{"id":"5a8c8d0d-fc76-46a2-8197-79b70c71775e","arxiv_id":"1908.05587","paper_version":3,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":4.0,"correctness_risk":"high","formal_verification":"none","parameter_count":0,"one_line_summary":"New sufficient conditions for irreducibility in Z[x] using a prime power dividing an outer coefficient and a zero-location bound, proved by an elementary coefficient lemma.","lead":"The paper proposes new divisibility conditions on integer polynomials whose roots lie outside a disc, and claims these conditions force irreducibility over the integers. It generalizes a known criterion of Murty and Girstmair from a prime constant to prime powers with extra coefficient conditions.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Lemma 3's proof in arXiv:1908.05587 contains a false divisibility step, so the main theorems are unproven as written; the lemma itself is not refuted, but no corrected proof is supplied.","rationale":"The central claim is the pair of irreducibility criteria, and the only route from the zero-location hypothesis to irreducibility in the case where both factor constants are divisible by p is Lemma 3. That lemma is not disproved by the quoted example: in the example p|a4, so the lemma's conclusion holds; what fails is a specific divisibility assertion in its proof. This makes the situation a genuine gap rather than a known counterexample to the theorem. The other defects (unhandled symmetric p|c0,p∤b0 case in Theorem 1; unhandled p|bm,p∤c_{n-m} case in Theorem 2; Example 1's root bound |zeta|>1 instead of |zeta|>p) are less load-bearing and appear patchable. Theorems are plausible and Theorem A's proof is clean, but as written the central proof is unsupported, so a conditional verdict is appropriate; I see no reason to move the reader's verdict.","tokens_in":7192,"tokens_out":17177,"duration_ms":149987,"concrete_test":"Enumerate small integer polynomials: for p in {2,3}, k=2..6, j=2..6, list f1,f2 with coefficients in [-50,50] and nonzero constant terms; retain cases with v_p(a0)=k, p^k|a1..a_{j-1}, p|b0, p|c0, and record whether p∤a_j occurs. If a counterexample appears, Lemma 3 and the theorems fall. If none appears, attempt a corrected induction on p-adic valuations, ideally machine-checked, and separately verify whether Example 1 can be strengthened to |zeta|>p for its stated parameters.","verdict_should_be":"UNCHANGED","load_bearing_attack":"Both theorems rest on Lemma 3, and the proof of Lemma 3 breaks at the displayed assertion after (4). In the even-j case the proof asserts that p^k divides a_{j-1} minus every term except b_{(j-2)/2}c_{j/2}+b_{j/2}c_{(j-2)/2}. To subtract b0c_{j-1} and b_{j-1}c0 one needs p^k|b0c_{j-1} and p^k|b_{j-1}c0, but the induction only established valuations for indices up to (j-2)/2. The reader's example (p=2,k=5,j=4, f1=2+4x+x^2, f2=16+16x+8x^2+8x^3) satisfies the lemma's hypotheses and even has p|a4, yet a3-b0c3-b3c0=48 is not divisible by 32, so the displayed equality is false. Because Theorem 1 and Theorem 2 both invoke Lemma 3 in the p|b0,p|c0 case, the main proofs are invalid as written; the lemma may be repairable, but no repair is supplied. Example 1 also proves only |zeta|>1 while the hypotheses require |zeta|>d=p, so it does not illustrate Theorem 1.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The manuscript proposes two irreducibility criteria for primitive integer polynomials whose zeros lie outside the closed disc |z|≤d. Theorem 1 assumes the constant term is ±p^k d, with p^k dividing an initial block a_0,...,a_{j-1}, gcd(k,j)=1, and p∤a_j for k>1; Theorem 2 is the analogous criterion at the leading coefficient. The proofs are built on an auxiliary divisibility lemma, Lemma 3, which is supposed to force p|a_j whenever a product polynomial has p^k dividing its first j coefficients and both factor constants are divisible by p. Several examples are