REVIEW 3 major objections 3 minor 6 references
Prime Quadruplets and Jump Conditions on Arithmetic Functions
T0 review · 3 major / 3 minor · reviewed 2026-08-02 · deepseek-v4-flash
Pith's one-line read For products of two distinct primes, the simultaneous 12-jump in φ and σ happens exactly for prime quadruplets.
desk verdict Theorem 2.2 (the semiprime characterization) is correct and worth citing; Theorem 3.1 (no prime powers) has a reversed inequality in Subcase 3.2 and is not proved as written. 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 central identity is u^2-v^2=48, where u=q-p and v=q'-p' are the prime gaps inside n and n+12. Because u and v share parity, the factorization 48=(u-v)(u+v) yields a finite list of candidates, and the only survivor is (u,v)=(8,4). That single pair forces q=p+8 and q'=p+4, and combined with p+q=p'+q' reproduces the quadruplet. For the prime-power exclusion, the machinery is the elementary bound φ(m)≤m-1 plus the observation that any repeated prime factor of n+12 must also divide n+12-φ(n+12), which contradicts the reduced equation.
What would settle it
Run an independent implementation of the same segmented sieve over n∈[2,10^12] and check that it returns exactly the same 166 solutions; any discrepancy would falsify the claimed completeness. Alternatively, a single composite n not of the form p(p+8) with (p,p+2,p+6,p+8) prime and satisfying both jump conditions would falsify the open conjecture, and a single prime power p^k satisfying φ(p^k+12)=φ(p^k)+12 would falsify Theorem 3.1.
Extended reading notes
Core claim
The main result, Theorem 2.2, states that if n=pq and n+12=p'q' are both products of two distinct primes with p<q and p'<q', then the two jump conditions φ(n+12)-φ(n)=12 and σ(n+12)-σ(n)=12 hold together if and only if p≥5, q=p+8, and {p',q'}={p+2,p+6}. The proof reduces both equations to the equality p+q=p'+q', then uses the discriminant relation (q-p)^2-(q'-p')^2=48. Factoring 48 gives only three possible pairs of square gaps; parity rules out all but (8,4), forcing the prime-quadruplet spacing. A separate theorem shows no prime power p^k with k≥2 can satisfy the first jump condition alone.
Load-bearing premise
The completeness of the computational search up to 10^12 rests on an unverified custom segmented-sieve program; if that program has an overflow or sieving bug, the reported 166 solutions and the absence of counterexamples could be wrong.
Editorial extensions
If this is right
- For squarefree semiprime n with n+12 semiprime, the simultaneous jump conditions hold only for the prime-quadruplet form n=p(p+8).
- Every such semiprime solution is congruent to 65 modulo 72; under the open characterization conjecture, all composite solutions would share this residue.
- No integer of the form p^k with k≥2 satisfies both jump conditions; in fact, the first condition alone fails for every prime power.
- The exhaustive search up to 10^12 found exactly 166 composite solutions, all congruent to 65 (mod 72) and all of the prime-quadruplet semiprime type.
- If the characterization conjecture is true and infinitely many solutions exist, then infinitely many prime quadruplets exist.
Reading between the lines
- The same discriminant trick likely adapts to other even offsets: replacing 12 by 2m would produce a finite list of possible prime gaps from the factorizations of 8m, giving a path to similar characterizations for shifts like 6 or 20.
- A natural testable prediction from the search is that any future solution beyond 10^12 will again be 65 (mod 72) and will factor as p(p+8); finding one that does not would refute the conjecture outright.
- The conditional implication ties the infinitude of these solutions to the prime k-tuple conjecture for k=4; any unconditional proof of infinitely many such n would settle that case.
- The elementary technique used to rule out prime powers might be extended to rule out numbers with exactly two prime factors of unequal size, though the semiprime case is already fully settled.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper studies composite integers n satisfying the simultaneous jump conditions φ(n+12)=φ(n)+12 and σ(n+12)=σ(n)+12. The main results are: (i) Theorem 2.2, a complete characterization when both n and n+12 are squarefree semiprimes, namely n=p(p+8) with (p,p+2,p+6,p+8) a prime quadruplet; (ii) Theorem 3.1, asserting that no prime power p^k satisfies even the φ-jump alone; and (iii) a computational search up to 10^12 reporting exactly 166 solutions, all of the predicted semiprime form and all congruent to 65 mod 72. The paper is motivated by Conjecture 1.1, attributed to Stephan and McCranie, and notes that a proof of Conjecture 1.1 together with infinitely many solutions would imply infinitely many prime quadruplets.
Significance. If the results stand, the paper makes a clean contribution to an old jump-condition problem: Theorem 2.2 is an elegant, fully self-contained proof that within the semiprime class the prime-quadruplet construction is necessary and sufficient. The computational extension of the OEIS data by three orders of magnitude is also useful evidence for Conjecture 1.1. However, the proof of Theorem 3.1 contains a reversed inequality in Subcase 3.2, so the non-existence of prime-power solutions is not established as written. The value of the paper therefore depends on a corrected proof of that theorem; the computational search does not repair the logical gap.
major comments (3)
- [§3, Subcase 3.2] The displayed inequality φ(N)/N ≤ (1−1/(p+2))(1−1/(p+4))(1−1/r3) has the wrong direction. Since r1≥p+2 and r2≥p+4, we have 1−1/r1 ≥ 1−1/(p+2) and 1−1/r2 ≥ 1−1/(p+4); the remaining factors are also at least 1−1/r3. Thus φ(N)/N is ≥, not ≤, that product (and is larger still when there are more than three prime factors). Consequently the later chain leading to 1/r3<0 is invalid. The correct inequality is compatible with φ(N)/N>(p−1)/p without contradiction. Theorem 3.1 is therefore not proved as written, and the computational search in Section 4 does not supply the missing argument.
