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Subject and independence

Role. Independent reviewer in a fresh context, commissioned for refutation with only the assignment text; the reviewer took no part in writing the page, had no contact with its author, and read no other review of it.

Subject. wiki/research/erdos_49/theorem_3_1_reconstruction.md as it stood on 2026-09-28T05:03:27Z, read whole.

Artifact. The 17-page author manuscript pollack_et_al_2013_sets_monotonicity_euler_totient_function.pdf under the primes card, compared byte for byte with the 17-page PDF under the arithmetic_functions card (identical; both cards' provenance paragraphs then gave the same SHA-256). Physical pages read: 5 (Theorem A, the definitions of $P(x;k)$, $P_0(x;k)$ and $P_1(x;k)$) and 6 (Theorem C, Theorem 3.1 and its proof sketch), each in full both as extracted text and as a page image rendered at 150 dpi; physical pages 16 and 17 (references [6] and [10]) as extracted text only. Physical page 7 was rendered but not read. The physical page numbers equal the printed page numbers.

Allowed material read. The Definitions and Statement sections of wiki/research/erdos_49/theorem_3_3_reconstruction.md in the same state (the Definitions include the corpus's verification of Theorem A; read because step 1 of the sketch turns on Theorem A's form); the provenance paragraph of each of the two card indexes; the Source paragraph and the statement of the arithmetic_functions card's theorem_3_1.md; the Statement paragraph of wiki/problems/primes/E0049/_index.md; in docs/verification.md the sections "Audit checklist — the canonical failure modes", "Whole-claim report" and "Audit checklist"; the "Source fidelity" section of docs/evidence.md; docs/math_authoring.md whole. The existence, not the content, of the page's link targets under the Graham--Holt--Pomerance card and the Erdős--Pomerance--Sárközy card was checked by listing the tree in that state; neither card was read. The third card folder named in the assignment (a 2024 source on monotone totient sequences) does not exist in the worktree.

Exposures. While locating the permitted sections by printing heading and bold-label lines, the first line of three excluded paragraphs was seen: the Status paragraph of E0049 ("Open for the remaining clause ...") and the Living verification paragraphs of the arithmetic_functions card index and of its theorem_3_1.md ("Needs review ..."). Those fragments were not used. No other excluded content, no workspace content, no other review and no web search reached this review.

Restatement

Conventions: natural numbers start at $1$; $\log$ is the natural logarithm; for $n\ge2$, $P^+(n)$ is the largest prime factor of $n$; $\gamma(j)$ is the product of the distinct primes dividing $j$. For real $x$ and a natural number $k$, $P(x;k)$ is the number of natural numbers $n\le x$ with $\varphi(n)=\varphi(n+k)$; $P_0(x;k)$ is the number of those $n$ that can be written as $n=j\bigl(\frac{j+k}{g}r+1\bigr)$ for some natural numbers $j$ and $r$ with $\gamma(j)=\gamma(j+k)$, $g=\gcd(j,j+k)$, and $\frac{j}{g}r+1$, $\frac{j+k}{g}r+1$ both prime and not dividing $j$ (Theorem A's shape); $P_1(x;k)=P(x;k)-P_0(x;k)$.

The theorem (source Theorem 3.1, physical p. 6): there is a real $x_0$, not depending on $k$, such that for every real $x>x_0$ and every natural number $k$ with $k\le\exp((\log x)^{1/3})$,

$$P_1(x;k)<\frac{x}{\exp((\log x)^{1/3})}.$$

Theorem C (source p. 6, quoting [10, Theorem 2]) is the same inequality for one fixed $k$ and $x>x_0(k)$. The page records the source's sketch and reconstructs exactly one deduction of it: with $l=\exp((\log x)^{1/3})$, for a solution $n=mp$, $p=P^+(n)$, and a prime $q'$ arising at [6, eq. (4.4)] with $q'\equiv1\pmod r$ for some modulus $r\ge l^4$, one has $q'\nmid m$, so $mp+k\equiv0\pmod{q'}$ holds for exactly one residue class of $p$ modulo $q'$.