given, including a family X_{j,k} claimed to satisfy Theorem 1.","tokens_in":7440,"tokens_out":22712,"duration_ms":212380,"significance":"If the theorems are correct, they give a natural prime-power extension of the Murty–Girstmair criterion and of Eisenstein–Dumas style arguments, and the paper is self-contained with no fitted parameters or circular dependencies. The intended statements are plausible and the elementary machinery could be of independent interest. However, the current text does not provide valid proofs of the main results: Lemma 3, which is load-bearing for both theorems, contains a false divisibility step, and the proofs of Theorems 1 and 2 have additional structural gaps. The paper therefore needs substantial revision before its claims can be accepted.","major_comments":[{"comment":"The displayed assertion that p^k divides (a_{j-1} - b_0c_{j-1} - ... - b_{j-1}c_0) = b_{(j-2)/2}c_{j/2} + b_{j/2}c_{(j-2)/2} does not follow from Eq. (4). Equation (4) only controls indices up to κ=(j-2)/2, so it cannot justify subtracting b_0c_{j-1} and b_{j-1}c_0; one would need p^k|b_0c_{j-1} and p^k|b_{j-1}c_0. The step is demonstrably false in a valid instance: take p=2, k=5, j=4, f1=2+4x+x^2, f2=16+16x+8x^2+8x^3. Then a0=32, a1=a2=96, a3=64, a4=40, so the hypotheses of Lemma 3 hold, but a3 - b_0c_3 - b_3c_0 = 64 - 16 - 0 = 48, which is not divisible by 32. Since Theorem 1 and Theorem 2 both invoke Lemma 3 in the p|b_0, p|c_0 case, the main proofs are incomplete as written.","section":"Proof of Lemma 3, Case I, Subcase I, even-j paragraph after Eq. (4)"},{"comment":"The equality a_m = b_m c_{n-m} is false in general: the coefficient of x^m in f1f2 is the sum Σ_i b_i c_{m-i}, not a single product of the leading coefficient of f1 with a coefficient of f2. The contradiction in the p∤c_0 case therefore does not follow from Eq. (13). A correct argument is available directly from a_0=b_0c_0: if p^k|b_0, then c_0 divides d, so |c_0|≤d, contradicting |c_0|>d from the zero-location assumption. The proof should be rewritten with this substitution; as it stands, the displayed reasoning is invalid.","section":"Proof of Theorem 1, Eq. (13)"},{"comment":"The case analysis in the proof of Theorem 2 omits the case p|b_m with p∤c_{n-m}. The proof treats p∤b_m and then jumps to the case p|b_m, p|c_{n-m}. Since a_n=b_m c_{n-m}=±p^k d, the mixed case p|b_m, p∤c_{n-m} is possible and must be handled separately. A symmetric version of the first paragraph, using |c_0|≥q instead of |b_0|≥q, should close the gap, but the argument is absent from the manuscript.","section":"Proof of Theorem 2, second paragraph"}],"minor_comments":[{"comment":"The polynomial X_{j,k} in Eq. (14) has a_0=p^{k+1}=p^k·p, so the parameter d in Theorem 1 is d=p, not d=1. The proof in Eqs. (15)-(16) only establishes |ζ|>1 for each zero, which is weaker than the required |ζ|>p. Thus the example does not instantiate Theorem 1 as written.","section":"Section 3, Example 1"},{"comment":"The statement that Y_{j,k,m} satisfies the hypotheses of Theorem 2 is asserted without verifying the zero-location condition |θ|>d for its zeros; an explicit verification, or a reference to a general lemma, is needed.","section":"Section 3, Example 2"},{"comment":"The sentence introducing ℓ with ℓ≤k−ℓ should state explicitly that the factors f1 and f2 may be swapped so that v_p(b_0)≤v_p(c_0); otherwise the inequality ℓ≤k−ℓ is not guaranteed by the hypotheses alone.","section":"Proof of Lemma 3, definition of ℓ"},{"comment":"The phrase 'in view of (4)' at the start of the proof of Theorem 1 is confusing because Eq. (4) belongs to the proof of Lemma 3 and concerns p-adic valuations of intermediate coefficients; the intended reference appears to be to the factorization a_0=b_0c_0.","section":"Proof of Theorem 1, reference to Eq. (4)"}],"recommendation":"major_revision","confidential_remarks":"I agree with the external stress-test assessment: the central difficulty is the even-j case of Lemma 3, and the false divisibility step after Eq. (4) is not a cosmetic issue. The lemma itself is not refuted by the counterexample, and a repair may exist, but the authors must supply a correct proof of that case. The other gaps (Eq. (13) in Theorem 1 and the omitted symmetric case in Theorem 2) are local and apparently repairable. If the authors return with only typographical changes, the paper should not be accepted."