- [§3, Subcase 3.1, k=2] The line 'p+1−(p+2) ≥ p+4' is not a valid substitution. From r2 = p+1−r1 and r1≥p+2 one obtains r2≤−1, which already contradicts r2≥p+4. The intended contradiction is recoverable, but the written derivation is algebraically wrong and should be corrected.
- [§4 and Abstract] The claim of a 'complete search up to 10^12' is not independently verifiable from the manuscript. The cited GitHub repository [6] is not pinned to a commit, and no build/run instructions, compiler flags, or checksums for the 166-solution table are given. Since the abstract's empirical claim depends on the correctness of the custom C++ segmented sieve, the authors should provide a permanent artifact (DOI/commit) and enough detail for the computation to be reproduced.
minor comments (3)
- [Abstract and §4] The abstract has '10 12' and §4 has '1012'; these should be typeset as 10^12.
- [References] Reference [6] is a GitHub repository without a version, commit hash, or access date; a permanent identifier and retrieval date should be given.
- [Abstract] The phrase 'no solutions can be composed of a single prime power' could be clarified: Theorem 3.1 actually shows that no prime power satisfies the φ-jump alone, which is stronger than needed for the simultaneous conditions.
Circularity Check
No significant circularity found; the proofs are self-contained and the empirical search is independent of the theorems.
full rationale
The paper's central derivation chain is not circular. Theorem 2.2 is a direct two-way proof from the multiplicative formulas for phi and sigma on squarefree semiprimes, reducing the simultaneous jump conditions to equality of prime sums and then to the integer equation u^2 - v^2 = 48; no target conclusion is assumed and no fitted parameter is renamed as a prediction. Theorem 3.1 likewise proceeds from the standard prime-power totient formula and attempts a contradiction by case analysis; its validity is a mathematical correctness question, not a circularity issue. The computational section is an external exhaustive search over n up to 10^12, not a parameter fit used inside the proofs, and it is clearly labeled as supporting evidence rather than as a derivation of Conjecture 1.1. The Hardy-Littlewood conjecture appears only in the conditional motivation ('If Conjecture 1.1 is true and there are infinitely many solutions...'), which is a straightforward logical implication, not an imported uniqueness theorem or ansatz. The references to OEIS entries and the prior conjecture by McCranie are cited as history and context, not as load-bearing proof steps, and no result is justified solely by a self-citation. The known construction n = p(p+8) for prime quadruplets is used as the backward direction of Theorem 2.2 and as the form asserted in Conjecture 1.1, but the forward direction is proven independently by the discriminant argument. There is no equation in the paper that reduces to its own input by construction, and no fitted quantity is later called a prediction. Therefore the paper receives a circularity score of 0.
Assumptions & free parameters
assumptions (4)
- standard math Multiplicativity formulas φ(p^k)=p^k−p^{k−1}, φ(pq)=(p−1)(q−1), σ(pq)=(p+1)(q+1)
- standard math φ(m) ≤ m−1 for every integer m>1
- standard math Chinese Remainder Theorem
- domain assumption The theorem's semiprime hypothesis n=pq, n+12=p'q' with distinct primes
Cite this review
Pith. "Pith review of Prime Quadruplets and Jump Conditions on Arithmetic Functions." pith.science (2026). https://pith.science/paper/B3Z2TCNT
@misc{pith2026260610331,
author = {Pith},
title = {Pith review of: Prime Quadruplets and Jump Conditions on Arithmetic Functions},
year = {2026},
howpublished = {\url{https://pith.science/paper/B3Z2TCNT}},
note = {Machine review of arXiv:2606.10331}
}
abstract
We provide progress on the characterization of composite integers $n$ that satisfy the jump conditions $\varphi(n+12)=\varphi(n)+12$ and $\sigma(n+12)=\sigma(n)+12$ simultaneously. While it is known that prime quadruplets $(p,p+2,p+6,p+8)$ generate solutions $n=p(p+8)$, the complete characterization remains an open conjecture. We prove that this characterization is complete when $n$ and $n+12$ are both squarefree semiprimes, and that no solution $n$ can be a prime power. Furthermore, a complete search up to $10^{12}$ resulted in no counterexamples to the conjecture. If this conjecture is proven true, and there are infinitely many such solutions, then it can be proved that there are infinitely many prime quadruplets.
Reference graph
Works this paper leans on
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[6]
com/bvrtoverfitprimes/integersequencetesting
bvrtoverfitprimes (Roy Choudhury),integersequencetesting, GitHub repository, 2026,https://github. com/bvrtoverfitprimes/integersequencetesting
2026
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[1]
G. H. Hardy and J. E. Littlewood,Some problems of ‘Partitio Numerorum’; III: On the expression of a number as a sum of primes, Acta Math.44(1923), 1–70
1923
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[2]
OEIS Foundation Inc.,Sequence A056777: Composite numbersksuch thatφ(k+ 12) =φ(k) + 12and σ(k+ 12) =σ(k) + 12, The On-Line Encyclopedia of Integer Sequences,https://oeis.org/A056777
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[3]
OEIS Foundation Inc.,Sequence A007530: Prime quadruplets: numbersksuch thatk,k+ 2,k+ 6,k+ 8 are all prime, The On-Line Encyclopedia of Integer Sequences,https://oeis.org/A007530
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[4]
R. K. Guy,Unsolved Problems in Number Theory, 3rd ed., Springer-Verlag, New York, 2004
2004
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[5]
R. Stephan,Prove or Disprove: 100 Conjectures from the OEIS, arXiv preprint math/0409509, 2004, https://arxiv.org/abs/math/0409509
arXiv 2004
Reviewed August 2, 2026 · model on record in the stance chip above.
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