gpt pro 5.4 attempt on Math-5859

Round-2 continuation for Erdős Problems 741, 749, 750

This is a genuine continuation of the previous round. I only use Round-1 material when explicitly indicated below, and I only re-prove statements when they are strengthened.

Conventions for this continuation: $\overline d(\cdot)$ and $\underline d(\cdot)$ denote upper and lower asymptotic density in $\mathbb N$, and when discussing Problem 750 it is enough to work with induced subgraphs, since deleting edges can only increase the independence number.

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Problem 741

1) ROUND-2 OBJECTIVE

Path (C): isolate a sharp structural obstruction and convert the Round-1 heuristic into a rigorous theorem.

More precisely, Round-1 identified the following missing input: a positive-density family of sums with many disjoint representations would force a good 2-colouring of $A$. In this round I prove that this mechanism is rigorous, and that in fact the "disjointness" requirement from Round-1 is automatic once one counts distinct unordered representations. This gives a genuine conditional theorem and shows what any counterexample must look like.

2) Round-1 FOUNDATION USED

I use exactly the following Round-1 points.

• Round-1 already handled the easy dense case $\underline d(A)>0$.
• Round-1 isolated the key missing lemma: if a positive-density set of sums has many pairwise disjoint representations, then a random 2-colouring should leave positive density in both monochromatic sumsets.

I do not re-prove those Round-1 facts here.

3) NEW INSIGHT / TOOL (ROUND-2)

For a fixed integer $n$, distinct unordered representations of $n$ are automatically disjoint.

Indeed, if \[ a+b=n,\qquad c+d=n,\qquad a\le b,\ c\le d, \] and $\{a,b\}\cap\{c,d\}\neq\varnothing$, then necessarily $\{a,b\}=\{c,d\}$. So the only real input needed is not "many disjoint representations", but simply "many distinct unordered representations".

That converts the Round-1 heuristic into a clean quantitative theorem.

4) ATTACK PLAN (ROUND-2)

The gap left by Round-1 was:

Can one rigorously deduce a good partition from a positive-density set of sums with many representations?

I will prove the following.

Theorem 741.1. Let \[ S_R:=\{n\in\mathbb N : r_A^{\ast}(n)\ge R\}, \] where $r_A^{\ast}(n)$ denotes the number of distinct unordered representations \[ n=a+b,\qquad a,b\in A,\qquad a\le b. \] If $\overline d(S_R)=\delta>0$ for some integer $R\ge 3$, then there exists a partition \[ A=A_1\sqcup A_2 \] such that \[ \overline d(A_1+A_1)\ge \left(1-2(3/4)^R\right)\delta>0, \qquad \overline d(A_2+A_2)\ge \left(1-2(3/4)^R\right)\delta>0. \]

This overcomes the precise obstacle named in Round-1, and it yields the structural corollary that any counterexample to Problem 741 must have \[ \overline d\left(\{n:r_A^{\ast}(n)\ge 3\}\right)=0. \]

5) WORK (ROUND-2)

For $n\in\mathbb N$, define \[ r_A^{\ast}(n):=\left|\{(a,b)\in A^2:a\le b,\ a+b=n\}\right|. \]

Lemma 741.2. If $(a,b)$ and $(c,d)$ are two distinct unordered representations of the same integer $n$, then \[ \{a,b\}\cap\{c,d\}=\varnothing. \]

Proof. Suppose $a+b=c+d=n$ and $(a,b)\neq(c,d)$ with $a\le b$ and $c\le d$. If $a=c$, then $b=d$, contradiction. If $a=d$, then $b=c$, and since $a\le b$ and $c\le d$, this again forces $(a,b)=(c,d)$. The other possibilities are symmetric. Hence distinct unordered representations are disjoint. $\square$

Now colour each element of $A$ independently red or blue with probability $1/2$ each. Let $A_1$ be the red elements and $A_2$ the blue elements.

Fix $n\in S_R$. By definition $n$ has at least $R$ distinct unordered representations \[ n=a_1+b_1=\cdots=a_R+b_R, \] and by Lemma 741.2 the pairs $\{a_i,b_i\}$ are pairwise disjoint. For each $i$, let $E_i^{(1)}$ be the event that the $i$-th representation is monochromatic red, and let $E_i^{(2)}$ be the event that it is monochromatic blue.

