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Record, attribution and exact subject

Grader, Claude Fable 5.1, distinct from the author of the reconstruction and from the reviewer, separately spawned with only this assignment; dated 2026-09-17. Grade: PASS for the report contract and for independence, with one required bookkeeping change to the review record (the landing commit, section 3). The graded report is the reconstruction review. The grader read the frozen subject directly: the byte-identical copies under evidence/assets/reviewed_pages/ of this card (lemme.md, theoreme.md, duverney_1993_theoreme_2.md), compared with cmp against the working pages lemme.md and theoreme.md of this card and theoreme_2.md of the 1993 card before any page was edited (identical); the two folder-name PDFs, rendered with pdftoppm -r 200 -png and read as page images (Duverney 1995 printed pp. 1287--1289 in full; Duverney 1993 printed pp. 175, 176, 178, 179), whose sizes and SHA-256 values agreed with the provenance lines the two cards then carried (each PDF was identified by its path and byte count); the card's _index.md, the Compiled proof coverage paragraph of Problem 250 and the evidence indexes as consequence sentences. Operating instructions read: the repository instructions, docs/anatomy.md, docs/evidence.md, docs/verification.md, docs/tools.md and docs/math_authoring.md, the commission's operating instructions and the reviewer's own summary of the review, neither retained. Not read: the prior-art dossiers compiled outside this repository (not retained), the note's cited books, Nesterenko's paper, any other review. No computation was needed; the sanity aids the review mentions were not rerun and carry no weight here.

Nothing in this record changes the status of Problem 250, which rests on the refereed publications; the grade decides only whether the reconstruction counts as independently accepted compilation proof coverage in the scope the review states.

1. Criteria

Subject and independence: PASS, one change required. The report names the three pages by repository path and section, the head commit at review time, the byte-identical snapshots as the reviewed bytes and their relation to the current pages, the consequence sentences read, the PDF pages read with their rendering command, the allowed operating reading, the exclusions and the one exposure. What it lacks is the landing commit, left as the placeholder "landing commit to be recorded at filing" for the committer; the snapshots pin the subject meanwhile, as docs/verification.md allows for uncommitted bytes, so the placeholder is a bookkeeping gap, not a subject gap. The report lists no per-file hashes of repository files; the two PDF lines are provenance for bytes the repository does not own.

Independence and exposure: PASS. The reviewer had not authored or built on the pages. The disclosed exposure is the reconstruction author's summary of the pages (not retained), a description of what the pages contain and of the author's own checks. It is not a dossier, a plan paraphrase, a sibling verdict or an argument absent from the pages, and every derivation in the report is traceable to a page step or a printed display, which I confirmed step by step. I therefore rule that independence stands. Recommendation for later commissions: pass reviewers the frozen subject and the contract only, without the author's summary. The frozen copies read whole also carried the pages' own standing wording, evidence/assets/reviewed_pages/lemme.md lines 203--214 and theoreme.md lines 160--174 ("author-recorded; independent review pending", with theoreme.md 171--174 recording the source result as a refereed publication) and duverney_1993_theoreme_2.md lines 78--79 ("No independent review exists here"), and the reviewer read Problem 250 whole with its frontmatter status: proved as it stood at 2026-09-17T07:01:06Z; a separately spawned grader, Claude Fable 5.1, ruled this exposure immaterial by the content test on 2026-09-18: the text states no answer to whether the reconstruction is faithful and complete, and the review's verdict rests on its rederivation of every step against the page images, not on that text.

Restatement: PASS. Both propositions are restated with every quantifier: qq over all integers with ∣q∣≥2|q|\ge2, negative included; the Lemme as the absence of a nontrivial rational relation and, equivalently, the irrationality of a f(1/q)+b (1/q)f′(1/q)a\,f(1/q)+b\,(1/q)f'(1/q) for all integer pairs (a,b)≠(0,0)(a,b)\ne(0,0); the Théorème with its three equal forms and the specialization to q=2q=2 and to every integer base.

Statement fidelity against the source: PASS. I checked on the page images: the Théorème (p. 1287) and the Lemme (p. 1288) as quoted; the displays (1)--(15), including the exponents n(3n±1)/2n(3n\pm1)/2, the signs (−1)n(-1)^n, the constant term aa in (9), the ranges n≥0n\ge0 in (10) and (12), the sign in (14) and the denominators qn−1q^n-1 in (15); the citations of (E) to [2] p. 124 and [6] p. 229, of T2 to "le théorème 2 de [3]", of (5) to [8] p. 257 and of the route to [1]; the two print slips ("si xq=η/δx_q=\eta/\delta" before (12); kk zeros in (11)); and on the 1993 paper the title page identity (Acta Arithmetica LXIV.2, 1993), Théorème 2 on p. 176 with (a), (b), (b1_1), (b2_2), (c), (c1_1), (c2_2) and (4), and the proof in section 2 on p. 178 with the Remarque running onto p. 179. The report's fidelity findings are exact and do not strengthen the source.

