REVIEW 3 major objections 3 minor 1 cited by
On the third largest prime divisor of an odd perfect number
T0 review · 3 major / 3 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read For an odd perfect number $N$ with three largest prime divisors $a<b<c$, the paper proves $abc < (2N)^{3/5}$ and hence $a < (2N)^{1/6}$.
desk verdict The a-bound is a real new result, but the headline abc bound rests on a linear combination error in Proposition 20 that the authors need to fix. 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 engine of the proof is the m-type argument: any proper divisor $M$ of a perfect number is deficient, so $\sigma(M)<2M$, and therefore at least one prime power $p^e$ in the top component set does not divide $\sigma(M)$; perfectness of $N$ then forces another component $m$ of $N$, outside $M$, with $p\mid\sigma(m)$ and $m>p/2$. Appending such $m$ to divisibility chains is what converts weak bounds into the theorems. The second tool is the complete classification of $\sigma_{2,2}$ quasisolutions by the identity $5pq=p^2+q^2+p+q+1$ and the recurrence above, which controls which two-cycles can occur in the divisor graph of an odd perfect number.
What would settle it
Run the quasisolution recurrence $t_{n+2}=(t_{n+1}^2+t_{n+1}+1)/t_n$ and search consecutive terms for a pair of primes $(t_n,t_{n+1})$ with $t_n^2\mid t_{n+1}^2+t_{n+1}+1$; finding one would violate Lemma 6, on which the proof of Theorem 14 depends. The paper already reports that the analogous search for $\sigma_{2,2}$ pairs reaches $10^{4000}$, so extending this search is a concrete test of the structural claims.
Extended reading notes
Core claim
The central discovery is a product bound for the tail of the prime factorization of an odd perfect number. Writing $a=p_{k-2}$, $b=p_{k-1}$, $c=p_k$ for the three largest prime divisors, Theorem 14 asserts $abc < (2N)^{3/5}$. The proof is a case analysis over which of $a,b,c$ is the special prime (the one raised to an odd exponent); in every case an m-type deficiency argument produces a new prime-power component of $N$ whose size forces the desired inequality. A separate thread classifies the $\sigma_{2,2}$ pairs $q\mid p^2+p+1$, $p\mid q^2+q+1$: they are consecutive terms of the recurrence $t_{n+2}=(t_{n+1}^2+t_{n+1}+1)/t_n$, and no such pair has $p^2\mid q^2+q+1$. These classifications rule out the divisibility configurations that would otherwise defeat the product bound.
Load-bearing premise
The proofs lean on the m-type step: whenever the largest prime-power components are collected into a proper divisor $M$, perfectness must supply a new component $m$ outside $M$ with $m>p/2$ that fits the needed divisibility chain, and every possible configuration must fall into one of the paper's case splits.
Editorial extensions
If this is right
- For every odd perfect number $N$, the three largest prime divisors satisfy $abc < (2N)^{3/5}$.
- The third largest prime divisor satisfies $a < (2N)^{1/6}$, and unless $a^2\|N$, $b^2\|N$, $c\|N$ the bound improves to $a < (2N)^{1/7}$.
- The second and third largest prime divisors satisfy either $bc < 4N^{4/9}$ or the exceptional condition $b^2\|N$, $c\|N$ with $c\mid\sigma(b^2)$ or $b\mid\sigma(c)$.
- Any two primes $p,q$ forming a $\sigma_{2,2}$ pair are consecutive terms of the quasisolution recurrence, and no such pair satisfies $p^2\mid q^2+q+1$; this restricts the possible divisor graph of an odd perfect number.
Reading between the lines
- The exceptional case $a^2\|N$, $b^2\|N$, $c\|N$ is the only obstruction to replacing the exponent $1/6$ by $1/7$ in the bound on $a$; if that configuration were ruled out, the main corollary would improve automatically.
