Planar Point Sets with Many Unit Distances

• p. 1 Section 1, display (1)

  ν(n) ≤n1+C/ log log n. (1)

• p. 4 Final Response from Internal Model, display (1)

  d(G) ≥ℓ−1. (1)

• p. 6 Section 2, display (2)

  s = 1, . . . , m. (2)

• p. 4 Final Response from Internal Model, display (2)

  r(G) ≤d(G) + C0 (2)

Judge's note: Checked pages 1, 4, 5 and 6 on the images: the reproduced block numbers its displays (1)-(11), while the paper proper has (1) on p1 and (2) on p6. The text layer shows the paper's own (3)-(4) and (5)-(7) further on, so numbers 1-7 each label two displays. The block is set off in italics and described as verbatim, so context resolves each reference; this is a minor ambiguity where separate numbering may be intended. The second reviewer note ("repeated log") is not a defect: "log log n" is the iterated logarithm.

Second check: another model agreed, code check

• p. 4 Final Response from Internal Model, display (2) and following line

  with an absolute constant C0

• p. 6 Final Response from Internal Model, after display (11)

  Given arbitrary C0 > 0 and N

• p. 6 Final Response from Internal Model, after (11)

  Given arbitrary C0 > 0 and N, choose j so large that nj ≥ N and C0/ log log nj < δ.

• p. 4 Final Response, eq. (2)

  r(G) ≤d(G) + C0

Judge's note: I checked p.4 (display (2) and the line after it) and p.6 (the text after (11)). The same symbol C0 is declared as the absolute Shafarevich constant, which feeds the choice of ℓ, t, γ, B and δ, and is later reintroduced as an arbitrary constant standing for the C of the conjecture. The meaning can be recovered from context, and the final sentence itself uses C. The passage is announced as verbatim model output (text line 134). This is a notation clash within one argument; the mathematics is not affected.

Second check: another model agreed

• p. 3 Final Response from Internal Model, first paragraph

  First, if (g1, . . . , gk ∈Φ(G)), then the quotient

• p. 3 model's final response, first standard fact

  if (g1, . . . , gk ∈ Φ(G))

Judge's note: Page 3 image prints the parentheses around the whole membership statement. They look like inline-math delimiters rendered literally. The meaning is clear, so this is cosmetic. The block is said to be reproduced verbatim, so the defect may be in the original output.

Second check: none (judge only)

• p. 4 Final Response from Internal Model, display (4)

  rd(Kj) ≤2, rd(F) =: Aℓ.

• p. 12 Proposition 3.8 (P5)

  rd(Kj) ≤2 rd(F)

• p. 14 Step 4

  gives rd(Kj) ≤2 rd(F)

• p. 4 Final Response from Internal Model, display (4)

  rd(Kj) ≤ 2,

Judge's note: I checked the crop and page 4. The comma and the wide gap appear in display (4) on the page image, so this is not an extraction artefact. The sentence before the display says the relative discriminant of adjoining i divides 4. That gives |D_K| ≤ |D_F|^2·4^f, and taking 2f-th roots gives rd(K) ≤ 2 rd(F). (P5) on p.12 states exactly this, and Step 4 on p.14 derives the same thing ('rd(Kj) ≤ 2 rd(F)'), so the stray comma looks like a thin space rendered as punctuation. Read literally, the display states the bound rd(Kj) ≤ 2 and defines Aℓ as rd(F). Line 134 of the text says the block is reproduced verbatim from the model output, so the slip may come from that output. It is still the printed equation. The detailed exposition has the correct form and the result is unaffected, so I rate it as a punctuation typo.

Second check: none (judge only)

• p. 4 Final Response from Internal Model, paragraph before the display |D_M| = …

  ai ∈0, 1, 2

• p. 5 Final Response from Internal Model, definition of the ideals

  For ϵ = (ϵs) ∈0, 1m

• p. 5 Final Response from Internal Model, valuations of u

  are in −2, 0, 2

• p. 5 paragraph before definition of Aϵ

  For ϵ = (ϵs) ∈ 0, 1m put

Judge's note: Pages 4 and 5 images confirm the braces are absent in the PDF in all three places, not lost in extraction. The paper's own proof of Proposition 2.2 prints "{0, 1}m" with braces (p7, from the text layer). Unescaped LaTeX braces were probably dropped when the output was typeset. The meaning is recoverable.

Second check: none (judge only)

• p. 5 Final Response from Internal Model, paragraph after the definition of u_ϵ

  are in −2, 0, 2, so

• p. 4 model's final response, characters of M

  ai ∈ 0, 1, 2

• p. 5 valuations of uϵ

  are in −2, 0, 2

Judge's note: Checked both page images. Page 4 prints "a_i ∈ 0, 1, 2" and page 5 prints "are in −2, 0, 2" with no braces. The same page 5 also prints "ϵ = (ϵ_s) ∈ 0, 1^m" for {0,1}^m, so the braces were lost systematically in this quoted block. Cosmetic; the meaning is recoverable.

Second check: none (judge only)

• p. 5 Final Response from Internal Model, display (9)

  2eγfj/2, nj.

• p. 9 Proof of Theorem 2.3

  ν(Pj) ≥1 2njeγfj/2

• p. 9 Proof of Theorem 2.3

  2njeγfj/2,

• p. 5 Final Response from Internal Model, display (9)

  ν(Pj) ≥ 1/2 e^{γfj/2}, nj.

Judge's note: I checked the crop of p.5 and the page image: the comma between e^{γfj/2} and nj is really printed there. The bound follows from (8): D_y ≥ e^{γfj/2}|X|, with each segment counted at most twice. Step (11) needs the factor nj. The proof of Theorem 2.3 on p.9 prints the correct form, ν(Pj) ≥ ½ nj e^{γfj/2}. This is the same comma-for-thin-space slip as in (4), inside a block announced as verbatim model output. It is a punctuation typo and the result is unaffected.

Second check: none (judge only)

• p. 5 Final Response from Internal Model, sentence before display (7)

  so the uϵ’s are distinct

• p. 5 Final Response from Internal Model, same paragraph

  Q2uϵ ∈OKj

Judge's note: Page 5 image: "uϵ's" is in running italic with ϵ on the baseline, while u_ϵ carries a subscript elsewhere in the block. "𝒪K_j" has K at full size after 𝒪, whereas three lines below "Λ_j = Q^{-2}𝒪_{K_j}" is set correctly. Typesetting slip only.

Second check: none (judge only)

• p. 13 Step 2, last paragraph

  G is infinite. Thus G is infinite, and the selected Frobenius classes are trivial

• p. 13 Proof of Proposition 3.8, end of Step 2

  is infinite, and the selected Frobenius classes are trivial

• p. 13 Step 2, last paragraph

  Ḡ is infinite. Thus Ḡ is infinite, and the selected Frobenius classes are trivial

• p. 13 Step 2, last paragraph

  Thus G is infinite

Judge's note: Page 13 image, end of Step 2: both occurrences carry the overline (Ḡ), so the second sentence restates the first as if it were a new deduction. An editing leftover with no effect on the argument.

Second check: none (judge only)