REVIEW 2 major objections 5 minor 4 references
Existence of primes between two consecutive squares
T0 review · 2 major / 5 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read Paper claims a proof that a prime lies between consecutive squares.
desk verdict 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. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
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.
What would settle it
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.
Extended reading notes
Core claim
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.
Load-bearing premise
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.
Editorial extensions
If this is right
- 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 $t<\lfloor\sqrt{p}\rfloor$.
Reading between the lines
- 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.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
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.
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 (2)
- [§2, Theorem 2.1, after Eq. (3)] 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.
- [§2, Corollary 2.2 and proof of Conjecture 1.3] 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.
minor comments (5)
- [§1, Conjecture 1.1] The statement contains typographical errors: 'between consecutive two perfect squares ?' should be reworded, and the question mark should be removed.
- [§2, Theorem 2.1, paragraph on the list of odd integers] 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.
- [§1, Theorem 1.6] Theorem 1.6 is labeled 'Pythagorean,' but the listed parity facts are not the Pythagorean theorem; the label is misleading.
- [§2, Theorem 2.1, proof] 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.
- [§2, Theorem 2.1, proof] 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.
Circularity Check
Theorem 2.1 imports an odd prime from the set of all odd integers and declares it to lie in the target interval, dropping the boundary constraint t′ = −n(n+1).
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other
[Proof of Theorem 2.1, final paragraph after Eq. (3), bullet list for t′ cases]
"Let X = { p ∈ Z | p is odd }. Since p is odd, there exist a odd prime p must be in X. Based on t′ = ±t − p, we conclude there exist a prime in ( n2, n(n + 1)) and ( n(n + 1),(n + 1)2)."
The parameter p is introduced as a fresh odd positive integer in the general solution t′ = ±t ± p of the LDE y − x = 2t. The earlier boundary condition t′ = −n(n+1), forced by Eq. (2), is never imposed on p, so p is free. The observation that the set of all odd integers contains an odd prime is a tautology about the integer line; it does not locate that prime inside (n^2, (n+1)^2). The conclusion therefore reduces to assuming that an arbitrary odd prime coincides with the interval element x or y, which is exactly the theorem to be proved. For example, n=5, t=3 forces p=27 or 33 for the actual x,y, neither of which is prime, showing that the free choice is not available.
full rationale
The derivation chain is not circular in the usual senses of fitted inputs or self-citation: Theorem 2.1 begins with arbitrary odd x,y in the interval, derives k+l=2n, and correctly obtains x=n^2+n−t and y=n^2+n+t with odd t. No external result from the same authors is invoked. The circularity (or at least conclusion-import) enters only in the final case analysis. The LDE y−x=2t has a free parameter t′, but once x,y are fixed by Eq. (2), t′ is fixed as t′=−n(n+1), not free. The proof instead writes t′=±t±p with p a new odd integer, ignores the fixed value, observes that the set of odd integers contains an odd prime, and then uses that prime as the p in the parametrization. That step does not derive the existence of a prime in (n^2,(n+1)^2); it assumes it by allowing p to be any prime. Hence the central claim is unsupported. Corollary 2.2 and the Andrica proof inherit this gap. Because this is a localized logical leap rather than a self-citation loop or a fitted parameter renamed as a prediction, the circularity score is moderate.
Assumptions & free parameters
free parameters (1)
- p =
an arbitrary odd prime, unspecified
assumptions (4)
- standard math The linear Diophantine equation ax+by=c is solvable iff d|c, with the stated general solution (Theorem 1.5).
- standard math The parity facts in Theorem 1.6 (sums and products of evens and odds).
- ad hoc to paper The even integer t' can be written as ±t ± p for the given odd t and an odd positive integer p.
- standard math There exists at least one odd prime among all odd integers.
Cite this review
Pith. "Pith review of Existence of primes between two consecutive squares." pith.science (2026). https://pith.science/paper/D32FAS3F
@misc{pith2026190808995,
author = {Pith},
title = {Pith review of: Existence of primes between two consecutive squares},
year = {2026},
howpublished = {\url{https://pith.science/paper/D32FAS3F}},
note = {Machine review of arXiv:1908.08995}
}
read the original abstract
Legendre's Conjecture is one of the most elegant open problems in Number Theory, which states that there is a prime between consecutive two perfect squares. In this note, we prove the conjecture holds true and also discuss the related results.
Reference graph
Works this paper leans on
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[1]
Wells, David, Prime Numbers: The Most Mysterious Figure s in Math, John Wiley & Sons, (2011)
work page 2011
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[2]
Om vor Kundskab om Primtallenes Mngde me llem givne Grndser
Oppermann, L., “Om vor Kundskab om Primtallenes Mngde me llem givne Grndser”, Oversigt over det Kongelige Danske Videnskabern es Selskabs Forhandlinger og dets Medlemmers Arbejder: 169179, (1882)
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[3]
Note on a conjecture in prime number theory
Andrica, D. “Note on a conjecture in prime number theory” . Studia Univ. BabesBolyai Math. 31 (4): 4448, (1986)
work page 1986
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[4]
Koshy, Elementary Number Theory with application , 2 nd Edition, Aca- demic Press(2007)
T. Koshy, Elementary Number Theory with application , 2 nd Edition, Aca- demic Press(2007). 3
work page 2007
Reviewed August 14, 2026 · model on record in the stance chip above.
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