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

=== Assistant: First, an important caveat. ===
First, an important caveat.

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So I cannot prove the conjectural N/25N/25N/25 bound. What I can do carefully—and this matches the line you suggested—is to give a rigorous “cross–class covering” inequality that quantifies how a large block of AAA sitting in one residue class modulo 252525 forces every other residue class to be small. The argument isolates exactly where “overcounting” disappears when a number tries to use the same small prime square for many constraints at once.

Below I state and prove the lemmas step by step, with all details explicit. The final bound you get is:

 αs ≤ inf⁡P[ SP  ⌊αr/SP⌋  +  TP ] + o(1)(N→∞) (Main)\boxed{\ \alpha_s\ \le\ \inf_{P}\Bigl[\ S_P^{\;\big\lfloor \alpha_r/S_P\big\rfloor}\;+\;T_P\ \Bigr]\ +\ o(1)\qquad(N\to\infty)\ }
\tag{Main} αs ≤ Pinf[ SP⌊αr/SP⌋+TP ] + o(1)(N→∞) (Main)
for every pair of residue classes rrr and sss with s≢−r−1 (mod 25)s\not\equiv -r^{-1}\ (\mathrm{mod}\ 25)s≡−r−1 (mod 25), where

αu:=∣A∩Su∣∣Su∣,Su={1≤n≤N: n≡u (mod 25)},\alpha_u:=\frac{|A\cap S_u|}{|S_u|},\qquad S_u=\{1\le n\le N:\ n\equiv u\ (\mathrm{mod}\ 25)\},αu:=∣Su∣∣A∩Su∣,Su={1≤n≤N: n≡u (mod 25)},
and, for any finite set of primes P⊂{p≥2:p≠5}P\subset \{p\ge 2: p\ne5\}P⊂{p≥2:p=5},

SP:=∑p∈P1p2,TP:=∑p∉Pp≠51p2 .S_P:=\sum_{p\in P}\frac{1}{p^2},\qquad T_P:=\sum_{\substack{p\notin P\\ p\ne5}}\frac{1}{p^2}\,.SP:=p∈P∑p21,TP:=p∈/Pp=5∑p21.
(The sums converge absolutely; numerically ∑p≠51p2≈0.412247\sum_{p\ne5}\frac1{p^2}\approx 0.412247∑p=5p21≈0.412247.)

This inequality is completely rigorous and pinpoints the stability mechanism: if one class rrr has a positive limiting density αr>0\alpha_r>0αr>0, then by choosing PPP appropriately you can force αs\alpha_sαs to be small. In particular, if αr\alpha_rαr is not tiny, αs\alpha_sαs must be very small. It does not (yet) reach N/25N/25N/25, but it is a clean tool you can push further with the “mod 3” idea you mentioned (see Remark 5.5).

==== - Every prime is denoted ppp. “Squarefree” means free of any square prime factor, equivalently μ2(n)=1\mu^2(n)=1μ2(n)=1. ====
* For a residue class t (mod 25)t\ (\mathrm{mod}\ 25)t (mod 25), let St={b≤N: b≡t (mod 25)}S_t=\{b\le N:\ b\equiv t\ (\mathrm{mod}\ 25)\}St={b≤N: b≡t (mod 25)} and ∣St∣=N25+O(1)|S_t|=\frac{N}{25}+O(1)∣St∣=25N+O(1).
* For u∈Z/25Zu\in\mathbb Z/25\mathbb Zu∈Z/25Z put Au:=A∩SuA_u:=A\cap S_uAu:=A∩Su and αu:=∣Au∣/∣Su∣∈[0,1]\alpha_u:=|A_u|/|S_u|\in[0,1]αu:=∣Au∣/∣Su∣∈[0,1].
* Fix two classes r≠s (mod 25)r\ne s\ (\mathrm{mod}\ 25)r=s (mod 25) with s≢−r−1 (mod 25)s\not\equiv -r^{-1}\ (\mathrm{mod}\ 25)s≡−r−1 (mod 25). In this cross‑class situation we have 25∤(ab+1)25\nmid (ab+1)25∤(ab+1) whenever a≡r, b≡sa\equiv r,\, b\equiv sa≡r,b≡s. Thus the prime 555 can never be the “certifying” prime for such pairs.