Checklist

• Quantifiers and scope. The Statement is faithful clause for clause: the source's unadorned $x_0$ together with "uniformly" is the page's "absolute $x_0$", and "natural numbers $k\le\exp((\log x)^{1/3})$" is verbatim. One boundary case is dropped in the sketch: step 1's consequence sentence quantifies over all solutions counted by $P_1(x;k)$ and fails for those with $p\mid m$ (F1). The solution $n=1$ of $\varphi(n)=\varphi(n+1)$ has no largest prime factor and is silently outside the decomposition $n=mp$ (F3, harmless).
• Circularity. None. The sketch reduces to [10] and to the argument of [6], neither of which is Theorem 3.1, and the reconstructed deduction is elementary divisibility.
• Model and convention changes. None. The page works with the source's objects; $P^+$ is the page's notation for the source's "largest prime factor", and $l$ is the source's own abbreviation.
• Finite and statistical overreach. Inapplicable: no finite check or heuristic is presented as proof. The reviewer's own small enumeration served only to find witnesses for F1 and proves nothing on the page.
• Uniformity. The only place where the range of $k$ enters the written sketch is the inequality $q'>l^4\ge l\ge k$, which uses $k\le l$ and nothing else, and the page says so; the dependence of $x_0$ on nothing is what the source claims. The unwritten "obvious minor changes" to [6] carry their own uniformity in $k$, which neither the source nor the page exhibits; the page's Gaps paragraph admits this (F2 asks for one qualifying word).
• Extremal conclusions. Inapplicable: no infimum, supremum, attained value or sharpness is claimed.
• Consequences and composition. The "So for the solutions counted by $P_1(x;k)$" sentence of step 1 is refuted by the witness $n=4$, $k=2$ (F1). Every "hence" of the written deduction is verified below. The composition inherits the unproved premises of [10] and [6]; the page states this in its Standing and Gaps paragraphs.
• Computation. Inapplicable: the page contains no computation.
• Reproduction. Inapplicable: the page states no rerun command and no coverage claim.
• Source and verdict fidelity. Locators verified: Theorem C, Theorem 3.1 and the sketch are on physical p. 6, Theorem A and the definitions on p. 5; [10] is Graham, Holt and Pomerance (1999), Theorem 2 of it is Theorem C, and [6] is the 1987 Acta Math. Hungar. paper the page names; the quoted phrase "goes through with obvious minor changes" is exact; the source labels its argument "Proof (sketch)". The Standing sentence claims only an author-recorded record and no tier. One strengthening: the source's "So we can assume that" became the page's universal "So for the solutions counted by $P_1(x;k)$" (F1). The page's description of the Graham--Holt--Pomerance card page ("records the statement and, likewise, only a proof pointer") could not be checked inside the read set.

