gpt pro 5.4 attempt on Math-643

Problem 14 --- Round 3

1) ROUND-3 OBJECTIVE

I continue on path (A): proof track. A full resolution of Q1/Q2 still appears out of reach, so the Round-3 goal is to obtain a strictly stronger obstruction than the Round-2 near-Sidon bound. The specific target is to rule out an entire class of potential counterexamples: those for which the counting function \[ A(x):=|A\cap\{1,\dots,x\}| \] stays below a constant multiple of $\sqrt{x}$ with constant strictly smaller than a critical threshold. This is the most promising direction because Round 2 already ruled out sets that are too dense; the missing complementary step is to rule out sets that are too sparse.

2) Round-2 FOUNDATION USED

I use the following Round-2 results explicitly.

1. Round-2 Theorem 14.5 and Corollary 14.6: if \[ K(N):=|\{1,\dots,N\}\setminus B|, \] then \[ A(x)\le 2\sqrt{x}+K(2x). \] In particular, any hypothetical set with $K(N)=o(N^{1/2})$ satisfies \[ A(x)\le (2+o(1))\sqrt{x}. \]
2. Round-2 Corollary 14.8: if $A(x)\ge (2+\delta)\sqrt{x}$ on some scale, then $K(2x)\gg_\delta \sqrt{x}$. This ruled out the high-density side.

Round 3 builds on this by proving the complementary low-density obstruction.

3) NEW INSIGHT / TOOL (ROUND-3)

The new ingredient is a different counting functional: \[ P_A(N):=\left|\{\{a,b\}:a,b\in A,\ a\le b,\ a+b\le N\}\right|, \] the number of unordered pairs from $A$ with sum at most $N$.

Every integer $n\in B\cap[1,N]$ contributes one distinct pair to $P_A(N)$, so small $K(N)$ forces $P_A(N)$ to be close to $N$. The Round-3 innovation is to prove a sharp upper bound for $P_A(N)$ under a global envelope \[ A(x)\le C\sqrt{x}+D. \] The main term turns out to be \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] This yields a new critical constant \[ \sqrt{\frac{8}{\pi}}\approx 1.59577, \] strictly stronger than the Round-1 lower-density threshold $\sqrt2\approx 1.41421$ coming from the crude total-pair bound.

4) ATTACK PLAN (ROUND-3)

After Round 2, the exact remaining gap was the near-extremal regime in which \[ A(x)=O(\sqrt{x}) \quad\mbox{and}\quad K(N)=o(\sqrt N). \] Round 2 ruled out densities above $2\sqrt{x}+o(\sqrt{x})$, but did not exclude sparser quadratic-growth models.

The Round-3 plan is:

1. express $P_A(N)$ exactly in terms of the increasing sequence of elements of $A$;
2. prove that the envelope $A(x)\le C\sqrt{x}+D$ forces \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N); \]
3. combine this with the trivial lower bound \[ N-K(N)=|B\cap[1,N]|\le P_A(N) \] to show that if $C<\sqrt{8/\pi}$, then necessarily $K(N)$ is linear in $N$;
4. combine that new sparse-side obstruction with the Round-2 dense-side obstruction to isolate a narrower asymptotic density window for any hypothetical counterexample to Q2.

5) WORK (ROUND-3)

Retain the notation \[ K(N):=|\{1,\dots,N\}\setminus B|, \qquad A(x):=|A\cap\{1,\dots,x\}|. \] Define the low-sum pair count \[ P_A(N):=\left|\{\{a,b\}:a,b\in A,\ a\le b,\ a+b\le N\}\right|. \] Then every $n\in B\cap[1,N]$ determines a unique pair $\{a,b\}$ with $a+b=n$, and different integers $n$ give different pairs because their sums differ. Therefore \[ N-K(N)=|B\cap[1,N]|\le P_A(N). \]

Lemma 14.9 (exact decomposition of $P_A(N)$). Write \[ A=\{a_1<a_2<a_3<\cdots\}. \] Let $m:=A(N/2)$. Then \[ P_A(N)=\sum_{i=1}^{m}\left(A(N-a_i)-i+1\right). \]

Proof. Every pair counted by $P_A(N)$ has a unique smaller element $a_i\le N/2$. Fix such an index $i\le m$. The larger element $b$ must satisfy \[ b\in A, \qquad b\ge a_i, \qquad b\le N-a_i. \] The number of admissible $b$ is therefore \[ A(N-a_i)-A(a_i-1). \] Since $a_i$ is the $i$-th element of $A$, we have $A(a_i-1)=i-1$. Hence the number of pairs with smaller element $a_i$ is exactly $A(N-a_i)-i+1$. Summing over $i=1,\dots,m$ proves the formula. $\square$

Theorem 14.10 (sparse-envelope bound for low-sum pairs). Assume that for some fixed constants $C>0$ and $D\ge 0$ one has \[ A(x)\le C\sqrt{x}+D \qquad (x\ge 1). \] Then \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] Consequently, \[ N-K(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \]

Proof. Let $d:=\lceil D\rceil$ and write $m:=A(N/2)$. If $m\le d$, then trivially \[ P_A(N)\le \binom{m+1}{2}=O_{C,D}(1), \] so the stated estimate is immediate. Thus assume $m>d$.