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Let me give you the short version. The paper extends the Murty–Girstmair irreducibility criterion from a prime constant term to a prime power times d, under explicit coefficient-block conditions involving gcd(k,j)=1. That is a real, publishable idea if the proof can be repaired. The zero-location argument in Theorem A is clean, and the applications (Examples 2 and 3) are sensible. The paper also cites the relevant literature, including Dumas, though it does not connect its conditions to Newton polygons, which is a missed opportunity.\n\nThe trouble is that both main theorems rest on Lemma 3, and the proof of that lemma is wrong in the even-j case. After equation (4), the authors assert that p^k divides a_{j-1} minus all terms except the two middle ones. That requires p^k to divide b0 c_{j-1} and b_{j-1} c0, but the induction only established valuations for indices up to (j-2)/2. The reader's concrete instance checks out: with p=2, k=5, j=4, f1=2+4x+x^2, f2=16+16x+8x^2+8x^3, the product satisfies the lemma's hypotheses, and the asserted expression a3 - b0 c3 - b3 c0 is 48, which is not divisible by 32. The lemma's conclusion happens to hold in that example, but the proof line is false. No repair is supplied.\n\nThere are two more gaps. Theorem 2's proof omits the case p|b_m with p∤c_{n-m}. And Example 1 only shows the zeros satisfy |ζ|>1, while Theorem 1 requires |ζ|>p; as written, the example is not an instance of the theorem. It might be fixable, but it is not a correct illustration.\n\nNone of this disproves the central theorems; the gaps look repairable. But as written, the proofs do not support the claims. This is a paper for specialists in polynomial irreducibility. It deserves a serious referee, not a desk reject, because the intended result is useful and the flaw is localized. I would ask the authors to fix Lemma 3, handle the missing case, and rework Example 1 before publication. For my own work, I would not cite it until the proof is corrected.","headline":"Plausible extension of Murty–Girstmair to prime powers, but the proof of the key lemma has a false divisibility step, so the main theorems are unproven as written.","tokens_in":7978,"tokens_out":5012,"would_cite":false,"duration_ms":41856,"reading_group":"maybe","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["12E05","11C08"],"pacs":[],"model":"deepseek-v4-flash","headline":"The paper proves that a primitive integer polynomial is irreducible if all its zeros lie outside a disc of radius $d$ and its constant term is a prime power times $d$, provided $p^k$ divides a consecutive block of coefficients whose…","keywords":["irreducible polynomial","integer coefficients","prime power","zero location","Eisenstein criterion","Newton polygon","coefficient divisibility","gcd(k,j)=1"],"falsifier":"Take $p=2$, $k=5$, $j=4$, $f_1=2+4x+x^2$, and $f_2=16+16x+8x^2+8x^3$. The product has its first four coefficients divisible by $2^5$, its constant term equals $2^5$, and both factor constants are even, so Lemma 3's hypotheses hold. Yet the proof's intermediate claim that $2^5$ divides $a_3 - b_0 c_3 - b_3 c_0$ is false: $a_3 - b_0 c_3 - b_3 c_0 = 64-16-0 = 48$, and $32$ does not divide $48$. Checking whether a different argument can still force $p \\mid a_j$ in this instance would resolve whether the lemma as stated is sound.","tokens_in":1717,"feed_emoji":"🔢","tokens_out":1684,"duration_ms":153190,"temperature":0.7,"pith_summary":"The