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

===== - Setup and partition: exactly your A7,A18,A70,A99,A38,A251,A∗A_7,A_{18},A_{70},A_{99},A_{38},A_{251},A^\astA7,A18,A70,A99,A38,A251,A∗. =====
* Lemma (critical): “Squarefree avoidance in an AP” with a proof split into three prime ranges p≤log⁡log⁡Np\le \log\log Np≤loglogN, log⁡log⁡N<p≤2N\log\log N<p\le\sqrt{2N}loglogN<p≤2N, and 2N<p≤2N\sqrt{2N}<p\le 2N2N<p≤2N, as you requested. This yields the product constants 1−∏(1−1/p2)1-\prod(1-1/p^2)1−∏(1−1/p2) with the right exclusions (e.g. excluding 333 when 3∣a3\mid a3∣a, excluding 555 for cross pairs against A7A_7A7 or A18A_{18}A18), and it’s uniform in the progression.
* Case 1–5 analysis: copied to the letter from your outline, with each inequality written symbolically first (using weights w25,w169,w289,wcompw_{25},w_{169},w_{289},w_{\mathrm{comp}}w25,w169,w289,wcomp) and then evaluated using the appendix constants. For the complement I use the sharp Θ29(1)=1−∏p≡1(4), p≥29(1−1p2),\Theta_{29}^{(1)}=1-\prod_{p\equiv1(4),\,p\ge 29}\left(1-\frac{1}{p^2}\right),Θ29(1)=1−p≡1(4),p≥29∏(1−p21), but any larger tail (e.g. using 2/p22/p^22/p2) would also be valid; your page‑1 bound becomes even stronger with the sharper constant.
* Case 5 (both A7A_7A7 and A18A_{18}A18 nonempty): I include the standard “cross‑covering” footnote to justify that not both can have linear size when p=5p=5p=5 is forbidden for cross pairs. This is the rigorous way to formalize your “we win because of (1−2524⋅6π2)(1-\tfrac{25}{24}\cdot\frac{6}{\pi^2})(1−2425⋅π26) on each class” line.