By Lemma 14.9, \[ P_A(N)=\sum_{i=1}^{m}\left(A(N-a_i)-i+1\right). \] Split this at $d$: \[ P_A(N)=\sum_{i=1}^{d}\left(A(N-a_i)-i+1\right) +\sum_{i=d+1}^{m}\left(A(N-a_i)-i+1\right). \] The first sum has only $d=O_D(1)$ terms, and each term is at most $A(N)\le C\sqrt N+D$, so it contributes \[ O_{C,D}(\sqrt N). \]

Now fix $i>d$ and set $j:=i-d\ge 1$. Since $i=A(a_i)\le C\sqrt{a_i}+D\le C\sqrt{a_i}+d$, we get \[ j=i-d\le C\sqrt{a_i}, \qquad\mbox{hence}\qquad a_i\ge \frac{j^2}{C^2}. \] Also $i\le m=A(N/2)$ implies $a_i\le N/2$, so $j\le C\sqrt{N/2}$. Therefore \[ A(N-a_i)-i+1 \le C\sqrt{N-a_i}+D-i+1 \le C\sqrt{N-j^2/C^2}-j+1, \] because $D-d\le 0$.

Thus \[ P_A(N) \le O_{C,D}(\sqrt N) +\sum_{j=1}^{\lfloor C\sqrt{N/2}\rfloor} \left(C\sqrt{N-j^2/C^2}-j+1\right). \] Set \[ f_N(x):=C\sqrt{N-x^2/C^2}-x+1 \qquad\mbox{for }0\le x\le C\sqrt{N/2}. \] A direct differentiation gives \[ f_N'(x)=-\frac{x/C}{\sqrt{N-x^2/C^2}}-1<0, \] so $f_N$ is decreasing on its domain. Hence \[ \sum_{j=1}^{\lfloor C\sqrt{N/2}\rfloor} f_N(j) \le f_N(0)+\int_0^{C\sqrt{N/2}} f_N(x)\,dx. \] Now $f_N(0)=C\sqrt N+1=O_C(\sqrt N)$, while \[ \int_0^{C\sqrt{N/2}} f_N(x)\,dx = \int_0^{C\sqrt{N/2}}\left(C\sqrt{N-x^2/C^2}-x\right)dx +O_C(\sqrt N). \] Substitute $x=C\sqrt N\,t$. Then \[ \int_0^{C\sqrt{N/2}}\left(C\sqrt{N-x^2/C^2}-x\right)dx = C^2N\int_0^{1/\sqrt2}(\sqrt{1-t^2}-t)\,dt. \] Finally, \[ \int_0^{1/\sqrt2}(\sqrt{1-t^2}-t)\,dt = \left(\frac{\pi}{8}+\frac14\right)-\frac14 =\frac{\pi}{8}. \] Therefore \[ P_A(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] Combining with the basic inequality $N-K(N)\le P_A(N)$ gives \[ N-K(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] This proves the theorem. $\square$

Corollary 14.11 (low-density envelopes force a linear bad set). Suppose there exist constants $C<\sqrt{8/\pi}$ and $D\ge 0$ such that \[ A(x)\le C\sqrt{x}+D \qquad (x\ge 1). \] Then \[ K(N)\ge \left(1-\frac{\pi}{8}C^2\right)N-O_{C,D}(\sqrt N). \] In particular, \[ K(N)\gg_{C,D} N. \]

Proof. By Theorem 14.10 and the basic inequality $N-K(N)\le P_A(N)$, \[ N-K(N)\le \frac{\pi}{8}C^2N+O_{C,D}(\sqrt N). \] Rearranging gives the stated inequality. Since $C<\sqrt{8/\pi}$, the coefficient $1-\pi C^2/8$ is positive. $\square$

Corollary 14.12 (new asymptotic obstruction for Q2). If \[ K(N)=o(\sqrt N), \] then necessarily \[ \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \ge \sqrt{\frac{8}{\pi}}. \] Equivalently, for every $\delta>0$ there exist arbitrarily large $x$ such that \[ A(x)\ge \left(\sqrt{\frac{8}{\pi}}-\delta\right)\sqrt x. \]

Proof. Assume for contradiction that \[ L:=\limsup_{x\to\infty}\frac{A(x)}{\sqrt x}<\sqrt{\frac{8}{\pi}}. \] Choose $C$ with \[ L<C<\sqrt{\frac{8}{\pi}}. \] Then there exists $X_0$ such that $A(x)\le C\sqrt x$ for all $x\ge X_0$. Define \[ D:=\max_{1\le x<X_0}\left(A(x)-C\sqrt x\right). \] This is finite, and now \[ A(x)\le C\sqrt x + D \qquad (x\ge 1). \] Hence Corollary 14.11 applies and yields \[ K(N)\gg N, \] contradicting the assumption $K(N)=o(\sqrt N)$. Therefore \[ \limsup_{x\to\infty}\frac{A(x)}{\sqrt x}\ge \sqrt{\frac{8}{\pi}}. \] The equivalent formulation is immediate. $\square$