Independent rederivation of every essential deduction: PASS. The report rederives Steps 0--7 of the Lemme, Steps 1--4 of the Théorème and the proof of T2, not by paraphrase: it supplies the gap arithmetic behind the zero runs, the threshold n≥∣a∣+∣b∣n\ge|a|+|b| for (b), the bound nk+k+1≤4k2n_k+k+1\le4k^2 for (c2_2), the exact split of the sum in (12), the constant 2∣δ∣(∣a∣+∣b∣)2|\delta|(|a|+|b|) in the growth step, the absolute-convergence bound 4∣q∣−n4|q|^{-n} behind (3) and (5), the two-sided bound on log⁡(1−xn)\log(1-x^n) behind (14) and the geometric sum inside T2's proof. My own rederivation (section 4) agrees at every step.

External premises with reading depth: PASS. (E) is recorded as an external statement, checked against the print and the classical identity, proof not inspected, books not consulted; T2 as statement and proof checked, with the unread sections of the 1993 paper excluded from coverage; the elementary analysis used without citation is itemized. No native L-claim is consumed and the report says so.

Weakest steps and strongest attack: PASS. Three weakest steps are named with the exact quantity each depends on (the gap pk+−pk−=kp_k^+-p_k^-=k, the printed strength of T2, the nonvanishing f(1/q)f(1/q)) and how each composes with its neighbors; the attack section tries six routes against the arithmetic contradiction and the analytic exchanges and reports why each fails. These are the routes I would have chosen; my one further probe is in section 4.

Checklist: PASS. All ten items of the audit checklist carry an explicit verdict with a reason, and the three inapplicable items (extremal conclusions, computation, reproduction of mathematics) say why.

Verdict warranted and scoped: PASS. "Refutation-failed" is written in full and follows from the evidence; the limitations exclude (E)'s proof, Nesterenko's theorem, the Bundschuh--Väänänen route, the card metadata, any other page, any status change and any native tier, and the report leaves the grade to a distinct grader. The consequence sentences it proposes for the pages match the verdict's scope.

Provenance (revision, paths, date, role, model): PASS after the change above. Reviewer named by role and model, dated, paths and sections named, no person, seat, session or tool harness named, no time cutoff, American spelling.

2. Overall grade

PASS. The review identifies its frozen subject, restates both propositions exactly, checks every statement and display against the page images, rederives every essential deduction of both proofs and of the external criterion's proof, records both external premises at their true reading depth, attacks the argument along the routes that could break it, gives every checklist item a verdict and scopes its verdict correctly. With this grade filed, the complete rewritten proofs on the Lemme and Théorème pages are independently accepted compilation proof coverage relative to Euler's pentagonal number theorem, consumed as an unproved external statement, and to Théorème 2 of Duverney 1993, whose statement and proof were checked. Nothing is established about (E) beyond its identity with the classical theorem, nothing about Nesterenko's transcendence proof, and no problem status or native tier changes.

3. Required change

Enter the landing commit in the review record's Revision paragraph in place of "landing commit to be recorded at filing", once the pages are committed, so that git diff <landing commit> HEAD -- <paths> separates later edits from the reviewed text as the paragraph promises. No other change is required. The reviewer's four presentation recommendations are optional; they were not applied at grading, so the mathematical text of the three pages is still the reviewed text. If they are applied later, each page's standing line must say that the current text differs from the reviewed copy under evidence/assets/reviewed_pages/.

4. Independent check of the mathematical core

Let q∈Zq\in\mathbb Z, ∣q∣≥2|q|\ge2, x=1/q∈(−1,1)∖{0}x=1/q\in(-1,1)\setminus\{0\}, and f(x)=∏n≥1(1−xn)f(x)=\prod_{n\ge1}(1-x^n).

