Without this, the p ≡ 2,5 (mod 9) case is not established.","section":"Lemma 3.1(a)"},{"comment":"The application of Theorem 1.1(b)(ii) requires the roles of the two primes to be interchanged relative to the statement as written: in the theorem the prime congruent to 8 mod 9 appears as ℓ, whereas in Lemma 3.2 it is the large prime p. The paper should state this relabeling explicitly and reproduce the exact hypotheses of [MS23, Thm. 1.1(b)(ii)], because if the published theorem carries any additional condition, the density-100% conclusion for primes p ≡ 8 (mod 9) is not justified.","section":"Lemma 3.2"},{"comment":"The sentence 'The set S is infinite' is asserted without proof; the infinitude of S = {ℓ prime : ℓ ≡ 4,7 (mod 9), 3 ∉ F_ℓ³} follows from Chebotarev's theorem but should be stated, as the density argument also relies on the fact that the congruence conditions p ∈ F_ℓ³ for distinct ℓ_i are independent with density 3^{-k}.","section":"Lemma 3.2"},{"comment":"The proof assumes rk_Z E_{−432}(Q(ζ3)) = 0 without proof; the curve E_{−432} has the rational point (12,36), which may be torsion, and the rank-zero claim needs a proof or citation. Additionally, the displayed equality 'rk_Z E_{−432p²}(Q) = rk_Z E_{−432p⁴}(Q)' should read that both ranks are zero; as written it is a tautology and does not support the subsequent conclusion.","section":"Proposition 3.4"}],"minor_comments":[{"comment":"There is a typo in the abstract: 'tha Hilbert's 10th Problem' should be 'that Hilbert's 10th problem'.","section":"Abstract"},{"comment":"The indexing of the density steps is off by one: after processing ℓ₁,...,ℓ_k, the covered proportion is 1 − 3^{−k}, not 1 − 3^{−(k+1)}; the limit conclusion is unaffected.","section":"Lemma 3.2"},{"comment":"Equation (2.1) would benefit from a brief explanation that the equality of ranks over L follows from the fact that the three curves E, E₁, E₂ become isomorphic over L; the formula is used repeatedly in the proofs.","section":"Section 2.1"},{"comment":"It would be useful to state that the SAGE/MAGMA rank computations are rigorous (e.g., via 2-descent) for the curves listed, so that the examples are verifiable and reproducible.","section":"Section 2.2"},{"comment":"In the final paragraph, the notation 'A = E(D)' is confusing; it should say 'A = E_{−432D³}' (the quadratic twist).","section":"Proposition 3.4"}],"recommendation":"major_revision","confidential_remarks":"The paper's main theorem is subsumed by the recent works [KP24] and [ABHS25]; the editor may wish to weigh whether the cubic-twist method justifies publication in this journal. Also, the central density claim depends on the exact statement of [MS23], which shares an author with the present paper; it would be prudent to have that theorem verified during revision. The computational examples are not needed for the main theorem and could be shortened if space is a concern."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Colleague,\n\nThis is a genuine method note, not a new frontier theorem. The paper's own references (KP24 and ABHS25) proved Hilbert's tenth problem unsolvable over the ring of integers of every number field, so the 5/6-density theorem is subsumed. Credit where due: the route through Mordell curves and the cube-sum problem is clean. Lemma 2.1 distills the Shlapentokh criterion into a very usable form: find exactly one positive-rank cubic twist, and H10 becomes unsolvable. The CRT/density argument in Lemma 3.2, which uses cube-sum primes to peel off successive thirds of the p≡8 mod 9 class, is elegant, and the Sage/Magma examples are a reasonable sanity check.\n\nThe soft spots are real but mostly fixable.\n\nFirst, the paper asserts that E_{-432·9^2} has rank 1 in Lemma 3.1(a), and Proposition 3.4 silently needs rank E_{-432}(Q)=0. These are checkable, but they are load-bearing. Show the computation or cite a source.