Weakest steps

1. Step 1, the reduction to $\varphi(m)/m\ne\varphi(m')/m'$. Own derivation, under the two hypotheses $p\nmid m$ and $p'\nmid m'$ that the page and the source leave unstated. Then $\varphi(n)=\varphi(m)(p-1)$ and $\varphi(n+k)=\varphi(m')(p'-1)$. Put $\theta=\varphi(m)/m=\varphi(m')/m'>0$. From $\varphi(n)=\varphi(n+k)$, $m(p-1)=m'(p'-1)$, that is $n-m=(n+k)-m'$, so $m'=m+k$. Equal ratios force equal prime sets: let $S$, $T$ be the sets of primes dividing $m$, $m'$; cancel the common primes in $\prod_{q\in S}(1-1/q)=\prod_{q\in T}(1-1/q)$ and clear denominators to get $\prod_{q\in S\setminus T}(q-1)\prod_{q\in T\setminus S}q=\prod_{q\in T\setminus S}(q-1)\prod_{q\in S\setminus T}q$; if $S\ne T$, the largest prime $Q$ of the symmetric difference, say $Q\in S\setminus T$, divides the right side but not the left, since $Q\notin T$ and $0<q-1<q\le Q$ for $q\in S\setminus T$. So $\gamma(m)=\gamma(m+k)$. Put $j=m$, $g=\gcd(j,j+k)$; then $\frac{j}{g}(p-1)=\frac{j+k}{g}(p'-1)$ with coprime cofactors gives $p-1=\frac{j+k}{g}r$ and $p'-1=\frac{j}{g}r$ for one natural number $r$; $p\nmid j$ by hypothesis, and $p'\nmid j+k=m'$ by hypothesis, hence $p'\nmid j$ because $j$ and $j+k$ have the same primes. So $n=mp=j\bigl(\frac{j+k}{g}r+1\bigr)$ has Theorem A's shape. Without the hypotheses the deduction fails: if $p\mid m$ and $p'\nmid m'$ then $\varphi(n)=p\,\varphi(m)$ and equal ratios give $mp=m'(p'-1)$, that is $m'=k$, which occurs ($n=4$, $k=2$); if $p'\mid m'$ and $p\nmid m$ they give $m=-k$, impossible; if both divide they give $n=n+k$, impossible. So the solutions that escape the reduction are exactly those with $P^+(n)^2\mid n$, and for them $n+k=k\,P^+(n+k)$. Composition: the sketch must count these solutions separately; neither the source's sketch nor the page says how. The gap is closable by a standard argument the page could supply and label: the number of $n\le x$ with $P^+(n)^2\mid n$ is at most $\Psi(x,y)+\sum_{q>y}x/q^2\le\Psi(x,y)+x/y$, and with $y=\exp((\log x)^{1/2})$ the elementary bound $\Psi(x,y)\ll x\,e^{-\log x/\log y}(\log y)^{O(1)}$ (weight each $y$-smooth $n\le x$ by $(x/n)^{\sigma}$ with $\sigma=1-1/\log y$) makes both terms $o(x/\exp((\log x)^{1/3}))$, independently of $k$. That is the reviewer's remark, not the source's text.

2. The written deduction. Own derivation. $r$ is a modulus, so a natural number, and $r\ge l^4\ge1$ because $l=\exp((\log x)^{1/3})\ge1$ for $x\ge1$. Since $q'$ is prime, $q'\ge2$, so $q'-1$ is a positive multiple of $r$ and $q'\ge r+1>r\ge l^4\ge l\ge k\ge1$; the last two inequalities are $l\ge1$ and the hypothesis $k\le l$. Hence $q'\nmid k$. If $q'\mid m$, then $q'\mid mp$, and with $q'\mid mp+k$ this gives $q'\mid k$, a contradiction; so $q'\nmid m$, $\gcd(m,q')=1$, $m$ is a unit modulo the prime $q'$, and $mp+k\equiv0\pmod{q'}$ is equivalent to $p\equiv-km^{-1}\pmod{q'}$, one residue class. The page's chain is exactly this and is sound; it is slightly weaker than the source's "$q'>l^4>k$" (which needs $l>1$) and needs only $l\ge1$. Composition: this is the one place where $k\le l$ is used in the written sketch; the rest of the sketch is a pointer to [6], as the page says. A side remark: even if $q'\mid m$ held, the congruence would have no solution $p$ at all, so for an upper bound the case split is not needed; the page follows the source's route, which is correct as it stands.

3. The uniform reading of $x_0$. The source writes "For $x>x_0$" without an argument and adds "uniformly for natural numbers $k\le\exp((\log x)^{1/3})$", where Theorem C two lines above writes $x_0(k)$. The page's "There is an absolute $x_0$" is the only reading under which "uniformly" has content, and the linked card page reads it the same way. No defect; the word "absolute" is a reading, not a quotation, and nothing on the page presents it otherwise.

