gpt pro 5.4 attempt on Math-146

Erdős Problem #52 --- Round 4 (upgrade to the strongest located exponent)

1) ROUND-4 OBJECTIVE

Path pursued: (A) proof-track (partial, but strictly stronger than Round 3).

Let \[ M(A):=\max\{|A+A|,|AA|\}. \] Round 3 established the self-contained lower bound \[ M(A)\gg \frac{|A|^{4/3}}{(\log(2|A|))^{1/3}} \qquad (A\subset\mathbb Z\text{ finite}). \] The exact remaining gap after Round 3 was to remove the logarithmic loss and obtain a fixed power improvement beyond the Solymosi exponent $4/3$.

In this round we import the strongest theorem located in the current literature search. This yields \[ M(A)\gg_{\varepsilon} |A|^{\frac43+\frac{10}{4407}-\varepsilon} \qquad (A\subset\mathbb Z\text{ finite},\ \varepsilon>0), \] which is a strict improvement over Round 3.

2) ROUND-3 FOUNDATION USED

We treat the Round 3 output as correct and use the following items without reproving them.

1. Round 3 Theorem 5.5: for every finite nonempty $A\subset\mathbb Z$, \[ M(A)\gg \frac{|A|^{4/3}}{(\log(2|A|))^{1/3}}. \]
2. Round 3 conclusion on the remaining gap: the Solymosi-type multiplicative-energy route reaches the $4/3$ threshold up to a logarithmic loss, but does not by itself produce a fixed power improvement above $4/3$.

These are used only as the benchmark from which Round 4 must strictly advance.

3) NEW INSIGHT / TOOL (ROUND-4)

The new ingredient is the following external theorem.

Theorem 3.1 (Cushman, Theorem 1.3).

For every $\varepsilon>0$ there exists $c_{\varepsilon}>0$ such that for every finite set $X\subset\mathbb R$, \[ \max\{|X+X|,|XX|\}\ge c_{\varepsilon}|X|^{\frac43+\frac{10}{4407}-\varepsilon}. \] Since \[ \frac43+\frac{10}{4407}=\frac{1962}{1469}=1.335602\ldots, \] this gives a genuine fixed gain above $4/3$.

For comparison with the exponent quoted in the Round 3 context, \[ \frac{1962}{1469}-\frac{1270}{951} =\frac{232}{1397019}>0, \] so Theorem 3.1 is also strictly stronger than Bloom's $1270/951$ exponent.

4) ATTACK PLAN (ROUND-4)

Exact gap left after Round 3. Round 3 achieved only \[ M(A)\gg |A|^{4/3}(\log|A|)^{-1/3}, \] so the missing step was a bound of the form \[ M(A)\gg_{\eta}|A|^{4/3+\delta-\eta} \] for some fixed $\delta>0$.

Claims to prove now.

1. (i) State the strongest located external theorem that applies to arbitrary finite subsets of $\mathbb R$.
2. (ii) Verify that its hypotheses apply verbatim to finite integer sets.
3. (iii) Derive a fixed-exponent corollary with an absolute constant.
4. (iv) Prove that the new bound strictly dominates the Round 3 bound.

Why this overcomes the Round 3 obstacle. The new theorem bypasses the Solymosi-energy bottleneck identified in Round 3 by importing a more refined control/energy argument from the recent literature. Since the theorem is stated for all finite real sets, no additional structural reduction is required.

5) WORK (ROUND-4)

5.1 The upgraded sum-product bound for integers

We first record the external theorem in the form needed for the present problem.

Theorem 5.1 (external input).

For every $\varepsilon>0$ there exists $c_{\varepsilon}>0$ such that every finite set $X\subset\mathbb R$ satisfies \[ M(X):=\max\{|X+X|,|XX|\}\ge c_{\varepsilon}|X|^{\frac{1962}{1469}-\varepsilon}. \]

Corollary 5.2.

For every $\varepsilon>0$ there exists $c_{\varepsilon}>0$ such that every finite set $A\subset\mathbb Z$ satisfies \[ M(A)\ge c_{\varepsilon}|A|^{\frac{1962}{1469}-\varepsilon}. \]

Proof.

A finite set of integers is, in particular, a finite set of reals. Therefore Theorem 5.1 applies directly to $X=A\subset\mathbb Z\subset\mathbb R$, yielding \[ M(A)\ge c_{\varepsilon}|A|^{\frac{1962}{1469}-\varepsilon}. \] No further reduction is needed. $\square$

5.2 An explicit fixed-exponent corollary

The $\varepsilon$-form above immediately yields an absolute-constant version.

Corollary 5.3.

There exists an absolute constant $c>0$ such that every finite set $A\subset\mathbb Z$ satisfies \[ M(A)\ge c|A|^{\frac43+\frac{5}{4407}}=c|A|^{\frac{5881}{4407}}. \]

Proof.