\n\nSecond, Table 1 claims every D from 1 to 100 satisfies the criterion, but the displayed examples do not cover all cube-free D — 1 and 16, for instance, are missing from the listed sets. Either the table is wrong or it is the output of an unshown exhaustive check. Fix it.\n\nThird — the substantive one — the proof leans on two external cube-sum results with overlapping authorship: [JMS23, Thm B], an arXiv preprint, for infinitely many cube-sum primes in a prescribed AP, and [MS23, Thm 1.1(b)(ii)], which powers both Lemmas 3.1(b) and 3.2. The stress-test note has the right worry: if [MS23] has an extra hypothesis, or if the cubic-residue direction is opposite to what is assumed after swapping p and ℓ in Lemma 3.2, then the density claim for p≡8 collapses and the argument only leaves the p≡2,5 class, i.e. density 1/3. I don't think the authors are hiding anything — [MS23] is published and the statement may well be exactly as quoted — but the paper should reproduce the relevant special case or give enough context for a referee to check without digging.\n\nThe citation pattern is honest: the authors flag KP24 and ABHS25 themselves and describe their own contribution as complementary method. Who is this for? Readers working on explicit H10, rank stabilization, and cube-sum problems. It deserves a serious referee: the core argument is coherent, the method is reusable, and the gaps are checkable rather than structural. I would send it out.","headline":"A clean method note on H10 via cubic twists and cube-sum theorems, but the headline theorem is already subsumed by KP24/ABHS25; worth refereeing once the external theorems and two rank computations are verified.","tokens_in":10588,"tokens_out":13534,"would_cite":false,"duration_ms":129959,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11G05","11U05","11D25"],"pacs":[],"model":"deepseek-v4-flash","headline":"For five out of six primes, Hilbert's 10th problem is unsolvable over the cubic field Q(ζ3, ∛p); infinitely many degree-12 fields inherit the failure.","keywords":["Hilbert's tenth problem","rings of integers","Mordell curves","cube-sum problem","cubic twists","rank stabilization","undecidability","number fields"],"falsifier":"Compute the ranks of $E_{-432(\\ell p)^2}$, $E_{-432(\\ell p^2)^2}$, and $E_{-432(p\\ell^2)^2}$ for $\\ell=7,13,31,\\ldots$ and primes $p\\equiv8\\pmod9$; any positive rank among these for infinitely many $p$ would directly refute the 100%-density lemma. Similarly, an independent computation showing $E_{-432\\cdot 9^2}(\\mathbb{Q})$ has rank zero would refute the $p\\equiv2,5\\pmod9$ case.","tokens_in":9399,"feed_emoji":"🔢","tokens_out":16342,"duration_ms":135375,"temperature":0.7,"pith_summary":"This paper aims to establish that for five out of every six primes $p$, Hilbert's 10th problem has a negative answer over the ring of integers of $L_p = \\mathbb{Q}(\\zeta_3, \\sqrt[3]{p})$: no algorithm can decide whether polynomial equations with coefficients in that ring have solutions. The proof links this undecidability question to the classical cube-sum problem, which asks which integers are sums of two rational cubes and is encoded by the Mordell-Weil ranks of curves $y^2=x^3-432D^2$. A positive answer matters because it supplies a positive-density family of number fields where the analogue of Hilbert's 10th problem fails, using a cubic-twist mechanism rather than the quadratic-twist mechanism used in many earlier approaches. The paper also establishes that an infinite set of square-free integers $D$, counted by $X^{1-\\varepsilon}$, produces the same failure over the larger field $\\mathbb{Q}(\\zeta_3, \\sqrt{D}, \\sqrt[3]{p})$ for every prime $p\\equiv2,5\\pmod9$.","feed_headline":"5/6 of primes: Hilbert's 10th problem