Strongest attack

The strongest attack was aimed at the one sentence of the sketch that the page states as a universal fact about a defined counting function: "So for the solutions counted by $P_1(x;k)$, $\varphi(m)/m\ne\varphi(m')/m'$." It succeeded as a textual defect. Witness: $k=2$, $n=4$. Then $\varphi(4)=2=\varphi(6)$, so $n$ is a solution; $p=P^+(4)=2$, $m=2$; $p'=P^+(6)=3$, $m'=2$; and $\varphi(m)/m=\varphi(m')/m'=1/2$. For $k=2$ the only $j$ with $\gamma(j)=\gamma(j+2)$ is $j=2$ (an odd $j$ is coprime to $j+2$, and for $j=2a$ the coprime numbers $a$ and $a+1$ would both have to be powers of two), so every solution of Theorem A's shape is $2(2r+1)\equiv2\pmod4$, and $n=4$ is not one. Hence $n=4$ is counted by $P_1(x;2)$ for every $x\ge4$ and violates the sentence. The same holds for $n=8$ and $n=32$ with $k=2$, and for $n=36$ with $k=6$. This does not refute Theorem 3.1: the escaping solutions all have $P^+(n)^2\mid n$, a set whose size is $o(x/\exp((\log x)^{1/3}))$ uniformly in $k$ (weakest step 1); it refutes the sentence as written and shows that the source's "we can assume" is a sketch's hedge that the page hardened into a claim.

Against the reconstructed deduction the attacks were: take $r$ not an integer (excluded, it is a modulus), take $l<1$ (excluded, $x>x_0\ge1$), take $k=0$ (excluded, $k$ is a natural number), and take the boundary $q'=r+1$ (still $q'>l^4\ge k$). All failed. Against the Statement, each clause was compared with physical p. 6; nothing is added beyond the word "absolute", which the source's "uniformly" entails.

Premises

• Definitions of $P$, $P_0$, $P_1$ and Theorem A. Source physical p. 5, read as text and image; the Theorem 3.3 reconstruction page's Definitions in that state, read whole, agree with the source (Theorem A is [10, Theorem 1], quoted by the source; the corpus's own verification of it on that page was read but is not needed here).
• Theorem C. Interface: for each fixed natural number $k$ there is $x_0(k)$ with $P_1(x;k)<x/\exp((\log x)^{1/3})$ for $x>x_0(k)$. Source p. 6, quoting [10, Theorem 2]; the page's version matches. The paper [10] is held on a card in the repository, outside this read set; not read.
• The reduction "as in [10]". Interface as actually needed: if $p\nmid m$, $p'\nmid m'$ and $\varphi(m)/m=\varphi(m')/m'$, then $n$ has Theorem A's shape with $j=m$. Source p. 6 states it without the first two hypotheses; the page copies that. Re-derived above; [10] not read.
• The argument of [6] and its uniform extension. Interface: the count of the solutions with unequal ratios follows the argument of [6] for $k=1$ up to [6, eq. (4.4)] with "obvious minor changes", and the prime $q'$ there satisfies $q'\equiv1\pmod r$ with $r\ge l^4$. Asserted by the source p. 6; the page labels the property as asserted by the source and not visible from it, and labels the whole as imported and not read. The paper [6] is held on a card in the repository, outside this read set; not read. Standing on the page: imported, unreconstructed.
• Explicit assumptions. $k\ge1$; $x>x_0\ge1$ so that $l\ge1$; $n\ge2$ so that $P^+(n)$ exists (see F3).

Findings

F1. Severity: required. Location: step 1 of "The source's sketch", "So for the solutions counted by $P_1(x;k)$, $\varphi(m)/m\ne\varphi(m')/m'$." Defect: the reduction that precedes it needs the hypotheses $p\nmid m$ and $p'\nmid m'$, which neither the source nor the page states, and the source's hedge "So we can assume that" is hardened into a universal statement about every solution counted by $P_1(x;k)$, which is false. Witness (source p. 6 for the sentence, p. 5 for the definitions): $k=2$, $n=4$, with $\varphi(4)=\varphi(6)=2$, $m=m'=2$, $p=2\mid m$, equal ratios $1/2$, and $n=4$ not of Theorem A's shape since for $k=2$ that shape is $2(2r+1)$. Proposed replacement for the two sentences of step 1:

