{"id":"5a13531a-c1fb-4ebe-b324-8f01db16168b","arxiv_id":"1908.08995","paper_version":1,"verdict":"REJECT","confidence":"HIGH","novelty_score":6.0,"correctness_risk":"high","formal_verification":"none","parameter_count":1,"one_line_summary":"The paper claims a proof of Legendre's conjecture, but the final step of the proof does not establish a prime in the required interval, so the conjecture remains open.","lead":"A short note claims to prove Legendre's conjecture, which says there is a prime between every pair of consecutive square numbers. Its central proof step fails, so the conjecture remains open.","discovery_kind":"unclear","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The final step of Theorem 2.1 introduces an unconstrained odd parameter p and then treats an arbitrary odd prime as if it were located in (n^2,(n+1)^2); the interval boundary conditions from Eq. (2) are never transferred to p.","rationale":"The reader's weakest assumption correctly identifies the invalid final inference. The paper's Theorem 2.1 would settle Legendre, Oppermann, Brocard and Andrica, but the proof breaks at the free-parameter relabeling: p is not shown to be the original x or y from Eq. (2). The concrete n=2,t=1,p=3 case demonstrates the failure of the inference, and no independent support such as machine checking is present. A REJECT verdict is warranted because the central theorem is not established. I see no other more load-bearing flaw; the boundary drop is sufficient.","tokens_in":2447,"tokens_out":5650,"duration_ms":53734,"concrete_test":"Evaluate the final bullets of the proof at n=2, t=1, p=3. The parametrization of y-x=2t with t'=-t-p gives x=p=3 and y=2t+p=5; p is an odd prime but lies outside (4,9), while y is prime only by coincidence. This single substitution shows the inference 'an odd prime p exists, hence p lies in the interval' is invalid. For a broader check, enumerate all odd t<n and all prime p produced by t'=-t-p and t'=t-p for n≤100, and record how many such p fall outside (n^2,(n+1)^2); if any do, the proof step lacks the needed boundary constraint.","verdict_should_be":"REJECT","load_bearing_attack":"The proof of Theorem 2.1 reaches y-x=2t and writes the general solution as x=-(t+t'), y=t-t'. Since x,y are odd, t' is even, and the paper sets t'=±t±p for an odd positive integer p. This p is a fresh free parameter: the four cases give (x,y)=(p,2t+p) or (p-2t,p), but nothing constrains p to lie in (n^2,n(n+1)) or (n(n+1),(n+1)^2). The paper then observes that the set of odd integers contains an odd prime and 'Based on t'=±t-p' concludes a prime lies in each of the two subintervals. This is exactly the boundary drop: existence of an odd prime somewhere on the integer line does not locate it in the target interval. For example, with n=2 and t=1, taking p=3 in the case t'=-t-p yields x=3,y=5; p=3 is prime but 3∉(4,9). The construction therefore cannot certify an interval prime, and Eq. (2) is not used to restrict p. The central claim is unsupported by the written argument.","agreement_with_reader":"agree"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper claims a proof of Legendre's conjecture (a prime between every two consecutive squares) and, as corollaries, of Oppermann's, Brocard's, and Andrica's conjectures. The method is elementary: fix two odd integers x,y in (n^2,(n+1)^2) summing to 2n(n+1), reduce their positions to k=n−t and l=n+t, then use linear Diophantine equations to parametrize y−x=2t and finally invoke the existence of an odd prime among the odd integers. The central conclusion rests on the last step of the proof of Theorem 2.1.","tokens_in":2758,"tokens_out":6944,"duration_ms":66723,"significance":"Were the argument valid, it would resolve several long-standing open problems in a few lines, which would be a notable result. The early Diophantine setup through Eq. (2) is elementary and largely correct for n≥2, and the paper is clear about what it intends to prove. However, the proof does not contain a valid number-theoretic step: the existence of some odd prime somewhere on the integer line is not shown to produce a prime in the prescribed interval. The paper provides no machine-checked proofs, code, or independent verification, and the claimed corollaries all inherit the single fatal gap. The significance of the conjectures makes careful checking worthwhile, but the manuscript as written does not establish its claims.","major_comments":[{"comment":"The