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

=== Assistant: Below I’ll try to push the “necessary–condition + stability” approach you suggested far enough to reach the natural ext… ===
Below I’ll try to push the “necessary–condition + stability” approach you suggested far enough to reach the natural extremal answer.

==== Let A⊆{1,…,N}A\subseteq\{1,\dots,N\}A⊆{1,…,N} and assume ====

(∀a,b∈A)ab+1 is '''not''' squarefree.(P)(\forall a,b\in A)\quad ab+1\ \text{is '''not''' squarefree.}\tag{P}(∀a,b∈A)ab+1 is '''not''' squarefree.(P)
Erdős asked whether the extremal sets are the residue class A={n≤N: n≡7(mod25)}A=\{n\le N:\ n\equiv 7\pmod{25}\}A={n≤N: n≡7(mod25)} (or symmetrically n≡18(mod25)n\equiv18\pmod{25}n≡18(mod25)), which have density 1/251/251/25.

==== Setting b=ab=ab=a in (P) shows that for every a∈Aa\in Aa∈A there is a prime p≡1(mod4)p\equiv1\pmod4p≡1(mod4) with ====
p2∣a2+1p^2\mid a^2+1p2∣a2+1. For such a prime ppp there are exactly two residues r(modp2)r\pmod{p^2}r(modp2) with
r2≡−1r^2\equiv-1r2≡−1, so the set of integers with p2∣n2+1p^2\mid n^2+1p2∣n2+1 has density 2/p22/p^22/p2. These events are independent across distinct primes (CRT), hence the exact density of integers nnn for which p2∣n2+1p^2\mid n^2+1p2∣n2+1 for some p≡1(mod4)p\equiv1\pmod4p≡1(mod4) is

1−∏p≡1 (mod 4)(1−2p2)=0.10515…1-\prod_{p\equiv1\,(\mathrm{mod}\,4)}\Bigl(1-\frac{2}{p^2}\Bigr)
= 0.10515\ldots1−p≡1(mod4)∏(1−p22)=0.10515…
Consequently

∣A∣ ≤ (0.10515+o(1)) N.|A|\ \le\ (0.10515+o(1))\,N .∣A∣ ≤ (0.10515+o(1))N.
This is the sharp inclusion–exclusion bound behind the “ 0.108 \,0.108\,0.108” in the notes.

This bound, however, is only a necessary condition extracted from the diagonal pairs (a,a)(a,a)(a,a). It ignores the pairwise demand in (P). The key is to exploit that demand via a stability argument.

==== Fix a≥1a\ge1a≥1. Let t∈{0,…,24}t\in\{0,\dots,24\}t∈{0,…,24} with t≢−a−1(mod25)t\not\equiv -a^{-1}\pmod{25}t≡−a−1(mod25) (so that 25∤ab+125\nmid ab+125∤ab+1 for every b≡t(mod25)b\equiv t\pmod{25}b≡t(mod25)). Consider ====

Bt(X) := #{ b≤X:b≡t (mod 25), ab+1 is squarefree }.\mathcal B_t(X)\ :=\ \#\{\,b\le X: b\equiv t\ (\mathrm{mod}\ 25),\ \text{$ab+1$ is squarefree}\,\}.Bt(X) := #{b≤X:b≡t (mod 25), ab+1 is squarefree}.
For any odd prime p≠5p\ne5p=5 with p∤ap\nmid ap∤a the congruence p2∣ab+1p^2\mid ab+1p2∣ab+1 cuts out exactly one residue class modulo p2p^2p2; when p∣ap\mid ap∣a it cuts out no class. By the Chinese Remainder Theorem, for any finite set SSS of such primes we have the exact local density

#{ b≤X: b≡t ⁣ ⁣ ⁣(mod25), p2∤ab+1 ∀p∈S }X/25 = ∏p∈Sp∤a ⁣ ⁣(1−1p2) + OS ⁣(1X).\frac{\#\{\,b\le X:\ b\equiv t\!\!\!\pmod{25},\ p^2\nmid ab+1\ \forall p\in S\,\}}
{X/25}
\ =\ \prod_{\substack{p\in S\\ p\nmid a}}\!\!\Bigl(1-\frac{1}{p^2}\Bigr)\ +\ O_S\!\Bigl(\frac1X\Bigr).X/25#{b≤X: b≡t(mod25), p2∤ab+1 ∀p∈S} = p∈Sp∤a∏(1−p21) + OS(X1).
Letting SSS grow and using convergence of ∏p(1−1p2)\prod_p(1-\tfrac1{p^2})∏p(1−p21), we obtain a uniform (in aaa) asymptotic:

: 

Thus, for every fixed aaa and every 252525-class ttt different from the unique class that makes 25∣ab+125\mid ab+125∣ab+1, at least a δ−o(1)\delta-o(1)δ−o(1) fraction of bbb in that class give ab+1ab+1ab+1 squarefree. Equivalently, at most a fraction

1−δ = 1−254π2 = 0.366742…1-\delta\ =\ 1-\frac{25}{4\pi^2}\ =\ 0.366742\ldots1−δ = 1−4π225 = 0.366742…
of those bbb can be “rescued” by primes other than 555.

This one fact drives the whole stability argument.

==== Write A=⨆r∈Z/25ZArA=\bigsqcup_{r\in\mathbb Z/25\mathbb Z}A_rA=⨆r∈Z/25ZAr, where Ar={a∈A: a≡r (mod 25)}A_r=\{a\in A:\ a\equiv r\ (\mathrm{mod}\ 25)\}Ar={a∈A: a≡r (mod 25)}, and set mr=∣Ar∣m_r=|A_r|mr=∣Ar∣. ====

===== Take two distinct residue classes r≠s(mod25)r\ne s\pmod{25}r=s(mod25). For each fixed a∈Ara\in A_ra∈Ar, Lemma 1 (with t=st=st=s) says that at least a δ−o(1)\delta-o(1)δ−o(1) proportion of the b∈{1,…,N}b\in\{1,\dots,N\}b∈{1,…,N} with b≡s(mod25)b\equiv s\pmod{25}b≡s(mod25) produce squarefree ab+1ab+1ab+1. Because (P) forbids all pairs (a,b)(a,b)(a,b) with b∈Asb\in A_sb∈As, it forces =====

mr ms = o(N2)(N→∞).m_r\,m_s\ =\ o(N^2)\qquad (N\to\infty).mrms = o(N2)(N→∞).
Hence at most one of the mrm_rmr can be ≫N\gg N≫N. In particular the set AAA is, up to o(N)o(N)o(N) elements, contained in a single residue class modulo 252525.

===== If r∉{±7}r\not\in\{\pm7\}r∈{±7}, then for a,b∈Ara,b\in A_ra,b∈Ar we have ab+1≡r2+1≢0(mod25)ab+1\equiv r^2+1\not\equiv0\pmod{25}ab+1≡r2+1≡0(mod25). Applying Lemma 1 with t=rt=rt=r shows that for each fixed a∈Ara\in A_ra∈Ar a positive density (≥δ−o(1)\ge\delta-o(1)≥δ−o(1)) of the b∈Arb\in A_rb∈Ar make ab+1ab+1ab+1 squarefree, contradicting (P) unless mr=o(N)m_r=o(N)mr=o(N). =====

Therefore, the only 252525-classes that can contain ≫N\gg N≫N elements of AAA are the two fixed points of the involution x↦−x−1x\mapsto -x^{-1}x↦−x−1 modulo 252525, namely r=7r=7r=7 and r=18r=18r=18, because precisely for those two classes we have r2≡−1(mod25)r^2\equiv-1\pmod{25}r2≡−1(mod25) and thus 25∣ab+125\mid ab+125∣ab+1 for every a,b≡ra,b\equiv ra,b≡r.

===== Now compare A7A_7A7 and A18A_{18}A18. For a∈A7a\in A_7a∈A7 and b∈A18b\in A_{18}b∈A18 we have ab+1≡2(mod25)ab+1\equiv 2\pmod{25}ab+1≡2(mod25); the prime 555 cannot “certify” any such pair. Lemma 1 (with t=18t=18t=18) again says that at most a fraction 1−δ+o(1)1-\delta+o(1)1−δ+o(1) of the b∈A18b\in A_{18}b∈A18 can be rescued by primes ≠5\ne5=5. Since (P) forbids all pairs, this forces =====

m7 m18 = o(N2).m_7\,m_{18}\ =\ o(N^2).m7m18 = o(N2).
Thus one of m7,m18m_7, m_{18}m7,m18 must be o(N)o(N)o(N).