paper gives new sufficient conditions for a polynomial with integer coefficients to be irreducible over the integers. The conditions combine a prime-power value of the constant or leading coefficient with a divisibility pattern across a block of coefficients: if $a_0 = \\pm p^k d$, if $p^k$ divides the first $j$ coefficients for some $j$ coprime to $k$, and if the next coefficient is not divisible by $p$, then no nontrivial factorization can exist provided all zeros lie outside a disc of radius $d$. The same pattern is mirrored at the leading end in a second theorem. These criteria matter because they turn irreducibility into a check of a few coefficients, extending an earlier criterion that only handled the case where $|f(0)|/d$ is prime.","feed_headline":"Prime-power coefficient patterns force irreducibility","feed_subtitle":"A block of coefficients sharing a prime-power factor can settle irreducibility, extending the prime-only criterion.","key_machinery":"The load-bearing device is Lemma 3, a divisibility lemma about the coefficient sequence of a product. It says: if $p^k$ divides the first $j$ coefficients of $f_1 f_2$, $p^{k+1}$ does not divide the constant term, $\\gcd(k,j)=1$, and $p$ divides both factor constants, then $p$ divides the next coefficient $a_j$. The proof works by recursively comparing highest powers of $p$ in the convolution formulas $a_t = \\sum_i b_i c_{t-i}$, using $\\gcd(k,j)=1$ to keep the block of preserved divisibility aligned. The zero-location hypothesis acts as a companion: it guarantees that in any factorization both factor constants exceed $d$ in absolute value, so the prime-power shape of $a_0$ cannot be split without violating Lemma 3 or the size bound.","core_discovery":"The central claim is Theorem 1 (and its leading-end analogue Theorem 2): a primitive $f \\in \\mathbb{Z}[x]$ with every zero $\\theta$ satisfying $|\\theta|>d$ is irreducible in $\\mathbb{Z}[x]$ if $a_0 = \\pm p^k d$ with $p \\nmid d$, and there is $j$ with $\\gcd(k,j)=1$ such that $p^k \\mid a_0, \\dots, a_{j-1}$ and (for $k>1$) $p \\nmid a_j$. Theorem 2 is the analogue at the leading coefficient end, with the additional hypothesis $|a_0/q| \\le |a_n|$ where $q$ is the smallest prime divisor of $a_0$. The proof supposes a factorization $f_1 f_2$ and uses the zero-location hypothesis to get $|b_0|>d$ and $|c_0|>d$ from the factor constants. Lemma 3 is then invoked to force $p \\mid a_j$ whenever both factor constants are divisible by $p$, contradicting the hypothesis; the remaining cases are excluded by the size comparison $|b_0|, |c_0| > d$. The same lemma also yields a Newton-polygon-free proof of the classical prime-power Eisenstein-type theorem (Theorem B).","pith_inferences":["A natural next step is to ask whether the prime power $p^k$ in Lemma 3 can be replaced by a product of powers of distinct primes, with the block length $j$ coprime to each exponent; the p-adic valuation chase would need a multivalued version.","Example 3 already shows that the root-location hypothesis can be certified by a coefficient-dominance inequality; combining that observation with Theorems 1 and 2 would yield a fully coefficient-based irreducibility test, which the paper does not spell out.","The role of the condition $\\gcd(k,j)=1$ suggests the criterion might extend to Eisenstein-type polynomials with higher ramification, where the $j$-block corresponds to a jump in a Newton polygon."],"forward_implications":["Theorems 1 and 2 provide explicit infinite families of irreducible polynomials, such as $X_{j,k}$ and $Y_{j,k,m}$ in the examples, all with zeros outside the unit disc.","The criterion extends the earlier constant-coefficient theorem from prime values of $|f(0)|/d$ to prime powers, under mild coefficient conditions.","Lemma 3 gives a proof of the classical prime-power Eisenstein-type theorem (Theorem B) that does not