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

===== - If αr\alpha_rαr is bounded below by a fixed constant c>0c>0c>0, then you can first choose PPP so that TP<ε/2T_P<\varepsilon/2TP<ε/2; for that PPP the quantity SPS_PSP is ≈0.412\approx 0.412≈0.412, hence ⌊αr/SP⌋\lfloor \alpha_r/S_P\rfloor⌊αr/SP⌋ is a positive integer and SP⌊αr/SP⌋S_P^{\lfloor \alpha_r/S_P\rfloor}SP⌊αr/SP⌋ is a fixed number <1<1<1. If ccc is not tiny you can ensure SP⌊αr/SP⌋<ε/2S_P^{\lfloor \alpha_r/S_P\rfloor}<\varepsilon/2SP⌊αr/SP⌋<ε/2 as well, which gives αs≤ε\alpha_s\le\varepsilonαs≤ε (for NNN large). In words: one class of appreciable density forces all incompatible classes to be negligible. =====
* This is the rigorous version of the “stability” heuristic: once many aaa’s sit in rrr, the same small primes cannot keep rescuing all cross pairs with sss, because the residue constraints modulo p2p^2p2 collide and annihilate unless different primes are used—hence the SPKS_P^{K}SPK factor.
* Note that in the exceptional pairs (r,s)(r,s)(r,s) with r∈{7,18}r\in\{7,18\}r∈{7,18} and s=rs=rs=r we do not apply Proposition 5.1: there the prime 555 alone takes care of all pairs, so the argument is not needed. For the cross pair (r,s)=(7,18)(r,s)=(7,18)(r,s)=(7,18) (or vice versa), we can apply the proposition (since ab+1≡2 (mod 25)ab+1\equiv 2\ (\mathrm{mod}\ 25)ab+1≡2 (mod 25), p=5p=5p=5 is excluded), which is the rigorous formulation of your “7 vs 18” observation.