- The proof's linear-programming combination of logarithmic inequalities is modular: each new restriction on $\sigma_{2,2}$ pairs can be fed back into the same inequality system to produce sharper constants for $abc$ and $bc$.
- Because the quasisolution recurrence has exponential growth, the heuristic that only finitely many $\sigma_{2,2}$ pairs exist could be tested computationally far beyond $10^{4000}$; any new pair would not by itself disprove the main theorem but would force the case analysis to be re-checked.
- If the paper's Conjecture 9 (squarefree $\sigma(p^2)$ along $\sigma_{2,2}$ pairs) is true, Lemma 6 follows immediately and several case splits in Propositions 17 and 20 can be simplified, likely tightening the $3/5$ exponent.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper studies the three largest prime divisors a, b, c of an odd perfect number N. It proves a general upper bound on the i-th largest prime divisor, obtaining a < (2N)^{1/6}, and claims the product bound abc < (2N)^{3/5}. The proof is case-based, organized around an m-type bootstrap argument and a linear-programming optimization of inequalities in logarithms of a, b, c, N. Substantial auxiliary results classify quasisolutions and constrain σ_{m,n} pairs, especially σ_{2,2} pairs.
Significance. If the main theorem were correctly proved, the abc bound would be a genuine improvement over the trivial Euler-based estimate and would give a nontrivial upper bound on the third largest prime divisor. The quasisolution classification and the structural lemmas about σ_{2,2} pairs are independently interesting and could be useful for future work. The linear-programming approach is a sound general method. However, the proof of the main theorem contains a load-bearing arithmetic error in Proposition 20, so the central claim is not established as written.
major comments (3)
- [Proposition 20, after Eq. (22)] The displayed combination "3/5(13)+22" is arithmetically incorrect. Adding 3/5 of (13), namely 3β ≤ (3/5)(log N + log 2), to (22), namely α+γ−2β < log 2, gives α+β+γ < (3/5)log N + (8/5)log 2, not (3/5)log N + (3/5)log 2. The claimed bound abc < (2N)^{3/5} therefore does not follow in the ac|σ(b²) subcase; the inequality actually obtained is weaker by a factor of 2. This is a load-bearing gap in the proof of the main theorem.
- [Proposition 20, Eq. (20) and Eq. (21)] The m-type inequality a³b²c < 2N should translate to 3α+2β+γ < log N + log 2, but Eq. (20) writes 3α+2β+3γ. Similarly, Eq. (21) writes "2α+4β+c" where the last term should be γ. These coefficient errors make it impossible to verify the subsequent linear combinations, and all combinations in this proposition need to be redone and checked.
- [Proposition 20, final paragraph] The final case asserts "b²cσ(c) < N" without proof. From b²||N and c||N one only immediately has b²c < N, and σ(c)=c+1 introduces an extra factor, so the assertion is not immediate and may depend on the remaining components of N. The line "a² 1/2 b³c ≤ N" is also malformed. This part of the proof requires a full divisibility argument before it can be accepted.
minor comments (3)
- [Corollary 3 and Abstract] Corollary 3 and the abstract state "a < 2N^{1/6}", but Theorem 2 gives a < (2N)^{1/6}. The missing parentheses change the claimed bound and should be corrected throughout.
- [Proposition 20, Eq. (21)] Eq. (21) uses the symbol c for a logarithm in one place: "2α+4β+c ≤ log N" should be "2α+4β+γ ≤ log N".
- [Proposition 17, proof] In the proof of Proposition 17, the sentence "Since we cannot have c² ⁄ |σ(a²)" appears to contain a double negation; it should presumably read "Since we cannot have c²|σ(a²)" or similar. Please clarify.
