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

==== The proof above used an arbitrary finite set P⊂{p≠5}P\subset\{p\ne5\}P⊂{p=5}; choosing PPP cleverly improves constants inside the argument: ====
* If, say, AAA has many elements ≡7(mod25)\equiv7\pmod{25}≡7(mod25) and the complementary part of AAA misses a residue class modulo 333, then in the covering lemma we may simply forbid p=3p=3p=3 from being used (for those bbb), which strictly decreases the tail TPT_PTP. This is exactly the observation you made.
* Likewise, if you restrict PPP to primes p≡1(mod4)p\equiv1\pmod4p≡1(mod4) (excluding 555), the sum ∑p∈P1p2\sum_{p\in P}\tfrac1{p^2}∑p∈Pp21 is smaller; the price is that the tail TPT_PTP is larger. You can then combine both tricks—exclude a few small p≡3(mod4)p\equiv3\pmod4p≡3(mod4) by a congruence‑class choice (e.g. mod 333), and let PPP contain many 1(mod4)1\pmod 41(mod4) primes—to achieve very small SPS_PSP and very small TPT_PTP. That only strengthens the inequality (2.2)(2.2)(2.2). (Numerically, ∑p≡1(4), p>51/p2\sum_{p\equiv1(4),\,p>5}1/p^2∑p≡1(4),p>51/p2 is indeed quite small; but the proof above never needs any numerical estimate beyond SP<1S_P<1SP<1 and TP→0T_P\to0TP→0.)