Lemme. By (E), f(x)=1+∑m≥1(−1)m(xpm++xpm−)f(x)=1+\sum_{m\ge1}(-1)^m(x^{p_m^+}+x^{p_m^-}) with pm±=m(3m±1)/2p_m^\pm=m(3m\pm1)/2; the series has coefficients in {0,±1}\{0,\pm1\}, so it converges on (−1,1)(-1,1) and may be differentiated termwise, giving xf′(x)=∑m(−1)m(pm+xpm++pm−xpm−)xf'(x)=\sum_m(-1)^m(p_m^+x^{p_m^+}+p_m^-x^{p_m^-}). Hence for integers (a,b)≠(0,0)(a,b)\ne(0,0), $\alpha_q=af(1/q)+b(1/q)f'(1/q)=a+\sum_m(-1)^m\big((a+bp_m^+)q^{-p_m^+} +(a+bp_m^-)q^{-p_m^-}\big)$, absolutely convergent since the coefficients are O(m2)O(m^2) and ∣q∣−pm−≤2−m|q|^{-p_m^-}\le2^{-m}. The exponents are 1,2,5,7,12,15,22,26,…1,2,5,7,12,15,22,26,\dots: pm+−pm−=mp_m^+-p_m^-=m and pm+1−−pm+=((3m2+5m+2)−(3m2+m))/2=2m+1p_{m+1}^--p_m^+=\big((3m^2+5m+2)-(3m^2+m)\big)/2=2m+1, so they increase strictly and αq=∑n≥0a(n)q−n\alpha_q=\sum_{n\ge0}a(n)q^{-n} with a(0)=aa(0)=a, a(pm±)=(−1)m(a+bpm±)a(p_m^\pm)=(-1)^m(a+bp_m^\pm) and a(n)=0a(n)=0 otherwise. Put nk=pk+n_k=p_k^+. Then a(nk+j)=0a(n_k+j)=0 for 1≤j≤2k1\le j\le2k and a(nk−j)=0a(n_k-j)=0 for 1≤j≤k−11\le j\le k-1, while a(nk−k)=a(pk−)a(n_k-k)=a(p_k^-) is in general nonzero; for k=1k=1 the second run is empty and n1−1=1=p1−n_1-1=1=p_1^-.

Hypotheses of T2 with r(n)=n2r(n)=n^2: (a) a(nk)=(−1)k(a+bnk)a(n_k)=(-1)^k(a+bn_k) vanishes for at most one kk if b≠0b\ne0 and never if b=0b=0; (b) for n≥c=∣a∣+∣b∣≥1n\ge c=|a|+|b|\ge1, a nonzero a(n)a(n) has $|a(n)|=|a+bn|\le(|a|+|b|)n \le n^2$, and n=0n=0 lies below the threshold; (b1_1) n2>0n^2>0 for n≥1n\ge1, and (b) is stated for nn large, so r(0)=0r(0)=0 is outside its scope; (b2_2) (1+1/n)2→1<∣q∣(1+1/n)^2\to1<|q|; (c) every k≥1k\ge1 with (c1_1) the first run and (c2_2) ((3k2+3k+2)/2)2/∣q∣k→0\big((3k^2+3k+2)/2\big)^2/|q|^k\to0. The series ∑a(n)q−n\sum a(n)q^{-n} that T2 requires converges by (b) and (b2_2). T2 gives k0k_0 with ηqnk=δ∑n=0nka(n)qnk−n\eta q^{n_k}=\delta\sum_{n=0}^{n_k}a(n)q^{n_k-n} for k≥k0k\ge k_0 when αq=η/δ\alpha_q=\eta/\delta. Isolating n=nkn=n_k: the indices nk−k<n<nkn_k-k<n<n_k contribute 00; the indices n≤nk−kn\le n_k-k carry qnk−nq^{n_k-n} with nk−n≥kn_k-n\ge k; and qnkq^{n_k} with nk≥kn_k\ge k. So qk∣δa(nk)q^k\mid\delta a(n_k) in Z\mathbb Z. But $|\delta a(n_k)|\le|\delta|(|a|+|b|)n_k\le 2|\delta|(|a|+|b|)k^2<2^k\le|q|^k$ for large kk, so δa(nk)=0\delta a(n_k)=0, that is a+bnk=0a+bn_k=0, for all large kk; two values of kk force b=0b=0 and then a=0a=0. Contradiction. The sign of qq never enters. Any nontrivial rational relation c0+c1f(1/q)+c2(1/q)f′(1/q)=0c_0+c_1f(1/q)+c_2(1/q)f'(1/q)=0 clears to integers with (c1,c2)≠(0,0)(c_1,c_2)\ne(0,0) and makes αq\alpha_q with (a,b)=(c1,c2)(a,b)=(c_1,c_2) rational, so the Lemme follows.