If $a_i\neq b_i$, then \[ \mathbb P(E_i^{(1)})=\mathbb P(E_i^{(2)})=\frac14. \] If $a_i=b_i$, then the probability is $1/2$ for each colour, which is only better for us. Therefore, in every case, \[ \mathbb P\left(n\notin A_1+A_1\right)\le \left(\frac34\right)^R, \qquad \mathbb P\left(n\notin A_2+A_2\right)\le \left(\frac34\right)^R. \] Hence, for \[ p_R:=1-\left(\frac34\right)^R, \] we have \[ \mathbb P\left(n\in A_1+A_1\right)\ge p_R, \qquad \mathbb P\left(n\in A_2+A_2\right)\ge p_R. \]

For $N\ge 1$, define \[ X_N:=\frac1N\sum_{\substack{n\le N\\ n\in S_R}} \left(1_{n\in A_1+A_1}+1_{n\in A_2+A_2}\right). \] Taking expectations and using the bound above, \[ \mathbb E[X_N]\ge \frac{2p_R}{N}|S_R\cap[1,N]|. \]

Choose a sequence $N_j\to\infty$ such that \[ \frac{|S_R\cap[1,N_j]|}{N_j}\to \delta=\overline d(S_R). \] Then \[ \limsup_{j\to\infty}\mathbb E[X_{N_j}]\ge 2p_R\delta. \] Because $0\le X_{N_j}\le 2$, reverse Fatou gives \[ \mathbb E\!\left[\limsup_{j\to\infty}X_{N_j}\right] \ge \limsup_{j\to\infty}\mathbb E[X_{N_j}] \ge 2p_R\delta. \] Therefore there exists at least one colouring for which \[ \limsup_{j\to\infty}X_{N_j}\ge 2p_R\delta. \]

Fix such a colouring. Set \[ R_j:=\frac1{N_j}\left|S_R\cap(A_1+A_1)\cap[1,N_j]\right|, \qquad B_j:=\frac1{N_j}\left|S_R\cap(A_2+A_2)\cap[1,N_j]\right|. \] Then $X_{N_j}=R_j+B_j$, so \[ \limsup_{j\to\infty}(R_j+B_j)\ge 2p_R\delta. \] Let \[ r:=\limsup_{j\to\infty}R_j,\qquad b:=\limsup_{j\to\infty}B_j. \] Since $\limsup(x_j+y_j)\le \limsup x_j+\limsup y_j$, we get \[ r+b\ge 2p_R\delta. \] Also $r\le \delta$ and $b\le \delta$, because $R_j,B_j\le |S_R\cap[1,N_j]|/N_j$. If, say, $r<(2p_R-1)\delta$, then \[ b>2p_R\delta-r>\delta, \] contradiction. Hence \[ r\ge (2p_R-1)\delta,\qquad b\ge (2p_R-1)\delta. \] Since $2p_R-1=1-2(3/4)^R$, we conclude \[ r\ge \left(1-2(3/4)^R\right)\delta, \qquad b\ge \left(1-2(3/4)^R\right)\delta. \] Finally, \[ \overline d(A_1+A_1)\ge r,\qquad \overline d(A_2+A_2)\ge b, \] because $S_R\cap(A_i+A_i)\subseteq A_i+A_i$. This proves Theorem 741.1. $\square$

Corollary 741.3. If Problem 741$Q1$ had a counterexample, then \[ \overline d\left(\{n:r_A^{\ast}(n)\ge 3\}\right)=0. \]

Proof. Apply Theorem 741.1 with $R=3$. $\square$

6) ADVERSARIAL VERIFICATION

• Diagonal representations. If $n=2a$ has the representation $(a,a)$, then the monochromatic probability is $1/2$, not $1/4$. I used only the weaker upper bound $3/4$ for the failure probability, so this causes no problem.
• Independence. The only independence used is across the $R$ chosen representations of a fixed $n$. Lemma 741.2 guarantees disjointness, so the colour events are independent.
• A single colouring for infinitely many $N_j$. I did not optimize separately for each $N_j$; the reverse-Fatou argument produces one colouring whose limsup is large.
• Upper density rather than lower density. The argument only uses a subsequence $N_j$ witnessing $\overline d(S_R)$, so upper density is the correct notion here.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round-2 proves a genuine new theorem:

If a positive-density set of sums has at least $3$ distinct unordered representations each, then Problem 741$Q1$ has a positive answer.

Equivalently, any counterexample must be essentially near-Sidon on a positive-density set of represented integers: the set of $n$ with at least $3$ unordered representations must have upper density $0$.

This is a substantive obstruction that was absent in Round-1.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

9) REFERENCES

• T. F. Bloom, Erdős Problem #741, erdosproblems.com.

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