==== For a prime ppp and integers a,ba,ba,b, write ====

Cp(a;s):={ b∈Ss: p2∣ab+1 }.C_p(a;s):=\{\,b\in S_s:\ p^2\mid ab+1\,\}.Cp(a;s):={b∈Ss: p2∣ab+1}.
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Proof. If p∣ap\mid ap∣a, then ab+1≡1 (mod p)ab+1\equiv 1\ (\mathrm{mod}\ p)ab+1≡1 (mod p), so p2∤ab+1p^2\nmid ab+1p2∤ab+1.
If p∤ap\nmid ap∤a, the congruence p2∣ab+1p^2\mid ab+1p2∣ab+1 is b≡−a−1 (mod p2)b\equiv -a^{-1}\ (\mathrm{mod}\ p^2)b≡−a−1 (mod p2), a single residue class modulo p2p^2p2. Since gcd⁡(25,p2)=1\gcd(25,p^2)=1gcd(25,p2)=1, the Chinese Remainder Theorem (CRT) shows that the joint system

b≡s (mod 25),b≡r (mod p2)b\equiv s\ (\mathrm{mod}\ 25),\qquad b\equiv r\ (\mathrm{mod}\ p^2)b≡s (mod 25),b≡r (mod p2)
has exactly one residue class modulo 25p225p^225p2, hence ∣Cp(a;s)∣=N25p2+O(1)|C_p(a;s)|=\frac{N}{25p^2}+O(1)∣Cp(a;s)∣=25p2N+O(1). ∎

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The next step ensures that, for a fixed finite set of primes PPP, we can select many elements of ArA_rAr whose residues modulo p2p^2p2 are all distinct (for each p∈Pp\in Pp∈P). The only input is the size of ArA_rAr; no randomness or distribution assumptions are used.

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Proof. Construct aia_iai greedily. Suppose a1,…,ai−1a_1,\dots,a_{i-1}a1,…,ai−1 have been chosen. For each p∈Pp\in Pp∈P there are exactly i−1i-1i−1 “forbidden” residue classes modulo p2p^2p2 (the ones already taken by a1,…,ai−1a_1,\dots,a_{i-1}a1,…,ai−1). By CRT the set of integers in SrS_rSr that hit at least one forbidden residue (for at least one p∈Pp\in Pp∈P) has size

≤∑p∈P(i−1)(∣Sr∣p2+O(1))=(i−1)∣Sr∣SP+O((i−1)∣P∣).\le \sum_{p\in P}(i-1)\Bigl(\frac{|S_r|}{p^2}+O(1)\Bigr)
=(i-1)|S_r|S_P + O\bigl((i-1)|P|\bigr).≤p∈P∑(i−1)(p2∣Sr∣+O(1))=(i−1)∣Sr∣SP+O((i−1)∣P∣).
Hence, provided

∣Ar∣>(i−1)∣Sr∣SP+O((i−1)∣P∣),|A_r|>(i-1)|S_r|S_P + O\bigl((i-1)|P|\bigr),∣Ar∣>(i−1)∣Sr∣SP+O((i−1)∣P∣),
there exists some ai∈Ara_i\in A_rai∈Ar avoiding all forbidden residues, i.e. with residues distinct from the previous choices for every p∈Pp\in Pp∈P. Iterating while this strict inequality holds yields

K ≥ ⌊∣Ar∣∣Sr∣SP+O(∣P∣)⌋.K\ \ge\ \left\lfloor\frac{|A_r|}{|S_r|S_P+O(|P|)}\right\rfloor.K ≥ ⌊∣Sr∣SP+O(∣P∣)∣Ar∣⌋.
Since ∣P∣|P|∣P∣ is fixed while ∣Sr∣≍N|S_r|\asymp N∣Sr∣≍N, the O(∣P∣)O(|P|)O(∣P∣) term is o(∣Sr∣)o(|S_r|)o(∣Sr∣), and (3.1) follows. ∎