Circularity Check
No circularity; the derivation uses independent external bounds and self-contained m-type arguments.
full rationale
The derivation chain is not circular. Theorem 2's bound on a is proved directly from the definition of perfect number via an m-type argument: a proper divisor M is deficient, some top prime-power component fails to divide sigma(M), perfectness forces an outside component p_l^{a_l} with p_l^{a_l} > p/2, and then N >= p_l^{a_l} M gives (1/2)p^{2i+2} < N. No target inequality is fed into that proof. The abc results assemble the external, parameter-free bounds of Acquaah-Konyagin (Inequality (1)/(8)) and of the third author's earlier published paper [11] (Inequalities (2)/(13) and (3)/(9)), together with m-type inequalities derived in the text. Although [11] is a self-citation, it is load-bearing but independent: the cited inequalities are stated with assumptions that do not include abc < (2N)^{3/5} or a < 2N^{1/6}, and they were published separately; the paper does not redefine them in terms of its own conclusions. The linear-programming combinations in Lemma 13 and the propositions choose coefficients for sums of already-proved inequalities; they do not fit parameters to the claimed output. There are no fitted inputs renamed as predictions and no uniqueness theorem imported from the authors to forbid alternatives. Per the reviewing rule, I flag two arithmetic defects in Proposition 20 as correctness issues, not circularity: the displayed conversion of a^3 b^2 c < 2N to (20) has a typo (3*gamma instead of gamma), and the stated combination '3/5(13)+(22)' actually yields alpha+beta+gamma <= (3/5)log N + (8/5)log 2, not (3/5)log N + (3/5)log 2. These undermine the proof of Theorem 14 as written but do not make it circular.
Assumptions & free parameters
assumptions (7)
- standard math Euler's theorem: every odd perfect number N has the form p^e m^2 with p a prime, p ≡ e ≡ 1 (mod 4).
- standard math Every proper divisor of a perfect number is deficient, i.e. sigma(M) < 2M.
- domain assumption Acquaah-Konyagin bound: p_k < (3N)^(1/3).
- domain assumption Zelinsky's earlier bounds p_{k-1} < (2N)^(1/5) and p_k p_{k-1} < 6^(1/4) N^(1/2).
- standard math Any sigma(p^a) for an even-exponent component, and any product of selected sigma(p^{a_i}) and p_i powers, divides 2N, with odd parts dividing N.
- domain assumption An odd perfect number has at least five distinct prime divisors.
- domain assumption Iannucci's lower bound p_{k-2} > 100.
Cite this review
Pith. "Pith review of On the third largest prime divisor of an odd perfect number." pith.science (2026). https://pith.science/paper/G5NXEVDL
@misc{pith2026190809420,
author = {Pith},
title = {Pith review of: On the third largest prime divisor of an odd perfect number},
year = {2026},
howpublished = {\url{https://pith.science/paper/G5NXEVDL}},
note = {Machine review of arXiv:1908.09420}
}
abstract
Let $N$ be an odd perfect number and let $a$ be its third largest prime divisor, $b$ be the second largest prime divisor, and $c$ be its largest prime divisor. We discuss steps towards obtaining a non-trivial upper bound on $a$, as well as the closely related problem of improving bounds $bc$, and $abc$. In particular, we prove two results. First we prove a new general bound on any prime divisor of an odd perfect number and obtain as a corollary of that bound that $$a < 2N^{\frac{1}{6}}.$$ Second, we show that $$abc < (2N)^{\frac{3}{5}}.$$ We also show how in certain circumstances these bounds and related inequalities can be tightened. Define a $\sigma_{m,n}$ pair to be a pair primes $p$ and $q$ where $q|\sigma(p^m)$, and $p|\sigma(q^n)$. Many of our results revolve around understanding $\sigma_{2,2}$ pairs. We also prove results concerning $\sigma_{m,n}$ pairs for other values of $m$ and $n$.
Forward citations
Cited by 1 Pith paper
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Certified Minimal-Prime Branch Closures for Odd Perfect Numbers
Relative to frozen certificate release C-small-2026-07, no odd perfect number has minimal prime divisor in {5,7,11,13,17}.
Reference graph
Works this paper leans on
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Reviewed August 14, 2026 · model on record in the stance chip above.
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