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

==== Fix a finite set of primes P⊂{2,3,5,7,… }∖{5}P\subset\{2,3,5,7,\dots\}\setminus\{5\}P⊂{2,3,5,7,…}∖{5} and set ====

SP:=∑p∈P1p2(note SP<1),TP:=∑p∉P, p≠51p2.S_P:=\sum_{p\in P}\frac{1}{p^2}\qquad(\text{note }S_P<1),\qquad
T_P:=\sum_{p\notin P,\ p\ne5}\frac{1}{p^2}.SP:=p∈P∑p21(note SP<1),TP:=p∈/P, p=5∑p21.
Given kkk integers a1,…,aka_1,\dots,a_ka1,…,ak and a 25‑class ttt, define

Ui:=⋃p∈PCp(ai;t)(1≤i≤k).\mathcal U_i:=\bigcup_{p\in P} C_p(a_i;t)\qquad(1\le i\le k).Ui:=p∈P⋃Cp(ai;t)(1≤i≤k).
By Fact 1–2, each Ui\mathcal U_iUi has relative density exactly SPS_PSP inside StS_tSt.

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Proof. For each function f:{1,…,k}→Pf:\{1,\dots,k\}\to Pf:{1,…,k}→P, set

Uf:=⋂i=1kCf(i)(ai;t).\mathcal U_f:=\bigcap_{i=1}^k C_{f(i)}(a_i;t).Uf:=i=1⋂kCf(i)(ai;t).
Then ⋂iUi⊆⋃fUf\bigcap_{i}\mathcal U_i\subseteq \bigcup_f \mathcal U_f⋂iUi⊆⋃fUf. For fixed fff and for each prime ppp, the intersection ⋂i: f(i)=pCp(ai;t)\bigcap_{i:\,f(i)=p}C_p(a_i;t)⋂i:f(i)=pCp(ai;t) is either empty (if two different residue classes are demanded) or a single residue class modulo p2p^2p2; in either case its relative density inside StS_tSt is ≤1/p2\le 1/p^2≤1/p2. By CRT these constraints across different ppp multiply, whence

∣Uf∣∣St∣≤∏p∈P(1p2)1f−1(p)≠∅=∏i=1k1(f(i))2.\frac{|\mathcal U_f|}{|S_t|}\le \prod_{p\in P}\left(\frac1{p^2}\right)^{\mathbf 1_{f^{-1}(p)\ne\varnothing}}
=\prod_{i=1}^k \frac1{(f(i))^2}.∣St∣∣Uf∣≤p∈P∏(p21)1f−1(p)=∅=i=1∏k(f(i))21.
Summing over all fff gives

∣⋃fUf∣∣St∣≤∑f∏i=1k1(f(i))2=(∑p∈P1p2)k=(SP)k,\frac{\Bigl|\bigcup_f \mathcal U_f\Bigr|}{|S_t|}
\le \sum_f \prod_{i=1}^k \frac{1}{(f(i))^2}=\left(\sum_{p\in P}\frac{1}{p^2}\right)^k=(S_P)^k,∣St∣⋃fUf≤f∑i=1∏k(f(i))21=p∈P∑p21k=(SP)k,
and the o(1)o(1)o(1) accounts for the O(1)O(1)O(1) boundary error when restricting to [1,N][1,N][1,N]. ∎

The primes outside PPP contribute at most TPT_PTP to the obstruction for each fixed aia_iai. Therefore the set of b∈Stb\in S_tb∈St for which “for every iii the square divisor that certifies (⋆)(\star)(⋆) comes from a prime outside PPP” has relative density at most TPT_PTP (indeed, it is ≤min⁡iTP=TP\le\min_i T_P=T_P≤miniTP=TP). Combining this with (2.1) yields the uniform covering inequality:

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Two important remarks.
* We did not assume any randomness; (2.2)(2.2)(2.2) is worst‑case, valid even if the residue classes Cp(ai;t)C_p(a_i;t)Cp(ai;t) line up adversarially.
* We excluded p=5p=5p=5 from PPP because in the cross‑class situations we analyse below the congruence 25∣ab+125\mid ab+125∣ab+1 is impossible; omitting it only makes the upper bound stronger.

Finally, since ∑p≠51/p2<1\sum_{p\ne5}1/p^2<1∑p=51/p2<1, we can make the “tail”
TPT_PTP as small as we please by enlarging PPP, while SP<1S_P<1SP<1 always.