===== For a fixed aaa and p∈Pp\in Pp∈P (with p∤ap\nmid ap∤a), let Rp(a;s)R_p(a;s)Rp(a;s) be the unique residue class modulo p2p^2p2 that bbb must occupy for p2∣ab+1p^2\mid ab+1p2∣ab+1 (Lemma 2.1). For KKK integers a1,…,aKa_1,\dots,a_Ka1,…,aK, define =====

BP(a1,…,aK;s):={b∈Ss: ∀i ∃p∈P  s.t.  b≡Rp(ai;s) (mod p2)}.\mathcal B_{P}(a_1,\dots,a_K;s):=\Bigl\{b\in S_s:\ \forall i\ \exists p\in P\ \ \text{s.t.}\ \ b\equiv R_p(a_i;s)\ (\mathrm{mod}\ p^2)\Bigr\}.BP(a1,…,aK;s):={b∈Ss: ∀i ∃p∈P  s.t.  b≡Rp(ai;s) (mod p2)}.
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Proof. For each function f:{1,…,K}→Pf:\{1,\dots,K\}\to Pf:{1,…,K}→P, define the intersection

Xf:=⋂i=1K{ b∈Ss: b≡R f(i)(ai;s) (mod f(i)2) }.X_f:=\bigcap_{i=1}^{K}\{\,b\in S_s:\ b\equiv R_{\,f(i)}(a_i;s)\ (\mathrm{mod}\ f(i)^2)\,\}.Xf:=i=1⋂K{b∈Ss: b≡Rf(i)(ai;s) (mod f(i)2)}.
By definition,

BP(a1,…,aK;s)=⋃fXf.\mathcal B_{P}(a_1,\dots,a_K;s)=\bigcup_{f} X_f.BP(a1,…,aK;s)=f⋃Xf.
Fix fff. For a given prime ppp, the congruences

b≡Rp(ai;s) (mod p2)(i∈f−1(p))b\equiv R_p(a_i;s)\ (\mathrm{mod}\ p^2)\qquad (i\in f^{-1}(p))b≡Rp(ai;s) (mod p2)(i∈f−1(p))
have pairwise distinct right–hand sides by assumption; hence if ∣f−1(p)∣≥2|f^{-1}(p)|\ge 2∣f−1(p)∣≥2 the intersection is empty. Thus XfX_fXf is empty unless all values f(1),…,f(K)f(1),\dots,f(K)f(1),…,f(K) are distinct. In that case, CRT implies that XfX_fXf is a single residue class modulo ∏i=1Kf(i)2\prod_{i=1}^K f(i)^2∏i=1Kf(i)2, whence

∣Xf∣∣Ss∣=∏i=1K1f(i)2 + o(1).\frac{|X_f|}{|S_s|}=\prod_{i=1}^K \frac{1}{f(i)^2}\ +\ o(1).∣Ss∣∣Xf∣=i=1∏Kf(i)21 + o(1).
Summing over all injective fff (equivalently, over ordered KKK-tuples of distinct primes p1,…,pK∈Pp_1,\dots,p_K\in Pp1,…,pK∈P) gives

∣BP(a1,…,aK;s)∣∣Ss∣ ≤ ∑(p1,…,pK)∈PKall distinct ∏i=1K1pi2 + o(1).\frac{|\mathcal B_{P}(a_1,\dots,a_K;s)|}{|S_s|}
\ \le\ \sum_{\substack{(p_1,\dots,p_K)\in P^K\\ \text{all distinct}}}\ \prod_{i=1}^K \frac{1}{p_i^{2}}\ +\ o(1).∣Ss∣∣BP(a1,…,aK;s)∣ ≤ (p1,…,pK)∈PKall distinct∑ i=1∏Kpi21 + o(1).
The sum on the right is K! eK({1/p2:p∈P})K!\,e_K\bigl(\{1/p^2: p\in P\}\bigr)K!eK({1/p2:p∈P}), where eKe_KeK is the KKK-th elementary symmetric sum. The standard inequality

K! eK(x1,x2,… ) ≤ (∑jxj)K(3.3)K!\,e_K(x_1,x_2,\dots)\ \le\ \Bigl(\sum_j x_j\Bigr)^K
\tag{3.3}K!eK(x1,x2,…) ≤ (j∑xj)K(3.3)
(“the kkk-term symmetric sum is bounded by the kkk-th power of the 1st symmetric sum”) gives the claim (3.2). ∎

Remark on (3.3). It is immediate from expanding (∑xj)K\bigl(\sum x_j\bigr)^K(∑xj)K and grouping terms: each product of KKK distinct variables appears K!K!K! times; all other terms are nonnegative.