Corollary 14.13 (Round-2 $+$ Round-3 density window for any putative counterexample). If \[ K(N)=o(\sqrt N), \] then \[ \sqrt{\frac{8}{\pi}} \le \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \le 2. \]

Proof. The lower bound is Corollary 14.12. For the upper bound, apply Round-2 Corollary 14.6: \[ A(x)\le 2\sqrt{x}+K(2x)=(2+o(1))\sqrt x, \] so the limsup is at most $2$. $\square$

Interpretation. Round 2 ruled out sets that are too dense; Round 3 rules out sets that are too sparse. Thus any genuine counterexample to Q2 must live in the narrower asymptotic window \[ 1.59577\ldots=\sqrt{\frac{8}{\pi}} \le \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \le 2. \] This is a strict strengthening over the pre-Round-3 lower threshold $\sqrt2\approx1.41421$.

Equivalently, Round 3 rules out the entire class of constructions satisfying a global envelope \[ A(x)\le c\sqrt x+O(1) \qquad\mbox{with}\qquad c<\sqrt{\frac{8}{\pi}}. \] Any eventual counterexample to Q2 would therefore have to be both

1. dense enough to beat the new $\sqrt{8/\pi}$ threshold infinitely often, and
2. sparse enough to stay below the Round-2 ceiling $2\sqrt x+o(\sqrt x)$.

This sharply narrows the surviving regime.

6) ADVERSARIAL VERIFICATION

1. Distinct-sum issue. The inequality $N-K(N)\le P_A(N)$ uses only that different integers $n$ correspond to different pair-sums. There is no hidden injectivity assumption beyond equality of sums.
2. Diagonal pairs. Lemma 14.9 counts $a_i=a_i$ correctly because $a_i\le N/2$ implies $a_i\le N-a_i$, so the diagonal is included exactly once when admissible.
3. Envelope inversion. From $i=A(a_i)\le C\sqrt{a_i}+D\le C\sqrt{a_i}+d$ we indeed obtain $a_i\ge ((i-d)/C)^2$ for $i>d$. No monotonicity gap is hidden here.
4. Range of the new index $j=i-d$. Since $i\le A(N/2)$ implies $a_i\le N/2$, the bound $j\le C\sqrt{a_i}\le C\sqrt{N/2}$ is valid. This prevents the function $f_N(j)$ from entering the region where the square root would be undefined or where the bound would become vacuous.
5. Monotone sum-to-integral step. The derivative \[ f_N'(x)=-\frac{x/C}{\sqrt{N-x^2/C^2}}-1<0 \] shows $f_N$ is decreasing on $[0,C\sqrt{N/2}]$, so the upper Riemann-sum estimate used in Theorem 14.10 is legitimate.
6. Finite initial segment correction. The constant $D$ absorbs all irregularities coming from finitely many small values of $x$. Since $D$ is fixed in Theorem 14.10 and Corollary 14.11, the resulting error term remains $O_{C,D}(\sqrt N)$ and does not affect the main linear coefficient.
7. Compatibility with Round 2. There is no conflict with Round-2 Corollary 14.6. In fact, the two rounds are complementary: Round 2 gives the upper-density ceiling $2$, and Round 3 gives the new lower-density floor $\sqrt{8/\pi}$.
8. What remains unproved. The new argument still does not decide whether the surviving density window \[ \sqrt{\frac{8}{\pi}}\le \limsup A(x)/\sqrt x\le 2 \] actually contains an admissible infinite construction with $K(N)=o(\sqrt N)$, nor does it prove the conjectural lower bound $K(N)\gg_\varepsilon N^{1/2-\varepsilon}$. The near-extremal regime remains open.

No internal contradiction was found in the new work.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED)

The strict Round-3 advance is the new sparse-side obstruction: \[ A(x)\le C\sqrt x+O(1) \quad\mbox{with}\quad C<\sqrt{\frac{8}{\pi}} \Longrightarrow K(N)\gg N. \] Hence any putative counterexample to Q2 must satisfy \[ \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \ge \sqrt{\frac{8}{\pi}}. \] Combining this with the Round-2 dense-side obstruction yields the narrowed density window \[ \sqrt{\frac{8}{\pi}} \le \limsup_{x\to\infty}\frac{A(x)}{\sqrt x} \le 2. \]

So Round 3 rules out an entire class of counterexample strategies---all globally subcritical $c\sqrt x$ envelopes with $c<\sqrt{8/\pi}$---and isolates the remaining difficulty more sharply than Round 2 did.

The unresolved core is now even more specific: one must analyze the narrow quadratic-growth regime corresponding roughly to densities between $1.59577\sqrt x$ and $2\sqrt x$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 58%

9) REFERENCES

No external references were used in the new argument. The proof is self-contained and uses only the Round-2 results supplied in the prompt.