Advancing Mathematics Research with AI-Driven Formal Proof Search

• p. 7 Table 1, row 741 (ii)

  Existence of a basis A of order 2 with A1 + A1, A2 + A2 having bounded gaps for all A1, A2 such that A = A1 ⊔ A2

• p. 37 Erdős #741-(ii), Question

  then A1 + A1 and A2 + A2 cannot both have bounded gaps?

• p. 7 Table 1, row 741 (ii)

  having bounded gaps for all 𝐴1, 𝐴2 such that 𝐴= 𝐴1 ⊔𝐴2

• p. 37 Erdős #741-(ii) question

  cannot both have bounded gaps

Judge's note: The Table 1 image on p7 and the p37 question and proof (Step 3: 'impossible to partition A so that both sumsets have bounded gaps') were compared. The table states the opposite property ('not both' is missing), so a reader takes away the negation of the proved result.

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• p. 10 Section 5, second paragraph

  they have K subagents by grouping the attempts into chunks of size 100/K

• p. 9 Figure 3 caption

  Numeric annotations denote the number of independent attempts K ∈ {1, 3, 6, 10} grouped together

• p. 10 Section 5

  grouping the attempts into chunks of size 100/𝐾

• p. 9 Fig. 3 caption

  𝐾∈{1, 3, 6, 10}

Judge's note: Read the p10 text and the p9 figure. Under the text's rule, K=1 gives one chunk of 100 attempts, so the solve rate could only be 0 or 1. Figure 3 instead shows fractional K=1 rates (e.g. about 0.34 for 12(i), about 0.16 for 152), and cost rises with K. So the figure follows the caption's definition (K attempts per chunk) and the methods sentence is inverted. Separately, 100/3 and 100/6 are not integers, so the chunking as described cannot be carried out exactly and the remainder rule is missing.

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• p. 32 Erdős #12-(ii), Theorem (ii) statement

  yet for any 𝜀> 0 and all sufficiently large 𝑁

• p. 32 Proof of Theorem (ii), first line

  Let 𝑐∈(0, 1), and let (𝑃𝑘) be a sequence of pairwise coprime integers

• p. 33 Proof of Theorem (ii), choice of m

  Choose an integer 𝑚≥2 satisfying 𝑚> (2𝑚+1)1−𝑐/2

Only the authors can say what is right here.

Judge's note: I read pp.32–34. The m in the proof depends on c, and the blocks A_k depend on m, so the proof shows ∀c ∃A, not ∃A ∀ε. My own check that a single constructed A fails for small ε: with m fixed, |A_j| ≤ m^{V_j−1} and V_j increases, so for N = max(A_k) < min(A_{k+1}) we get |A∩[1,N]| ≤ (k+1)m^{V_k−1}. Also N ≥ m(2m+1)^{V_k−1} ≥ (2m+1)^{V_k−1}. The ratio to N^{1−ε} is then at most (k+1)(m/(2m+1)^{1−ε})^{V_k−1}, which tends to 0 whenever (2m+1)^{1−ε} > m. That holds for all sufficiently small ε > 0, because m < 2m+1. So the printed quantifier order is not established by this proof. The 'Consequently' sentence needs only the weaker ∀c∃A form. Whether the Lean statement uses the stronger or the weaker form cannot be checked from the files.

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• p. 12 Reference [18]

  Paul Erdős and Alice Sárközi. On the divisibility properties of sequences of integers.

• p. 4 Section 3

  posed by Erdős and Sárközy in 1970

• p. 12 Reference [18]

  Paul Erdős and Alice Sárközi

• p. 12 Reference [18]

  Paul Erdős and Alice Sárközi.

Judge's note: Checked on the p12 and p4 images. The surname spelling within the paper is inconsistent, which can be settled from the files alone. Whether the first name should be 'András' rather than 'Alice' rests on outside knowledge (the 1970 PLMS paper's co-author is A. Sárközy), and the verdict does not depend on it.

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• p. 26 B.1 Natural Language Proof Discovery

  Other provers include FullProof [11] and DeepThink [26].

• p. 12 reference [11]

  The motivic class of the space of genus 0 maps to the flag variety.

Only the authors can say what is right here.

Judge's note: Rests on outside knowledge: whether a system called 'FullProof' (or 'FirstProof') exists and which work describes it. On p.26 and in the p.12 reference list, [11] is a research paper in algebraic geometry and is cited a few lines later as a result in that field, not as a prover. No reference entry names 'FullProof'. [20] is titled 'Aletheia tackles firstproof autonomously'. The name and the citation appear mismatched, and only the authors can give the intended reference.

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• p. 38 paragraph after #741-(ii) proof

  The Lean proof for part (ii) is available at https://github.com/google-deepmind/alphaproof-nexus-results/blob/main/APNOutputs/ErdosProblems/erdos_12 .parts.ii.lean

• p. 34 #12 section

  ErdosProblems/erdos_12.par ts.ii.lean

• p. 38 end of Erdős #741-(ii)

  APNOutputs/ErdosProblems/erdos_12 .parts.ii.lean

• p. 38 end of Erdős #741-(ii) section

  erdos_12 .parts.ii.lean

Only the authors can say what is right here.

Judge's note: I checked p.38 and p.34. The #741(ii) link is identical to the #12(ii) link on p.34. The #741(i) link on p.37 is erdos_741.parts.i.lean, and the next sentence on p.38 points to forum thread /741. The pointer is almost certainly a copy slip, probably for erdos_741.parts.ii.lean. The actual repository file name cannot be checked from the files, so this is a query for the authors.