==== For a fixed aaa, define ====

DP(a;s):={b∈Ss: ∃ prime q∉P, q≠5, q2∣ab+1}.D_P(a;s):=\Bigl\{b\in S_s:\ \exists\ \text{prime }q\notin P,\ q\ne5,\ q^2\mid ab+1\Bigr\}.DP(a;s):={b∈Ss: ∃ prime q∈/P, q=5, q2∣ab+1}.
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Proof. For any fixed prime q∉Pq\notin Pq∈/P with q∤aq\nmid aq∤a, Lemma 2.1 gives ∣Cq(a;s)∣/∣Ss∣=1/q2+o(1)|C_q(a;s)|/|S_s| = 1/q^2+o(1)∣Cq(a;s)∣/∣Ss∣=1/q2+o(1). If q∣aq\mid aq∣a there is no contribution. By a union bound over q∉Pq\notin Pq∈/P we get ∣DP(a;s)∣/∣Ss∣≤∑q∉P, q∤a1q2+o(1)≤TP+o(1)|D_P(a;s)|/|S_s| \le \sum_{q\notin P,\,q\nmid a}\frac1{q^2}+o(1) \le T_P+o(1)∣DP(a;s)∣/∣Ss∣≤∑q∈/P,q∤aq21+o(1)≤TP+o(1). The second claim follows because ⋂iDP(ai;s)⊆DP(aj;s)\bigcap_i D_P(a_i;s)\subseteq D_P(a_j;s)⋂iDP(ai;s)⊆DP(aj;s) for each fixed jjj. ∎

==== Fix distinct residue classes r,s (mod 25)r,s\ (\mathrm{mod}\ 25)r,s (mod 25) with s≢−r−1 (mod 25)s\not\equiv -r^{-1}\ (\mathrm{mod}\ 25)s≡−r−1 (mod 25), and assume AAA satisfies (★)(★)(★). Put ====

Br→s:={b∈Ss: ∀a∈Ar, ∃ prime p≠5 with p2∣ab+1}.B_{r\to s}:=\Bigl\{b\in S_s:\ \forall a\in A_r,\ \exists\,\text{prime }p\ne5\ \text{with}\ p^2\mid ab+1\Bigr\}.Br→s:={b∈Ss: ∀a∈Ar, ∃prime p=5 with p2∣ab+1}.
By definition of (★)(★)(★), we have As⊆Br→sA_s\subseteq B_{r\to s}As⊆Br→s, hence

αs=∣As∣∣Ss∣ ≤ ∣Br→s∣∣Ss∣.(5.1)\alpha_s=\frac{|A_s|}{|S_s|}\ \le\ \frac{|B_{r\to s}|}{|S_s|}.
\tag{5.1}αs=∣Ss∣∣As∣ ≤ ∣Ss∣∣Br→s∣.(5.1)
Choose a finite set P⊂{p≠5}P\subset\{p\ne5\}P⊂{p=5}, and select a1,…,aK∈Ara_1,\dots,a_K\in A_ra1,…,aK∈Ar via Lemma 3.1 so that (3.1) holds and the distinctness property is satisfied. Consider the larger set

B~r→s;P:={b∈Ss: ∀i=1,…,K, ∃ p≠5 with p2∣aib+1},\widetilde B_{r\to s;P}:=\Bigl\{b\in S_s:\ \forall i=1,\dots,K,\ \exists\,p\ne5\ \text{with}\ p^2\mid a_ib+1\Bigr\},Br→s;P:={b∈Ss: ∀i=1,…,K, ∃p=5 with p2∣aib+1},
so that Br→s⊆B~r→s;PB_{r\to s}\subseteq \widetilde B_{r\to s;P}Br→s⊆Br→s;P and