Reduction (imported from Graham, Holt and Pomerance, with two hypotheses supplied here): if $p\nmid m$, $p'\nmid m'$ and $\varphi(m)/m=\varphi(m')/m'$, then $n$ has the shape of Theorem A with $j=m$. The source states this without the two hypotheses and then says "we can assume" that the ratios differ; the solutions with $p\mid m$ (for instance $n=4$, $k=2$, where $\varphi(4)=\varphi(6)=2$ and $m=m'=2$) satisfy the equality without having Theorem A's shape, are counted by $P_1(x;k)$, and must be disposed of separately; the source's sketch does not say how, and this page does not close that gap. For the remaining solutions counted by $P_1(x;k)$, $\varphi(m)/m\ne\varphi(m')/m'$.

F2. Severity: note. Location: end of "The written deduction", "This is the only place where the size of $k$ enters the source's sketch". Defect: true of the written text, but a reader may take it as a statement about the proof; the unwritten "obvious minor changes" to [6] may also depend on $k$. Witness: source p. 6, the sentence "The argument of [6] goes through with obvious minor changes until [6, eq. (4.4)]". Proposed replacement: "This is the only place where the size of $k$ enters the written sketch; whether the unwritten changes to the argument of [6] use the range of $k$ is not visible from the source."

F3. Severity: note. Location: Definitions, "$P^+(n)$ denotes the largest prime factor of $n>1$", and the sketch's "Write $n=mp$". Defect: for $k=1$ the number $n=1$ solves $\varphi(n)=\varphi(n+1)$ ($\varphi(1)=\varphi(2)=1$), is counted by $P_1(x;1)$ for $x\ge1$ (Theorem A needs $k$ even), and has no largest prime factor; the source and the page pass over it in silence. It contributes at most $1$ and is harmless. Witness: source p. 5, the definition of $P(x;k)$ and Theorem A's "(so that $k$ is even)". Proposed addition after "Write $n=mp$ ...": "(for $n\ge2$; the single solution $n=1$, $k=1$ is ignored by the sketch and changes the count by at most one)".

F4. Severity: note. Location: Definitions, "For odd $k$ no $j$ has $\gamma(j)=\gamma(j+k)$, so $P_0(x;k)=0$ and $P_1(x;k)=P(x;k)$." Defect: a correct supplied remark, not marked as supplied and not used by the sketch; the first clause is the source's parenthetical "(so that $k$ is even)" in Theorem A (p. 5), the rest is the page's. Proposed replacement: prefix the sentence with "(Supplied, not used below.)" or move it to the page that uses it.

Verdict

Source fidelity: faithful with corrections. The Statement, Theorem C, the locators, the reference identities and the quoted phrase all match the artifact; one sentence of the sketch (F1) strengthens the source's "we can assume" into a false universal and omits two hypotheses, and needs the correction above.

The argument as reconstructed: the one deduction the page reconstructs (that $q'\nmid m$ and that $mp+k\equiv0\pmod{q'}$ fixes $p$ modulo $q'$) is sound and uses the range $k\le\exp((\log x)^{1/3})$ exactly where the page says. The sketch as presented is defective at step 1's consequence sentence, whose exceptional solutions ($P^+(n)^2\mid n$) are not disposed of on the page or in the source's sketch; the remaining steps are an unreconstructed proof pointer to [10] and [6], as the page's Standing and Gaps paragraphs state.

Limitations: the papers [10] and [6] were not read, so the imported reduction, Theorem C, and the property $q'\equiv1\pmod r$ with $r\ge l^4$ were not checked against their sources; the whole theorem was not verified, and no statement about its truth is made here beyond the witnessed defect and the verified deduction. The reviewer's remark that the escaping solutions form a set of size $o(x/\exp((\log x)^{1/3}))$ is offered as a route to close the gap, not as part of the page's record.

This focused review assigns no tier and changes no status.