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• p. 43 OEIS Conjectures, first Theorem

  (A conjecture of OEIS A051293, 2002)

• p. 48 paragraph after A228143 proof

  These OEIS problems were added as conjectures in 2022 (https://oeis.org/A0512 93) and 2018

• p. 48 paragraph after OEIS proofs

  These OEIS problems were added as conjectures in 2022

• p. 48 paragraph after the A228143 proof

  These OEIS problems were added as conjectures in 2022 (https://oeis.org/A0512

Only the authors can say what is right here.

Judge's note: Rests on outside knowledge: which year the conjecture was actually added to OEIS A051293. I checked the crops for p.43 and p.48. The theorem header says 2002 and the closing paragraph says 2022 for the same event (the conjecture being added). The A228143 date reads 2018 in both places (p.45 header and p.48). Nothing in the text explains the difference. The paper cannot tell us which year is right, so only the authors can resolve it.

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• p. 7 Table 1, row 12 (i)

  > δ > 0 infinitely often

• p. 4 Section 3, Erdős Problems paragraph

  dition lim infN→∞

• p. 31 Theorem (i)

  lim inf

• p. 4 Section 3, problem #12(i)

  lim inf

Judge's note: Page images p4, p7 and p31 were compared. 'Infinitely often' is strictly weaker than lim inf > 0. The table therefore misdescribes the condition the agent proved. Fix: 'for all sufficiently large N'.

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• p. 7 Table 1, row 741 (i)

  If A + A has upper density, there exists a decomposition

• p. 36 Erdős #741-(i), Question

  Let A ⊆ ℕ be such that A+A has positive upper density.

• p. 7 Table 1, row 741 (i)

  If 𝐴+ 𝐴has upper density, there

• p. 36 Erdős #741-(i) question

  has positive upper density

Judge's note: The p7 table and the p36 question were checked. Every set has an upper density, so without 'positive' the hypothesis is empty. The conclusion in the same table cell says 'positive upper density'. The missing word changes the meaning.

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• p. 7 Table 1, row 26

  upper den- sity of A + k is < 1 − 1/4 for all naturals k

• p. 38 Erdős #26 Theorem

  d(M(A + k)) < 1 − ϵ

• p. 40 paragraph after proof

  make the upper density of A + k small, for all k

Judge's note: Page 38 Theorem: M(S) denotes the set of multiples of S, and the claim is d̄(M(A+k)) < 1−ε with ∑1/A(i)=∞. The p40 proof bounds 'the total upper density of the multiples of A + k'. Table 1 and the closing remark on p40 omit 'multiples of' (and the table omits the divergence condition), so they state a different and much weaker claim.

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• p. 7 Graph theory paragraph below Table 1

  relating it to the maximum number, over all vertices v ∈ G, of independent sets in the neighborhood of v

• p. 59 Written on the Wall, Conjecture 2 definitions

  ℓ(G) := 1/n ∑v α(v) be the average of the local independence numbers

• p. 59 Supplement, Written on the Wall Conjecture 2

  be the average of the local independence numbers

• p. 59 Written on the Wall, Conjecture 2 theorem

  𝐿𝑆(𝐺) ≥2(ℓ(𝐺) −1)

Judge's note: Page 59 defines α(v) as the size of the largest independent set in G[N(v)] and ℓ(G) as their average. The main text differs in both aggregation (maximum, not average) and quantity (number of independent sets, not the independence number). A bound in terms of the maximum would be a different statement.

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• p. 37 Erdős #741-(ii), Step 1

  In the middle of this zone, we leave a single “oasis”

• p. 38 Erdős #741-(ii), Step 1/Step 2, third bullet

  If 𝑛∈𝑍𝑘and 𝑛lies in the upper half, we use the oasis: 𝑛= 𝑥𝑘+ (𝑛−𝑥𝑘).

• p. 38 Step 2

  The difference 𝑛−𝑥𝑘is small enough that it falls safely before the forbidden zone 𝑍𝑘.

Only the authors can say what is right here.

Judge's note: I read pp.37–38. The zone midpoint is about 8.25P_k, not 10P_k. My own counterexamples: k=1, P_1=100, Z_1≈[550,1101], x_1=1000. n=900 is in the upper half, but n−x_1=−100 is not a natural number. k=2: n=100800 is in the upper half of Z_2≈[55000,110002], and n−x_2=800 lies in Z_1 and is not x_1, so 800 ∉ A. The basis claim can be repaired: halving works for n in Z_k up to about 11P_k, and the oasis is needed only near 11P_k..11P_k+k, where n−x_k ≈ P_k..P_k+k is in A. So the case split in Step 2 is faulty as printed but the stated result stands.

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• p. 45 OEIS A051293 proof, bound on E_j

  Substituting this back into the error term yields

Only the authors can say what is right here.

Judge's note: I checked p.45 and reworked the step. E_j = 2^j A(j)/(j^6(j−1)^6). With |A(j)| ≤ 100000 j^5 this gives |E_j| ≤ 100000·2^j/(j(j−1)^6), and that exceeds 100000·2^j/j^7 because (j−1)^6 < j^6. At j=2 it gives 200000 against the printed 3125. So the inference fails as printed. I checked the printed polynomial at j=2: it gives A(2) = −38978, which matches 2P(2)−64P(1)−32 with P(2)=815 and P(1)=634. The printed final bound still holds for small j (j=2: |A|·j/(j−1)^6 = 77956 ≤ 100000; j=3: about 20851), and asymptotically the constant is 4683. The O(2^n/n^7) result and the theorem therefore stand.

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