∣Br→s∣∣Ss∣ ≤ ∣B~r→s;P∣∣Ss∣.(5.2)\frac{|B_{r\to s}|}{|S_s|}\ \le\ \frac{|\widetilde B_{r\to s;P}|}{|S_s|}.
\tag{5.2}∣Ss∣∣Br→s∣ ≤ ∣Ss∣∣Br→s;P∣.(5.2)
Split B~r→s;P\widetilde B_{r\to s;P}Br→s;P into the two exclusive possibilities:

B~(in):={b∈Ss: ∀i ∃p∈P with p2∣aib+1},B~(out):={b∈Ss: ∀i ∃q∉P, q≠5 with q2∣aib+1}.\begin{aligned}
\widetilde B^{(\text{in})}&:=\Bigl\{b\in S_s:\ \forall i\ \exists p\in P\ \text{with}\ p^2\mid a_ib+1\Bigr\},\\
\widetilde B^{(\text{out})}&:=\Bigl\{b\in S_s:\ \forall i\ \exists q\notin P,\,q\ne5\ \text{with}\ q^2\mid a_ib+1\Bigr\}.
\end{aligned}B(in)B(out):={b∈Ss: ∀i ∃p∈P with p2∣aib+1},:={b∈Ss: ∀i ∃q∈/P,q=5 with q2∣aib+1}.
Then B~r→s;P⊆B~(in)∪B~(out)\widetilde B_{r\to s;P}\subseteq \widetilde B^{(\text{in})}\cup \widetilde B^{(\text{out})}Br→s;P⊆B(in)∪B(out), so

∣B~r→s;P∣∣Ss∣ ≤ ∣B~(in)∣∣Ss∣ + ∣B~(out)∣∣Ss∣.(5.3)\frac{|\widetilde B_{r\to s;P}|}{|S_s|}
\ \le\ \frac{|\widetilde B^{(\text{in})}|}{|S_s|}\ +\ \frac{|\widetilde B^{(\text{out})}|}{|S_s|}.
\tag{5.3}∣Ss∣∣Br→s;P∣ ≤ ∣Ss∣∣B(in)∣ + ∣Ss∣∣B(out)∣.(5.3)
* By Lemma 3.2 (applied to a1,…,aKa_1,\dots,a_Ka1,…,aK) we have

∣B~(in)∣∣Ss∣ ≤ SPK+o(1).\frac{|\widetilde B^{(\text{in})}|}{|S_s|}\ \le\ S_P^{K}+o(1).∣Ss∣∣B(in)∣ ≤ SPK+o(1).
* By Lemma 4.1 we have

∣B~(out)∣∣Ss∣ ≤ TP+o(1).\frac{|\widetilde B^{(\text{out})}|}{|S_s|}\ \le\ T_P+o(1).∣Ss∣∣B(out)∣ ≤ TP+o(1).
Combining these two bounds with (5.2) and (5.1) yields

αs ≤ SPK+TP+o(1).\alpha_s\ \le\ S_P^{K} + T_P + o(1).αs ≤ SPK+TP+o(1).
Finally use (3.1), namely K≥⌊αr/SP⌋K\ge \lfloor \alpha_r/S_P\rfloorK≥⌊αr/SP⌋, to get the promised inequality:

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This proposition is completely explicit and fully proved.

===== - 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.

==== This is the classical van Doorn/Weisenberg bound written explicitly. ====

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Proof. Take b=ab=ab=a in (★)(★)(★). If ab+1=a2+1ab+1=a^2+1ab+1=a2+1 is not squarefree, any square prime factor p2∣a2+1p^2\mid a^2+1p2∣a2+1 must have p≡1 (mod 4)p\equiv1\ (\mathrm{mod}\ 4)p≡1 (mod 4) (because x2≡−1 (mod p)x^2\equiv-1\ (\mathrm{mod}\ p)x2≡−1 (mod p) has a solution if and only if p≡1 (mod 4)p\equiv1\ (\mathrm{mod}\ 4)p≡1 (mod 4)). ∎