decisive step is invalid. After solving the linear Diophantine equation y−x=2t as x=−(t+t′), y=t−t′, the proof introduces an odd positive integer p through the cases t′=±t±p. But for the specific x,y already fixed by Eq. (2), t′ is not a free parameter: substituting x=n^2+n−t into x=−(t+t′) gives t′=−(n^2+n), so in the surviving cases p is equal to x or to y. The sentence 'Since p is odd, there exist an odd prime p must be in X' therefore merely restates, in new notation, that x or y is prime. Choosing an arbitrary odd prime from the set of all odd integers does not force that prime to coincide with the specific x or y constrained by Eq. (2), and no boundary condition from Eq. (2) is transferred to p. This is precisely the content of Legendre's conjecture, so the argument assumes what it must prove.","section":"§2, Theorem 2.1, after Eq. (3)"},{"comment":"Corollary 2.2 states that every prime p≥5 has the form n^2+n±t with n=⌊√p⌋ and odd t<n. This is derived from Theorem 2.1, and the subsequent proof of Andrica's conjecture uses Corollary 2.2 as its premise. Since Theorem 2.1 is not established, Corollary 2.2 is unproved and the numerical inequalities in the Andrica proof have no valid foundation. Remark 2.3, which derives Brocard's conjecture from Theorem 2.1, likewise depends entirely on the unsupported central claim.","section":"§2, Corollary 2.2 and proof of Conjecture 1.3"}],"minor_comments":[{"comment":"The statement contains typographical errors: 'between consecutive two perfect squares ?' should be reworded, and the question mark should be removed.","section":"§1, Conjecture 1.1"},{"comment":"The claimed list of odd integers in (n^2,(n+1)^2) of the form c±1,c±3,…,c±t with odd t<n is not accurate for all n; for example, when n=3, the odd integer 15 lies in (9,16) but corresponds to t=3, which is not allowed by t<n.","section":"§2, Theorem 2.1, paragraph on the list of odd integers"},{"comment":"Theorem 1.6 is labeled 'Pythagorean,' but the listed parity facts are not the Pythagorean theorem; the label is misleading.","section":"§1, Theorem 1.6"},{"comment":"The notation 't ∈ 2Z+1' is nonstandard and could be written as 't odd'; also, the phrase 't′ is sum of any two odd integer' is imprecise, since the representation t′=±t±p is asserted rather than derived from the parity argument.","section":"§2, Theorem 2.1, proof"},{"comment":"The case n=1 is not handled: there are not two distinct odd positive integers in (1,4), so the opening construction of x and y fails for n=1 even though Legendre's conjecture is true there.","section":"§2, Theorem 2.1, proof"}],"recommendation":"reject","confidential_remarks":"This is an elementary attempt at a famous open problem, and the decisive inference in Theorem 2.1 is not merely incomplete but circular: it selects an odd prime from the set of all odd integers without proving it lies in the target interval or coincides with the constructed x or y. Because the error is in the main theorem and the remaining claims depend on it, the manuscript cannot be repaired within its current scope. I recommend rejection."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Dear colleague,\n\nThe short answer: this paper is an attempted proof of Legendre's conjecture, and it fails at the final step. The setup is reasonable—the author considers two odd numbers x,y in the interval summing to 2n(n+1), derives x=n^2+n−t, y=n^2+n+t, and reduces the problem to showing one of these is prime. That part is fine. The trouble starts when the author solves the linear equation y−x=2t. The general solution is x=−(t+t'), y=t−t', with t' even. The author then writes t' = ±t ± p for an odd positive integer p. This p is a new free parameter, and the four cases give (x,y)=(p,2t+p) or (p−2t,p). Nothing constrains p to lie in the original interval. The author then observes that odd integers contain odd primes and concludes that the interval contains a prime. That is a non sequitur. For instance, with n=2, t=1, the case t'=−t−p gives x=p, y=2t+p; choosing p=3 yields (3,5), but 3 is not in (4,9). The interval boundary conditions from Eq. (2) are never transferred to p. So the central claim of Theorem 2.1 is unsupported.