T2 itself (p. 178). If βx=α\beta x=\alpha then $\alpha q^{n_k}-\beta\sum_{n\le n_k}a(n)q^{n_k-n}=\beta q^{n_k} \sum_{n\ge n_k+k+1}a(n)q^{-n}$ by (c1_1). Choose η<∣q∣\eta<|q| with r(n+1)/r(n)≤ηr(n+1)/r(n)\le\eta for n≥Nn\ge N; for k≥Nk\ge N every index n≥nk+k+1≥k+1n\ge n_k+k+1\ge k+1 passes the thresholds of (b) and of the ratio bound, whether or not nk→∞n_k\to\infty, so ∣a(n)∣≤r(nk+k+1)ηn−(nk+k+1)|a(n)|\le r(n_k+k+1)\eta^{n-(n_k+k+1)} and the geometric sum bounds the right side by ∣β∣ r(nk+k+1)/(∣q∣k(∣q∣−η))|\beta|\,r(n_k+k+1)/\big(|q|^k(|q|-\eta)\big), which tends to 00 by (c2_2). An integer of absolute value less than 11 is 00.

Théorème. For t=q−nt=q^{-n}, 0<∣t∣≤1/20<|t|\le1/2: qn((q−1)/(qn−1))2=(q−1)2t/(1−t)2=(q−1)2∑j≥1jtjq^n\big((q-1)/(q^n-1)\big)^2=(q-1)^2t/(1-t)^2=(q-1)^2\sum_{j\ge1}jt^j, and ∑jj∣t∣j=∣t∣/(1−∣t∣)2≤4∣q∣−n\sum_j j|t|^j=|t|/(1-|t|)^2\le4|q|^{-n}, so the double series converges absolutely. Summing over nn first, ∑nq−nj=1/(qj−1)\sum_nq^{-nj}=1/(q^j-1), gives (3); grouping by m=njm=nj with ∑j∣mj=σ(m)\sum_{j\mid m}j=\sigma(m) gives (5). On [−ρ,ρ][-\rho,\rho], 0<ρ<10<\rho<1: 1−xn∈[1−ρn,1+ρn]1-x^n\in[1-\rho^n,1+\rho^n], so ∣log⁡(1−xn)∣≤−log⁡(1−ρn)≤ρn/(1−ρ)|\log(1-x^n)|\le-\log(1-\rho^n)\le\rho^n/(1-\rho) (using log⁡(1+t)≤t≤−log⁡(1−t)≤t/(1−t)\log(1+t)\le t\le-\log(1-t)\le t/(1-t) on [0,1)[0,1)), and ∣(log⁡(1−xn))′∣=n∣x∣n−1/(1−xn)≤nρn−1/(1−ρ)|(\log(1-x^n))'|=n|x|^{n-1}/(1-x^n)\le n\rho^{n-1}/(1-\rho); both bounds are summable, so g=∑log⁡(1−xn)g=\sum\log(1-x^n) is differentiable with termwise derivative, f=eg>0f=e^g>0 on (−1,1)(-1,1), and xf′/f=xg′=−∑nxn/(1−xn)xf'/f=xg'=-\sum nx^n/(1-x^n), which is (14). This ff is the function (E) equates with the series, so its derivative is the Lemme's f′f'. At x=1/qx=1/q, dividing by f(1/q)>0f(1/q)>0, (1/q)f′(1/q)/f(1/q)=−∑n/(qn−1)=−Dq(1/q)f'(1/q)/f(1/q)=-\sum n/(q^n-1)=-D_q, and ζ(q;2)=(q−1)2Dq\zeta(q;2)=(q-1)^2D_q with (q−1)2(q-1)^2 a nonzero integer. If ζ(q;2)∈Q\zeta(q;2)\in\mathbb Q then Dq∈QD_q\in\mathbb Q and Dq⋅f(1/q)+1⋅(1/q)f′(1/q)=0D_q\cdot f(1/q)+1\cdot(1/q)f'(1/q)=0 is a rational relation with a nonzero coefficient among 11, f(1/q)f(1/q), (1/q)f′(1/q)(1/q)f'(1/q), against the Lemme; so ζ(q;2)\zeta(q;2) and ∑σ(n)/qn=ζ(q;2)/(q−1)2\sum\sigma(n)/q^n=\zeta(q;2)/(q-1)^2 are irrational, and at q=2q=2 the factor is 11. The Théorème uses only the independence of f(1/q)f(1/q) and (1/q)f′(1/q)(1/q)f'(1/q) over Q\mathbb Q together with 11; the Lemme supplies it.

Further probe. I tried to weaken the divisibility by taking b=0b=0, when a(n)=±aa(n)=\pm a on every pentagonal index and the zero runs are the only structure: qk∣δaq^k\mid\delta a for all large kk then forces δa=0\delta a=0 directly, so the case is if anything easier, and the argument is uniform in (a,b)(a,b). I found no step that depends on anything beyond (E), T2 and the elementary analysis the pages list.