unsolvable in new fields","feed_subtitle":"Mordell curves tied to sums of two rational cubes push undecidability into degree-6 and degree-12 rings of integers.","key_machinery":"The carrying object is the Mordell curve $E_a:y^2=x^3+a$, together with its cubic twists $E_{aD^2}$ and $E_{aD^4}$, which are isomorphic to $E_a$ over $\\mathbb{Q}(\\zeta_3,\\sqrt[3]{D})$. For the cube-sum connection, the relevant family is $E_{-432D^2}$, since for cube-free $D>2$ the torsion of $E_{-432D^2}(\\mathbb{Q})$ vanishes and $D$ is a sum of two rational cubes exactly when this curve has positive rank. The argument is carried by the identity $$\\operatorname{rk} E(L)=\\operatorname{rk} E(K)+\\operatorname{rk} E_{$D^{2}$}(K)+\\operatorname{rk} E_{$D^{4}$}(K),\\quad K=\\mathbb{Q}(\\zeta_3),\\ L=K(\\sqrt[3]{D}),$$ together with $\\operatorname{rk} E_a(K)=2\\operatorname{rk}E_a(\\mathbb{Q})$. When exactly one of the three curves has positive rank over $\\mathbb{Q}$, these identities force $\\operatorname{rk}E(L)=\\operatorname{rk}E(K)>0$, and the rank-retention criterion transfers the known unsolvability of Hilbert's 10th problem over $\\mathbb{Q}(\\zeta_3)$ up to $L$.","core_discovery":"On its own terms, the paper's central discovery is a transfer from rank statements in the cube-sum problem to undecidability statements. The authors prove that Hilbert's 10th problem is unsolvable over the ring of integers of $\\mathbb{Q}(\\zeta_3,\\sqrt[3]{p})$ for every prime $p\\equiv2,4,5,7\\pmod9$, and for 100% of primes $p\\equiv8\\pmod9$ with respect to natural density, hence for $5/6$ of all primes. The transfer runs through the cluster of curves $E_a:y^2=x^3+a$, $E_{aD^2}$, and $E_{aD^4}$: when exactly one of these has positive rank over $\\mathbb{Q}$, the rank of $E_a$ over $\\mathbb{Q}(\\zeta_3)$ is positive and is unchanged after adjoining $\\sqrt[3]{D}$; a known rank-retention criterion then makes the integers definable by polynomial equations inside the larger ring, carrying unsolvability upward. For the degree-12 result, the proof shows that for an infinite square-free set $S$ with $\\#(S\\cap[-X,X])\\gg X^{1-\\varepsilon}$, the quadratic twist $E_{-432D^3}$ has positive rank, so the rank of $E_{-432}$ is positive over $\\mathbb{Q}(\\sqrt{D})$ and zero over $\\mathbb{Q}(\\zeta_3,\\sqrt[3]{p})$ for $p\\equiv2,5\\pmod9$, and the quadratic-extension criterion delivers unsolvability over the compositum.","pith_inferences":["Editorial inference: the sieve argument in the density proof should work for any finite list of auxiliary primes with the same rank-zero property, so the 100% density for $p\\equiv8\\pmod9$ likely extends to other residue classes as soon as the corresponding two-prime cube-sum theorems are available.","Editorial inference: combining the root-number computation mentioned in Remark 3.5 with a 3-Selmer parity condition would upgrade the lower bound for $S$ from $X^{1-\\varepsilon}$ to a positive natural density, making the degree-12 family quantitative rather than sparse.","Editorial inference: the computational data in Section 2 point toward the stronger statement that for every cube-free $D$ some Mordell curve has exactly one positive-rank cubic twist; if true, unsolvability would hold over $\\mathbb{Q}(\\zeta_3,\\sqrt[3]{D})$ for every cube-free $D$, not just a density-$5/6$ set.","Editorial inference: the same rank-transfer should iterate through higher layers of Kummer towers, turning a single-field undecidability result into undecidability for infinite towers of number fields."],"forward_implications":["For a set of primes of natural density $5/6$, the ring of integers of $\\mathbb{Q}(\\zeta_3,\\sqrt[3]{p})$ has an undecidable Diophantine problem, meaning no algorithm can decide solvability of polynomial equations there.","For every $D$ in an infinite square-free set $S$ with $\\#(S\\cap[-X,X])\\gg X^{1-\\varepsilon}$ and every prime $p\\equiv2,5\\pmod9$, the degree-12 field $\\mathbb{Q}(\\zeta_3,\\sqrt{D},\\sqrt[3]{p})$ also has an undecidable Diophantine problem.","Any new theorem saying that a family of integers is not a sum of two rational cubes can be fed into the same machine to produce new fields where Hilbert's 10th problem is unsolvable.","The paper supplies explicit congruence classes where the desired rank configuration provably occurs, so the undecidability statements are not merely existential."],"supporting_citations":[{"why":"Establishes that $p$, $p^2$, $9p$, and $9p^2$ are not sums of two rational cubes for $p\\equiv2,5\\pmod9$, supplying the zero ranks in the base case.","marker":"[Syl79]"},{"why":"Shows $p$ is a cube-sum when $3$ is not a cubic residue modulo $p$, producing the positive-rank auxiliary prime $\\ell$ used in the $p\\equiv4,7$ case.","marker":"[DV18]"},{"why":"Gives the two-prime-factor rank-zero theorem ($\\ell p$, $\\ell p^2$, $p\\ell^2$ not cube-sums for $\\ell\\equiv8\\pmod9$ with $\\ell$ not a cubic residue modulo $p$), the key input for the density-100% argument.","marker":"[MS23]"},{"why":"Produces infinitely many cube-sum primes in prescribed congruence classes, used to choose the auxiliary prime in Lemma 3.1(b).","marker":"[JMS23]"},{"why":"Provides the rank-retention criterion that turns equal positive ranks over a field and an extension into the Diophantine transfer needed for unsolvability.","marker":"[Shl08]"},{"why":"Supplies the corollary for quadratic extensions used in the degree-12 result, transferring unsolvability to the compositum.","marker":"[GFP20]"},{"why":"Gives the lower bound on the number of twists with analytic rank one which, together with the analytic-to-algebraic rank-one theorem, yields the infinite set $S$ in Proposition 3.4.","marker":"[PP97]"}],"fun_headline_variants":["Hilbert's 10th fails for 5/6 of primes in cubic fields","Mordell curves prove unsolvability in new number fields","Cube-sum curves show Hilbert's 10th unsolvable for most primes","5/6 primes: Hilbert's 10th unsolvable via Mordell curves"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The load-bearing premise is that the cited theorem saying certain products of two primes are not sums of two rational cubes still applies after the roles of the two primes are swapped, and that the fixed curve $y^2=x^3-432\\cdot 9^2$ really has rank 1 over $\\mathbb{Q}$; if the cubic-residue condition does not survive the swap, the density claim drops from 5/6 to 2/3, and if the rank-one computation fails, the $p\\equiv2,5$ case breaks.","fun_headline_variants_meta":{"raw":{"variants":["Hilbert's 10th fails for 5/6 of primes in cubic fields","Mordell curves prove unsolvability in new number fields","Cube-sum curves show Hilbert's 10th unsolvable for most primes","5/6 primes: Hilbert's 10th unsolvable via Mordell curves"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000178,"raw_usage":{"total_tokens":1337,"prompt_tokens":1023,"completion_tokens":314,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":639,"completion_tokens_details":{"reasoning_tokens":228}},"tokens_in":639,"tokens_out":314,"duration_ms":3618,"temperature":1.0,"reasoning_tokens":228,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-11T21:38:08.001601+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Compute the ranks of $E_{-432(\\ell p)^2}$, $E_{-432(\\ell p^2)^2}$, and $E_{-432(p\\ell^2)^2}$ for $\\ell=7,13,31,\\ldots$ and primes $p\\equiv8\\pmod9$; any positive rank among these for infinitely many $p$ would directly refute the 100%-density lemma. Similarly, an independent computation showing $E_{-432\\cdot 9^2}(\\mathbb{Q})$ has rank zero would refute the $p\\equiv2,5\\pmod9$ case.","supporting_citations":[],"review_version":1}