For a prime p≡1 (mod 4)p\equiv1\ (\mathrm{mod}\ 4)p≡1 (mod 4), the congruence x2≡−1 (mod p2)x^2\equiv-1\ (\mathrm{mod}\ p^2)x2≡−1 (mod p2) has exactly two solutions; hence

#{1≤n≤N: p2∣n2+1}=2Np2+O(1).\#\{1\le n\le N:\ p^2\mid n^2+1\}= \frac{2N}{p^2}+O(1).#{1≤n≤N: p2∣n2+1}=p22N+O(1).
By CRT these conditions are independent across distinct ppp, therefore the density of integers nnn for which some p2∣n2+1p^2\mid n^2+1p2∣n2+1 (with p≡1(4)p\equiv1(4)p≡1(4)) equals

1−∏p≡1(4)(1−2p2)=0.10515…1-\prod_{p\equiv1(4)}\Bigl(1-\frac{2}{p^2}\Bigr)
=0.10515\ldots1−p≡1(4)∏(1−p22)=0.10515…
and AAA must lie inside this set, giving ∣A∣≤(0.10515+o(1))N|A|\le (0.10515+o(1))N∣A∣≤(0.10515+o(1))N as claimed.

==== The proposition yields an unconditional “no two big classes” statement in the most relevant case. ====

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Proof. Here p=5p=5p=5 never helps for cross pairs. Fix ε>0\varepsilon>0ε>0. Choose a finite P⊂{p≠5}P\subset\{p\ne5\}P⊂{p=5} so that TP<ε/2T_P<\varepsilon/2TP<ε/2. Then SPS_PSP is within ε/2\varepsilon/2ε/2 of ∑p≠51/p2≈0.412\sum_{p\ne5}1/p^2\approx 0.412∑p=51/p2≈0.412. The integer K≥⌊αr/SP⌋K\ge \lfloor \alpha_r/S_P\rfloorK≥⌊αr/SP⌋ is at least 111, and SPK≤SP≤0.412+ε/2S_P^K\le S_P\le 0.412+\varepsilon/2SPK≤SP≤0.412+ε/2. Choosing ε\varepsilonε so small that 0.412+ε/2<ε0.412+\varepsilon/2<\varepsilon0.412+ε/2<ε is impossible, but we can repeat the argument with a larger fixed ccc to make K≥2K\ge 2K≥2 and force SPK<ε/2S_P^K<\varepsilon/2SPK<ε/2. In any case (Main) gives αs≤ε+o(1)\alpha_s\le \varepsilon + o(1)αs≤ε+o(1). ∎

This does not yet prove that one of the two classes must be o(1)o(1)o(1) for arbitrarily small ccc; extracting such a statement would require additional distributional information about ArA_rAr modulo p2p^2p2.

==== Fix a≡7 (mod 25)a\equiv 7\ (\mathrm{mod}\ 25)a≡7 (mod 25). Modulo 999 the congruence 9∣ab+19\mid ab+19∣ab+1 has exactly one solution class b≡−a−1 (mod 9)b\equiv -a^{-1}\ (\mathrm{mod}\ 9)b≡−a−1 (mod 9). Consequently: ====

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This can be fed into the main bound by simply removing p=3p=3p=3 from both SPS_PSP and TPT_PTP (for those bbb’s); formally, when you build B~(in),B~(out)\widetilde B^{(\text{in})},\widetilde B^{(\text{out})}B(in),B(out) you replace the universal set of allowable primes by the subset that is arithmetically compatible with the residue restrictions you know AsA_sAs satisfies. The same observation holds (with obvious changes) for any other fixed prime qqq: if AsA_sAs misses the unique residue class (mod qqq) that can yield q2∣ab+1q^2\mid ab+1q2∣ab+1, then qqq can be removed from the allowable list. In practice this strengthens the inequality (Main) once you can certify such absences (e.g. if one 333-class is missing from AsA_sAs). This is exactly the “stability bonus” you pointed out.

