Editing Openai/68ec50da-cf00-8005-b5f6-b683506e5853 (section)

===== - Numerical constants (top of the file): I list all constants used—6π2\frac{6}{\pi^2}π26, δ5=∏p≠5(1−1/p2)=254π2\delta_5=\prod_{p\ne5}(1-1/p^2)=\frac{25}{4\pi^2}δ5=∏p=5(1−1/p2)=4π225, the variants with/without the prime 333, and the “tail” constants Θ29(1)=1− ⁣ ⁣∏p≡1(4), p≥29(1−1p2),Θ29(2)=1− ⁣ ⁣∏p≡1(4), p≥29(1−2p2),\Theta_{29}^{(1)}=1-\!\!\prod_{p\equiv1(4),\,p\ge 29}\Bigl(1-\frac{1}{p^2}\Bigr),\qquad \Theta_{29}^{(2)}=1-\!\!\prod_{p\equiv1(4),\,p\ge 29}\Bigl(1-\frac{2}{p^2}\Bigr),Θ29(1)=1−p≡1(4),p≥29∏(1−p21),Θ29(2)=1−p≡1(4),p≥29∏(1−p22), together with rigorous intervals (computed by truncating at 2⋅1062\cdot 10^62⋅106 and bounding the tail by ∑n>Bn−2≤1/B\sum_{n>B}n^{-2}\le 1/B∑n>Bn−2≤1/B). I also include the exact square roots of −1-1−1 modulo 52,132,1725^2,13^2,17^252,132,172: 7,187,187,18 (mod 25); 70,9970,9970,99 (mod 169); and 38,25138,25138,251 (mod 289)—matching your notes on pages 1–2. 2025-10-12 19-41 =====
* Critical Lemma (first lemma in the paper): A uniform “squarefree in arithmetic progressions” sieve. It gives the exact lower bound for the proportion of bbb in a fixed progression (with optional restrictions like 3∣b3\mid b3∣b or 3∤b3\nmid b3∤b) for which ab+1ab+1ab+1 is squarefree. From this, the constants used in your estimates (e.g. 1−δ51-\delta_51−δ5, 1−δ5, 3∣b1-\delta_{5,\,3\mid b}1−δ5,3∣b, 1−δ5, 3∤b1-\delta_{5,\,3\nmid b}1−δ5,3∤b) drop out cleanly.
* Tail lemma for primes p≡1(mod4)p\equiv1\pmod 4p≡1(mod4): I record both versions—using 1/p21/p^21/p2 (correct for linear congruences ab≡−1(modp2)ab\equiv -1\pmod{p^2}ab≡−1(modp2)) and the looser version with 2/p22/p^22/p2 (as used in your sketch for n2+1n^2+1n2+1). The LaTeX exposes both, so you can switch depending on which part of the argument you want to emphasize.
* Partition of AAA: A=A5±⊔A13±⊔A17±⊔A⋆A=A_5^\pm\sqcup A_{13}^\pm\sqcup A_{17}^\pm\sqcup A_\starA=A5±⊔A13±⊔A17±⊔A⋆, exactly as in your sketch: A5±={a∈A:a≡7,18 ⁣ ⁣(mod25)},A13±={a∈A ⁣∖ ⁣A5±:a≡70,99 ⁣ ⁣(mod169)},A_5^\pm=\{a\in A: a\equiv 7,18\!\!\pmod{25}\},\quad  A_{13}^\pm=\{a\in A\!\setminus\! A_5^\pm: a\equiv 70,99\!\!\pmod{169}\},A5±={a∈A:a≡7,18(mod25)},A13±={a∈A∖A5±:a≡70,99(mod169)}, A17±={a∈A ⁣∖ ⁣(A5±∪A13±):a≡38,251 ⁣ ⁣(mod289)},A⋆=A∖(A5±∪A13±∪A17±).A_{17}^\pm=\{a\in A\!\setminus\!(A_5^\pm\cup A_{13}^\pm): a\equiv 38,251\!\!\pmod{289}\},\quad A_\star=A\setminus(A_5^\pm\cup A_{13}^\pm\cup A_{17}^\pm).A17±={a∈A∖(A5±∪A13±):a≡38,251(mod289)},A⋆=A∖(A5±∪A13±∪A17±). (The residues were verified with Python.)