Apply Corollary 5.2 with the fixed choice \[ \varepsilon:=\frac{5}{4407}. \] Then there exists a constant $c=c_{5/4407}>0$ such that \[ M(A)\ge c|A|^{\frac43+\frac{10}{4407}-\frac{5}{4407}} =c|A|^{\frac43+\frac{5}{4407}} =c|A|^{\frac{5881}{4407}}. \] This is exactly the claimed bound. $\square$

5.3 Strict comparison with Round 3

We now prove that the Round 4 theorem is asymptotically stronger than the Round 3 bound.

Proposition 5.4.

Fix any $\eta$ with \[ 0<\eta<\frac{10}{4407}. \] Then, as $n\to\infty$, \[ n^{\frac43+\frac{10}{4407}-\eta} \gg \frac{n^{4/3}}{(\log(2n))^{1/3}}. \] In particular, the Round 4 bound is a strict asymptotic improvement over the Round 3 bound.

Proof.

Consider the ratio \[ R(n):= \frac{n^{\frac43+\frac{10}{4407}-\eta}}{n^{4/3}(\log(2n))^{-1/3}} = n^{\frac{10}{4407}-\eta}(\log(2n))^{1/3}. \] Because $\frac{10}{4407}-\eta>0$, the power of $n$ tends to infinity. Hence $R(n)\to\infty$ as $n\to\infty$. Therefore \[ n^{\frac43+\frac{10}{4407}-\eta} \gg \frac{n^{4/3}}{(\log(2n))^{1/3}}, \] as claimed. $\square$

5.4 Precise statement of the Round 4 advance

Combining Corollary 5.2 with Proposition 5.4, we may summarise the new outcome as follows.

Theorem 5.5 (Round 4 upgrade).

For every $\varepsilon>0$ there exists $c_{\varepsilon}>0$ such that every finite set $A\subset\mathbb Z$ satisfies \[ \max\{|A+A|,|AA|\}\ge c_{\varepsilon}|A|^{\frac{1962}{1469}-\varepsilon}. \] Equivalently, \[ \max\{|A+A|,|AA|\}\gg_{\varepsilon}|A|^{\frac43+\frac{10}{4407}-\varepsilon}. \] This strictly strengthens the Round 3 bound.

6) ADVERSARIAL VERIFICATION

We check carefully that no hidden issue has been introduced.

1. Applicability of the external theorem. Theorem 5.1 is stated for every finite set $X\subset\mathbb R$. Our target sets satisfy $A\subset\mathbb Z\subset\mathbb R$, so the hypothesis is satisfied verbatim.

2. Dependence of constants on $\varepsilon$. The theorem only gives a constant $c_{\varepsilon}$ depending on $\varepsilon$. This is why Corollary 5.3 fixes one explicit value of $\varepsilon$ before claiming an absolute constant. No illicit uniformity in $\varepsilon$ is assumed.

3. Exponent arithmetic. We verified \[ \frac43+\frac{10}{4407}=\frac{1962}{1469}, \qquad \frac{1962}{1469}-\frac{1270}{951}=\frac{232}{1397019}>0. \] Thus the new exponent is genuinely larger than both $4/3$ and Bloom's $1270/951$ exponent.

4. Comparison with Round 3. The superiority over Round 3 is not merely numerical at a single point: Proposition 5.4 proves an asymptotic domination statement.

5. No hidden use of positivity or non-zeroness. Unlike the Round 2 and Round 3 self-contained arguments, the imported theorem is already stated for arbitrary finite real sets. Therefore no separate handling of signs or of the element $0$ is necessary.

The only genuine caveat is bibliographic rather than logical: the improvement comes from a very recent preprint. Within this round we have used it exactly as an external theorem, with its statement checked directly from the cited source.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 4 upgrades the proven lower bound from \[ \max\{|A+A|,|AA|\}\gg \frac{|A|^{4/3}}{(\log(2|A|))^{1/3}} \] from Round 3 to the stronger statement \[ \max\{|A+A|,|AA|\}\gg_{\varepsilon}|A|^{\frac{1962}{1469}-\varepsilon} \] for all finite $A\subset\mathbb Z$ and all $\varepsilon>0$. In particular, there is an absolute constant $c>0$ such that \[ \max\{|A+A|,|AA|\}\ge c|A|^{\frac{5881}{4407}}. \] The original Erdős--Szemerédi conjecture with exponent $2-\varepsilon$ remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 74%.

9) REFERENCES

Only references actually used or checked in this round are listed.

1. Adam Cushman, A Note on the Sum-Product Problem and the Convex Sumset Problem, arXiv:2512.13849v2, 2026. In particular, Theorem 1.3.
2. Thomas F. Bloom, Control and its applications in additive combinatorics, arXiv:2501.09470, 2025. In particular, Theorem 4.
3. T. F. Bloom, Erdős Problem #52, ErdosProblems.com, accessed 2026-03-07.