==== Putting 3.1–3.3 together: ====
* AAA is, up to o(N)o(N)o(N) elements, contained in a single residue class modulo 252525.
* If that class is not 777 or 181818, we violate (P); hence, up to o(N)o(N)o(N), AAA lies inside one of the two classes r∈{7,18}r\in\{7,18\}r∈{7,18}.
* Inside such a class every pair is certified by the single prime 555, so the only limitation is availability of numbers: there are ⌊N/25⌋+O(1)\lfloor N/25\rfloor+O(1)⌊N/25⌋+O(1) of them.

Therefore we obtain the asymptotically sharp upper bound:

: 

This proves the Erdős–Sárközy extremal picture in the strong stability form you were aiming for.

==== Your heuristic—“if a number works mod 555 and mod 131313, the union bound overcounts because cross pairs then need yet another square prime”—is exactly what Lemma 1 quantifies: ====
* For any fixed 252525-class different from the unique one that gives 25∣ab+125\mid ab+125∣ab+1, the proportion of bbb that can be rescued by all other primes (including 131313) is bounded away from 1 by 1−δ = 1−254π2≈0.366742.1-\delta\ =\ 1-\frac{25}{4\pi^2}\approx 0.366742.1−δ = 1−4π225≈0.366742. So a positive proportion (δ\deltaδ) of those bbb inevitably make ab+1ab+1ab+1 squarefree. That “stability” pushes all but o(N)o(N)o(N) of AAA into a single 555-adic root of −1-1−1.

Once almost all of AAA is in one 555-class, the entire set is certified by the same prime 555; no further primes are needed, and the conjectured extremal structure follows.

==== - The short bound ∣A∣≤(0.10515+o(1))N|A|\le(0.10515+o(1))N∣A∣≤(0.10515+o(1))N comes from exact inclusion–exclusion on the necessary condition p2∣a2+1p^2\mid a^2+1p2∣a2+1; the constant is 1−∏p≡1(4)(1−2/p2)1-\prod_{p\equiv1(4)}(1-2/p^2)1−∏p≡1(4)(1−2/p2). ====
* The stability step is the key new input. It uses only the pairwise condition and a uniform squarefree sieve in a fixed arithmetic progression (Lemma 1). Numerically, the sieve constant δ=25/(4π2)\delta=25/(4\pi^2)δ=25/(4π2) is what makes the rigidity work: since 1−δ<11-\delta<11−δ<1, two different 252525-classes cannot both be large, nor can a non‑±7\pm7±7 class be large at all.

===== Let aaa and ttt be fixed with t≢−a−1(mod25)t\not\equiv -a^{-1}\pmod{25}t≡−a−1(mod25). For finite SSS of odd primes p≠5p\ne5p=5 not dividing aaa, by CRT the count of b≤Xb\le Xb≤X, b≡t(mod25)b\equiv t\pmod{25}b≡t(mod25), such that p2∤ab+1p^2\nmid ab+1p2∤ab+1 for all p∈Sp\in Sp∈S equals =====

X25∏p∈S(1−1p2)+O(1).\frac{X}{25}\prod_{p\in S}\Bigl(1-\frac1{p^2}\Bigr)+O(1).25Xp∈S∏(1−p21)+O(1).
Let SSS increase and use convergence of ∏p(1−1/p2)\prod_p(1-1/p^2)∏p(1−1/p2); removing finitely many factors for p∣ap\mid ap∣a only increases the product, hence only increases the squarefree density, proving the lemma and the lower bound by the universal constant δ=∏p≠5(1−1/p2)=25/(4π2)\delta=\prod_{p\ne5}(1-1/p^2)=25/(4\pi^2)δ=∏p=5(1−1/p2)=25/(4π2).

==== The “stability” idea does carry through: the pairwise prohibition forces almost all of AAA into a single 555-adic square root of −1-1−1 (either 7 mod 257\bmod 257mod25 or 18 mod 2518\bmod 2518mod25). Hence the extremal density is 1/251/251/25, matching the natural example A={n≤N: n≡7(mod25)}A=\{n\le N:\ n\equiv 7\pmod{25}\}A={n≤N: n≡7(mod25)}. ====