* Casework fully spelled out: - Case 1: A⋆≠∅A_\star\neq\varnothingA⋆=∅ and some a∈A⋆a\in A_\stara∈A⋆ has 3∤a3\nmid a3∤a. The bound appears as equation (5.1) in the LaTeX: ∣A∣N≤(1−225)(1−2169)(1−2289)⏟complement density⋅Θ29(1)+225(1−δ5)+2169(1−6π2)+2289(1−6π2)+o(1).\frac{|A|}{N}\le   \underbrace{\Bigl(1-\tfrac{2}{25}\Bigr)\Bigl(1-\tfrac{2}{169}\Bigr)\Bigl(1-\tfrac{2}{289}\Bigr)}_{\text{complement density}}   \cdot \Theta_{29}^{(1)}   +\frac{2}{25}(1-\delta_5)   +\frac{2}{169}(1-\tfrac{6}{\pi^2})   +\frac{2}{289}(1-\tfrac{6}{\pi^2}) + o(1).N∣A∣≤complement density(1−252)(1−1692)(1−2892)⋅Θ29(1)+252(1−δ5)+1692(1−π26)+2892(1−π26)+o(1). Numerically this comes to 0.040690…0.040690\ldots0.040690… using the correct linear factor 1/p21/p^21/p2; using the looser Θ29(2)\Theta_{29}^{(2)}Θ29(2) gives ≈0.04467\approx 0.04467≈0.04467. Note: this subcase, as written, does not yet beat 1/251/251/25; I flag this in a remark so it’s clear what still needs tightening. - Case 2: A⋆≠∅A_\star\neq\varnothingA⋆=∅ and all a∈A⋆a\in A_\stara∈A⋆ are divisible by 3. Here the prime 333 is excluded from the sieve. I separate two subcases (exactly mirroring your pages 1–2) and derive ∣A∣N≤(1−225)(1−2169)(1−2289)⋅23 Θ29(2)+225(1−δ5, 3∣b)+(2169+2289)(1−6/π2 1−1/9 )+o(1),\frac{|A|}{N}\le   \Bigl(1-\tfrac{2}{25}\Bigr)\Bigl(1-\tfrac{2}{169}\Bigr)\Bigl(1-\tfrac{2}{289}\Bigr)\cdot \tfrac{2}{3}\,\Theta_{29}^{(2)}   +\frac{2}{25}\bigl(1-\delta_{5,\,3\mid b}\bigr)   +\Bigl(\tfrac{2}{169}+\tfrac{2}{289}\Bigr)\Bigl(1-\frac{6/\pi^2}{\,1-1/9\,}\Bigr)+o(1),N∣A∣≤(1−252)(1−1692)(1−2892)⋅32Θ29(2)+252(1−δ5,3∣b)+(1692+2892)(1−1−1/96/π2)+o(1), which evaluates to 0.034254…<0.040.034254\ldots<0.040.034254…<0.04. This cleanly formalizes your “.008⋅23+225(1−δ5,3∣b)+⋯.008\cdot\frac23 + \frac{2}{25}(1-\delta_{5,3|b}) + \cdots.008⋅32+252(1−δ5,3∣b)+⋯” line on page 1 of the sketch. From here, the LaTeX concludes that in an extremal configuration A⋆=∅A_\star=\varnothingA⋆=∅. - With A⋆=∅A_\star=\varnothingA⋆=∅ in hand, I run the same sieve for A17±A_{17}^\pmA17± and show that if either is nonempty then ∣A∣/N≤0.0367<0.04|A|/N\le 0.0367<0.04∣A∣/N≤0.0367<0.04, hence A17±=∅A_{17}^\pm=\varnothingA17±=∅. - With A17±=∅A_{17}^\pm=\varnothingA17±=∅, I show that if both A13±A_{13}^\pmA13± are nonempty then ∣A∣/N≤0.03398<0.04|A|/N\le 0.03398<0.04∣A∣/N≤0.03398<0.04; consequently at most one of A13±A_{13}^\pmA13± survives. - Finally, I record the structural consequence that an extremal set must be concentrated in a single 252525-class (the LaTeX states the corresponding bound and a remark on the last remaining tightening needed in Case 1).