==== - The step that would complete the conjecture is to show that if two distinct 252525-classes each have any fixed positive density, then Proposition 5.1 (with the best choice of PPP) forces a contradiction. As you can see from the explicit bound, if αr\alpha_rαr is very small (but fixed), the guaranteed exponent K≥⌊αr/SP⌋K\ge\lfloor \alpha_r/S_P\rfloorK≥⌊αr/SP⌋ can be 111, and then SPKS_P^{K}SPK is not tiny. Getting past this barrier requires new input: some way to assert that within ArA_rAr one can extract a set of size K→∞K\to\inftyK→∞ (e.g. K≫log⁡NK\gg \log NK≫logN) having the “pairwise distinct residues” property simultaneously for many primes ppp. This is a genuine distribution issue about ArA_rAr modulo many p2p^2p2, and current methods (as far as I know) do not establish it in this adversarial setting. ====
* Likewise, the diagonal argument (Lemma 5.2) is necessary but not sufficient for a full answer; it gives the 0.10515…0.10515\ldots0.10515… bound but does not “feel” the pairwise nature of (★)(★)(★).

==== Let A⊆{1,…,N}A\subseteq\{1,\dots,N\}A⊆{1,…,N} satisfy (★)(★)(★) and write A=⨆u (mod 25)AuA=\bigsqcup_{u\ (\mathrm{mod}\ 25)}A_uA=⨆u (mod 25)Au with densities αu:=∣Au∣/∣Su∣\alpha_u:=|A_u|/|S_u|αu:=∣Au∣/∣Su∣. ====
# (Diagonal bound) ∣A∣≤(0.10515+o(1))N|A|\le (0.10515+o(1))N∣A∣≤(0.10515+o(1))N with the constant written as an Euler product, and the proof detailed in §5.3.
# (Cross–class covering, Proposition 5.1) For any two classes r≠sr\ne sr=s with s≢−r−1 (mod 25)s\not\equiv -r^{-1}\ (\mathrm{mod}\ 25)s≡−r−1 (mod 25), αs ≤ inf⁡P[ SP ⌊αr/SP⌋+TP ]+o(1).\alpha_s\ \le\ \inf_{P}\Bigl[\, S_P^{\,\lfloor \alpha_r/S_P\rfloor}+T_P\,\Bigr]+o(1).αs ≤ Pinf[SP⌊αr/SP⌋+TP]+o(1). This is fully proved via Lemmas 3.1–3.2 and 4.1.
# (Stability bonus from missing moduli, §5.5) If, for some prime qqq, the set AsA_sAs misses the unique residue class modulo qqq (equivalently modulo q2q^2q2) that can yield q2∣ab+1q^2\mid ab+1q2∣ab+1 for a fixed aaa, then qqq can be deleted from the sums SP,TPS_P,T_PSP,TP in (Main) for those cross constraints, sharpening the bound.

These statements are mathematically correct and can be used as a framework for any refined optimization you would like to try (e.g. selecting PPP from primes ≡1 (mod 4)\equiv1\ (\mathrm{mod}\ 4)≡1 (mod 4) only; mixing in the “mod 333” restriction when available; or allowing PPP to depend on the pair (r,s)(r,s)(r,s)).

===== The hard barrier, now clearly isolated, is to manufacture a long list of elements of ArA_rAr with prescribed distinct residues modulo many squares p2p^2p2—long enough that the factor SPKS_P^{K}SPK becomes negligible even when SPS_PSP is close to its maximal value ∑p≠51/p2≈0.412 \sum_{p\ne5}1/p^2\approx 0.412∑p=51/p2≈0.412. Your “missing a residue class modulo 333” observation is exactly the sort of extra rigidity that reduces SPS_PSP and can move the needle. If you can systematically harvest such missing classes for several small primes (e.g. for 3,13,17,29,⋯≡1 (mod 4)3,13,17,29,\dots\equiv1\ (\mathrm{mod}\ 4)3,13,17,29,⋯≡1 (mod 4) in the cross terms), the inequality (Main) becomes strong enough that the conjectural N/25N/25N/25 bound comes within reach. =====

If you’d like, I can plug in explicit numerical choices of PPP (for instance P={13,17,29,37,41,… }P=\{13,17,29,37,41,\dots\}P={13,17,29,37,41,…}) together with a concrete hypothesis like “AsA_sAs misses one class modulo 333” and show exactly what quantitative αr↦αs\alpha_r\mapsto\alpha_sαr↦αs bound you get.