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

=== Assistant: Great—I've turned your sketch into a structured LaTeX document with the casework and the key lemmas spelled out, and I … ===
Great—I've turned your sketch into a structured LaTeX document with the casework and the key lemmas spelled out, and I verified every numerical constant with Python. You can download the .tex source here:

Download the LaTeX file<ref>{{cite web|title=Download the LaTeX file|url=https://chatgpt.com/share/Download%20the%20LaTeX%20file|publisher=Download the LaTeX file|access-date=2025-11-25}}</ref>

===== - 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).

===== All numerical values used in the document were computed in Python and embedded directly into the LaTeX: =====
* 6π2=0.6079271018540267\frac{6}{\pi^2}=0.6079271018540267π26=0.6079271018540267
* δ5=254π2=0.6332573977646111\delta_5=\frac{25}{4\pi^2}=0.6332573977646111δ5=4π225=0.6332573977646111 and 1−δ5=0.36674260223538891-\delta_5=0.36674260223538891−δ5=0.3667426022353889
* δ5, 3∣b=0.7124145724851875\delta_{5,\,3\mid b}=0.7124145724851875δ5,3∣b=0.7124145724851875 (so 1−δ5, 3∣b=0.287585427514812531-\delta_{5,\,3\mid b}=0.287585427514812531−δ5,3∣b=0.28758542751481253)
* δ5, 3∤b=0.5936788104043229\delta_{5,\,3\nmid b}=0.5936788104043229δ5,3∤b=0.5936788104043229 (so 1−δ5, 3∤b=0.40632118959567711-\delta_{5,\,3\nmid b}=0.40632118959567711−δ5,3∤b=0.4063211895956771)
* Θ29(1)∈[0.00442735, 0.00442835]\Theta_{29}^{(1)}\in[0.00442735,\,0.00442835]Θ29(1)∈[0.00442735,0.00442835], Θ29(2)∈[0.00883767, 0.00883967]\Theta_{29}^{(2)}\in[0.00883767,\,0.00883967]Θ29(2)∈[0.00883767,0.00883967] (rigorous intervals)
* Complement density (1−225)(1−2169)(1−2289)=0.9028209905612089(1-\frac{2}{25})(1-\frac{2}{169})(1-\frac{2}{289})=0.9028209905612089(1−252)(1−1692)(1−2892)=0.9028209905612089
* 225=0.08\frac{2}{25}=0.08252=0.08, 2169≈0.01183432\frac{2}{169}\approx 0.011834321692≈0.01183432, 2289≈0.00692042\frac{2}{289}\approx 0.006920422892≈0.00692042

(The two square roots of −1-1−1 modulo 52,132,1725^2,13^2,17^252,132,172 are 7,187,187,18; 70,9970,9970,99; 38,25138,25138,251 respectively—verified computationally and used to define A5±,A13±,A17±A_5^\pm, A_{13}^\pm, A_{17}^\pmA5±,A13±,A17±.)

===== Your pages make the same partition of AAA and then treat three situations for A⋆A_\starA⋆ (nonempty with some 3∤a3\nmid a3∤a; nonempty with all 3∣a3\mid a3∣a, split further depending on where 333 divides outside A5±A_5^\pmA5±). The LaTeX formalizes each into an inequality with named constants and separates the roles of: =====
* the prime 555 (giving δ5\delta_5δ5 and its 333-variants in Lemma 1’s corollaries);
* the prime 333 (either excluded or handled via the 3∣b3\mid b3∣b / 3∤b3\nmid b3∤b versions); and
* the large primes p≡1(mod4), p≥29p\equiv1\pmod 4,\ p\ge29p≡1(mod4), p≥29 (handled by Θ29(1)\Theta_{29}^{(1)}Θ29(1) or Θ29(2)\Theta_{29}^{(2)}Θ29(2)).

Where your handwritten constants (.008, .02301, .00374, .00218) appear, the LaTeX shows the exact symbolic expression they correspond to and plugs in the Python‑checked values (matching to the digits you wrote).

If you’d like me to compile this to PDF or tweak the macros (e.g., switch everywhere to the sharp Θ29(1)\Theta_{29}^{(1)}Θ29(1) or keep the looser Θ29(2)\Theta_{29}^{(2)}Θ29(2) to mirror the handwritten sketch), just say the word and I’ll update the source accordingly.