rely on Newton polygons.","The conditions are purely coefficient-based, so irreducibility can be checked without computing roots or factoring the polynomial.","Both the constant-coefficient and leading-coefficient versions are covered, and the second theorem requires only an additional divisibility-size comparison."],"supporting_citations":[{"why":"Supplies the prime-case criterion (Theorem A) that Theorems 1 and 2 generalize from prime to prime power.","marker":"[6]"},{"why":"Originates the connection between primes and irreducible polynomials that motivates the paper's sufficient conditions.","marker":"[5]"},{"why":"Provides the standard Newton-polygon context for Theorem B, which the paper reproves via Lemma 3.","marker":"[4]"},{"why":"Supplies the classical zero-location perspective used to translate root bounds into coefficient inequalities.","marker":"[7]"}],"fun_headline_variants":["Prime-power coefficient blocks force irreducibility","A single prime-power block settles irreducibility","Zeros outside a disc plus prime-power coefficients force irreducibility","Prime-power divisibility pattern forces irreducibility","Irreducibility from prime-power coefficient runs"],"cache_read_input_tokens":10112,"weakest_assumption_plain":"The whole proof rests on Lemma 3: that whenever $p^k$ divides the first $j$ coefficients of a product (with $p^{k+1}$ not dividing the constant term and $\\gcd(k,j)=1$), any factorization whose constant terms are both divisible by $p$ must have $p$ dividing the next coefficient; if this lemma fails, the arguments for both main theorems collapse.","fun_headline_variants_meta":{"raw":{"variants":["Prime-power coefficient blocks force irreducibility","A single prime-power block settles irreducibility","Zeros outside a disc plus prime-power coefficients force irreducibility","Prime-power divisibility pattern forces irreducibility","Irreducibility from prime-power coefficient runs"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000841,"raw_usage":{"total_tokens":3610,"prompt_tokens":839,"completion_tokens":2771,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":455,"completion_tokens_details":{"reasoning_tokens":2697}},"tokens_in":455,"tokens_out":2771,"duration_ms":20595,"temperature":1.0,"reasoning_tokens":2697,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T13:14:58.520971+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Take $p=2$, $k=5$, $j=4$, $f_1=2+4x+x^2$, and $f_2=16+16x+8x^2+8x^3$. The product has its first four coefficients divisible by $2^5$, its constant term equals $2^5$, and both factor constants are even, so Lemma 3's hypotheses hold. Yet the proof's intermediate claim that $2^5$ divides $a_3 - b_0 c_3 - b_3 c_0$ is false: $a_3 - b_0 c_3 - b_3 c_0 = 64-16-0 = 48$, and $32$ does not divide $48$. Checking whether a different argument can still force $p \\mid a_j$ in this instance would resolve whether the lemma as stated is sound.","supporting_citations":[{"cited_title":"Girstmair, On an irreducibility criterion of M","cited_arxiv_id":null,"evidence_quote":"Supplies the prime-case criterion (Theorem A) that Theorems 1 and 2 generalize from prime to prime power."},{"cited_title":"Ram Murty, Prime numbers and irreducible polynomials, Amer","cited_arxiv_id":null,"evidence_quote":"Originates the connection between primes and irreducible polynomials that motivates the paper's sufficient conditions."},{"cited_title":"Dumas, Sur quelques cas d’irr´ eductibilit´ e des polynomes ´ a coefficients rationnels, J","cited_arxiv_id":null,"evidence_quote":"Provides the standard Newton-polygon context for Theorem B, which the paper reproves via Lemma 3."},{"cited_title":"Perron, Neue kriterien f¨ ur die irreduzibilit¨ at algebraischer g leichungen, J","cited_arxiv_id":null,"evidence_quote":"Supplies the classical zero-location perspective used to translate root bounds into coefficient inequalities."}],"review_version":1}