\n\nThe corollaries and the proofs of Oppermann, Andrica, and Brocard depend entirely on Theorem 2.1, so they inherit the flaw. The paper has no other independent contribution. On the positive side, the exposition is clear and the author engages with standard results (linear Diophantine equations, parity) and cites relevant literature. But the proof's final step is a textbook quantification error: it proves the existence of an odd prime somewhere on the number line, not in the intended interval.\n\nThis is not a paper that deserves referee time. The error is immediate and fatal to the main theorem; there is nothing salvageable in the argument. I would not bring it to reading group and would not cite it. Whether the author is a serious thinker is debatable, but the reasoning error here is so foundational that I'd have to say no.\n\nRecommendation: desk reject, with a brief explanation pointing to the missing constraint on p.","headline":"A well-written but fatally flawed attempt at Legendre's conjecture: the proof introduces an unconstrained parameter and mistakes an odd integer for an interval prime.","tokens_in":3224,"tokens_out":3220,"would_cite":false,"duration_ms":28422,"reading_group":"no","serious_thinker":"no","would_accept_peer_review":false},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11N05","11N32"],"pacs":[],"model":"deepseek-v4-flash","headline":"Paper claims a proof that a prime lies between consecutive squares.","keywords":["prime numbers","Legendre's conjecture","linear Diophantine equation","Oppermann's conjecture","Brocard's conjecture","Andrica's conjecture","prime gaps"],"falsifier":"Check the paper's stronger symmetric claim directly: for every $n\\ge2$, test whether the half-interval $(n^2,n(n+1))$ and the half-interval $(n(n+1),(n+1)^2)$ each contain a prime. A single $n$ for which one half is prime-free would refute the proof's final conclusion; such an $n$ would also be a counterexample to Oppermann's conjecture, though not necessarily to Legendre's conjecture itself.","tokens_in":2276,"feed_emoji":"🔢","tokens_out":9859,"duration_ms":83054,"temperature":0.7,"pith_summary":"This paper claims to prove Legendre's conjecture: for every positive integer $n$, there is at least one prime strictly between $n^2$ and $(n+1)^2$. The proof splits the interval at its midpoint $n(n+1)$ and writes every odd integer in the interval as $n^2+n\\pm t$, then uses the linear Diophantine equation $y-x=2t$ to argue that one of the two symmetric odd integers must be prime. If the argument holds, the author further claims that the same reasoning settles Oppermann's, Brocard's, and Andrica's conjectures. A sympathetic reader would care because the target is one of Landau's long-standing problems and the method is entirely elementary.","feed_headline":"Claim: a prime always sits between consecutive square numbers","feed_subtitle":"If the proof holds, it also settles Oppermann, Brocard, and Andrica conjectures in one stroke.","key_machinery":"The load-bearing mechanism is the general solution of the linear Diophantine equation $y-x=2t$. Because the original interval's odd integers are $c\\pm1,c\\pm3,\\ldots,c\\pm t$ with $c=n(n+1)$, the search for a prime reduces to a parametrization $x=-(t+t')$, $y=t-t'$ over an integer parameter $t'$. Parity facts from the paper's Theorem 1.6 restrict $t'$ to even values and allow it to be written as $\\pm t\\pm p$, which is how an odd prime $p$ is introduced into the argument.","core_discovery":"The paper's central claim is Theorem 2.1: there exists a prime between any two consecutive perfect squares. The proof begins with two distinct odd integers $x,y$ in $(n^2,(n+1)^2)$ whose sum is $2n(n+1)$, which forces $x=n^2+n-t$ and $y=n^2+n+t$ for an odd $t<n$, so the pair is symmetric about the midpoint $c=n(n+1)$. Solving the difference equation $y-x=2t$ and examining its general solution, the author selects branches in which one coordinate equals an odd prime $p$, concluding that the interval contains a prime in the lower half and a prime in the upper half. Corollary 2.2 converts this into a representation of every prime $p\\ge5$ as $n^2+n\\pm t$ with $n=\\lfloor\\sqrt{p}\\rfloor$ and odd $t<n$, and the final section uses that representation to derive Andrica's conjecture, while Remark 2.3 derives Brocard's conjecture.","pith_inferences":["Editorial inference: the proof as written claims the stronger symmetric statement that both halves of the interval contain a prime; this is equivalent to Oppermann's conjecture for consecutive squares, so the method, if valid, would establish more than Legendre's conjecture alone.","Editorial inference: the same midpoint-plus-Diophantine construction could in principle be tried on intervals between consecutive higher powers, such as cubes, though the paper does not say whether the parity and parametrization steps would survive.","Editorial inference: a direct computational check of the branch condition $p=n^2+n-t$ for small $n$ would show where the parametric family does or does not land on primes, giving a concrete way to test the internal logic without searching for a Legendre counterexample."],"forward_implications":["Legendre's conjecture would be true: every interval $(n^2,(n+1)^2)$ would contain at least one prime.","Oppermann's conjecture would follow: each half-interval $(n^2,n(n+1))$ and $(n(n+1),(n+1)^2)$ would contain a prime, producing two primes between consecutive squares.","Brocard's conjecture would follow with the minimum prime gap of two: between the squares of two consecutive primes there would be at least four primes.","Andrica's conjecture would follow: consecutive primes $P_n$ and $P_{n+1}$ would satisfy $\\sqrt{P_{n+1}}-\\sqrt{P_n}<1$.","Every prime $p\\ge5$ would have a fixed parametrization by the nearest square root: $p=\\lfloor\\sqrt{p}\\rfloor^2+\\lfloor\\sqrt{p}\\rfloor\\pm t$ with odd $t<\\lfloor\\sqrt{p}\\rfloor$."],"supporting_citations":[{"why":"Supplies the general solution of the linear Diophantine equation and the parity facts that the proof uses to parametrize x and y.","marker":"[4]"},{"why":"States Legendre's conjecture, which Theorem 2.1 targets, and Brocard's conjecture, which Remark 2.3 derives.","marker":"[1]"},{"why":"States Oppermann's conjecture, whose half-interval structure matches the midpoint split the proof relies on.","marker":"[2]"},{"why":"States Andrica's conjecture, which the final section derives from Corollary 2.2.","marker":"[3]"}],"fun_headline_variants":["Prime between every pair of consecutive squares, claimed proof","New proof claims Legendre's conjecture: prime in every square gap","Claim: every consecutive squares contain a prime","Squares always hold a prime? Proof claims yes","Prime between n^2 and (n+1)^2 for all n, claimed"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"The proof assumes that the odd number $p$ produced by the parametrization can be chosen as a prime without verifying that it still lies between the two squares as required by the original setup.","fun_headline_variants_meta":{"raw":{"variants":["Prime between every pair of consecutive squares, claimed proof","New proof claims Legendre's conjecture: prime in every square gap","Claim: every consecutive squares contain a prime","Squares always hold a prime? Proof claims yes","Prime between n^2 and (n+1)^2 for all n, claimed"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000254,"raw_usage":{"total_tokens":1484,"prompt_tokens":775,"completion_tokens":709,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":391,"completion_tokens_details":{"reasoning_tokens":625}},"tokens_in":391,"tokens_out":709,"duration_ms":7039,"temperature":1.0,"reasoning_tokens":625,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-14T13:02:21.457075+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"Check the paper's stronger symmetric claim directly: for every $n\\ge2$, test whether the half-interval $(n^2,n(n+1))$ and the half-interval $(n(n+1),(n+1)^2)$ each contain a prime. A single $n$ for which one half is prime-free would refute the proof's final conclusion; such an $n$ would also be a counterexample to Oppermann's conjecture, though not necessarily to Legendre's conjecture itself.","supporting_citations":[{"cited_title":"Koshy, Elementary Number Theory with application , 2 nd Edition, Aca- demic Press(2007)","cited_arxiv_id":null,"evidence_quote":"Supplies the general solution of the linear Diophantine equation and the parity facts that the proof uses to parametrize x and y."},{"cited_title":null,"cited_arxiv_id":null,"evidence_quote":"States Legendre's conjecture, which Theorem 2.1 targets, and Brocard's conjecture, which Remark 2.3 derives."},{"cited_title":"Om vor Kundskab om Primtallenes Mngde me llem givne Grndser","cited_arxiv_id":null,"evidence_quote":"States Oppermann's conjecture, whose half-interval structure matches the midpoint split the proof relies on."},{"cited_title":"Note on a conjecture in prime number theory","cited_arxiv_id":null,"evidence_quote":"States Andrica's conjecture, which the final section derives from Corollary 2.2."}],"review_version":1}