Erdős Problem #52

Erdős Problem #52

Definitions and mathematical statement

For a finite set $A\subset{\mathbb{Z}}$, write $A+A=\{a+b:a,b\in A\}$ and $A\cdot A=\{ab:a,b\in A\}$. The conjecture is \[ \forall\varepsilon>0\ \exists c_\varepsilon>0\ \forall A\subset{\mathbb{Z}}\text{ finite}:\qquad \max(|A+A|,|A\cdot A|)\ge c_\varepsilon|A|^{2-\varepsilon}. \] The constant is uniform in the elements and cardinality of $A$. The assertion says that addition and multiplication cannot both produce exceptionally small image sets. It does not require both sets to have nearly quadratic size.

Formulation and concept references: [S1].

Short English statement

Must a finite integer set expand almost quadratically under either addition or multiplication?

Research attempt

Current frontier and source audit.

The original source's introduction states the integer conjecture, while most of that paper investigates sparse-graph variants. We retain the full sum and product sets here. The real-number version is not interchangeable with this target. [S1]

Cushman's December 2025 manuscript, Theorem 1.3, gives the general-real exponent $4/3+10/4407-\epsilon$, hence also a lower bound for integer sets. Bloom--Sawin--Schildkraut--Zhelezov's May 2026 manuscript, Theorem 1.1, constructs counterexamples over real number fields whose degrees grow with the set size; its introduction explicitly leaves the integer case separate. These are identified as research manuscripts, not independently verified by this notebook. No part of either proof is needed for the lemmas below. [E1][E2]

Attack route.

A purely real-geometric route must now distinguish itself from the real counterexample construction. We instead combine exact integer divisibility with an analysis of how a large translation affects product collisions. A first height-dependent divisor bound becomes a bound uniform in both the translation and the step of a containing arithmetic progression. The remaining issue is to replace that one-dimensional progression hypothesis by consequences of small additive doubling without destroying multiplicative relations.

Attempt: the sharp translated-interval threshold.

For a finite positive set $A$, call its products injective on unordered pairs if $ab=cd$ with $a,b,c,d\in A$ implies $\{a,b\}=\{c,d\}$ as multisets; repetitions such as $a=b$ are allowed.

Lemma 1. For integers $D\geq1$ and $M>0$, every subset of $[M,M+D]\cap{\mathbb{Z}}$ has injective unordered products whenever \begin{equation}\label{wsref-0-65-71-3a-31-39-30-34-3a-74-68-72-65-73-68-6f-6c-64} M>\left\lfloor\frac{(D-1)^2}{4}\right\rfloor. \end{equation} For every $D\geq3$ the integer threshold is sharp for the full interval.

Proof. More generally consider the real translate $t+B$, where $t>0$ and $B\subseteq\{0,\ldots,D\}$. A collision implies \[ t(x+y-z-w)=zw-xy. \] If the sums agree, so do the products, and the roots of the corresponding quadratic show that the unordered pairs agree. Otherwise exchange the pairs so that $x+y=s$, $z+w=s+\delta$, with integer $\delta\geq1$. Then $t\delta=xy-zw>0$. The elementary bounds \[ xy\leq\frac{s^2}{4},\qquad zw\geq\max\{0,D(s+\delta-D)\} \] show that no such positive collision is possible when $\delta>D$. For $1\leq\delta\leq D$, the difference of these bounds is largest at $s=D-\delta$: below that value it is increasing, and above it its derivative is $s/2-D\leq0$. Consequently \[ t\leq\frac{(D-\delta)^2}{4\delta} \leq\frac{(D-1)^2}{4}. \] For integer $t=M$ this proves (eq:1904:threshold). If $D=2k+1$, the collision at $M=k^2$ is \[ (M+k)^2=M(M+2k+1). \] If $D=2k$, the collision at $M=k(k-1)$ is \[ (M+k-1)(M+k)=M(M+2k). \] Both have different unordered pairs, proving sharpness. $\square$

The lemma gives $|AA|=|A|(|A|+1)/2$ in that regime. It also shows why large absolute height, by itself, is not a plausible source of counterexamples: a sufficiently remote short interval has maximal unordered product growth.

Attempt: eliminate the step size by a gcd dichotomy.

Lemma 2. Fix $C>0$ and $\epsilon>0$. There is a constant $c_{C,\epsilon}>0$ such that every $n$-element integer set contained in \[ a+g\{0,1,\ldots,D\},\qquad g\in{\mathbb{Z}}\setminus\{0\},\quad 0\leq D\leq n^C, \] satisfies $|AA|\geq c_{C,\epsilon}n^{2-\epsilon}$. The constant is independent of $a$, $g$, and the elements of $A$.

Proof. Discard zero and take either the positive or the negative part with at least $(n-1)/2$ elements. Negating the latter does not change its product cardinality. This produces a positive subset $A_0$, with $m\geq(n-1)/2$, which can be written \[ A_0\subseteq M+g\{0,\ldots,D\},\qquad M>0,\quad g>0, \] after shrinking and reindexing the containing progression if necessary. The finitely small cardinalities can be absorbed into the constant. Divide all elements by $d=\gcd(M,g)$, putting $M'=M/d$, $g'=g/d$; this preserves product cardinality and gives $\gcd(M',g')=1$.

A product collision among $M'+g'x$ and its companions gives \begin{equation}\label{wsref-0-65-71-3a-31-39-30-34-3a-64-69-76-69-73-69-62-69-6c-69-74-79} M'(x+y-z-w)=g'(zw-xy). \end{equation} If $g'>2D$, divisibility forces $x+y-z-w=0$, because its absolute value is at most $2D$; then the unordered pairs are equal. If $M'/g'>(D-1)^2/4$, the real-translation version of Lemma 1 also gives injectivity. In either case $|A_0A_0|=m(m+1)/2$.

In the only remaining case, \[ g'\leq2D,\qquad M'\leq g'(D-1)^2/4, \] all normalized elements are positive integers at most $3D^3$ for $D\geq1$. Their products are at most $9D^6$. For completeness, for every $\delta>0$ the divisor function satisfies \begin{equation}\label{wsref-0-65-71-3a-31-39-30-34-3a-64-69-76-69-73-6f-72} d(v)\leq C_\delta v^\delta\qquad(v\geq1). \end{equation} Indeed, in $v=\prod p^{e_p}$, for sufficiently large primes $e+1\leq2^e\leq p^{\delta e}$; for each of the finitely many smaller primes the supremum of $(e+1)p^{-\delta e}$ is finite. Multiplying these bounds proves (eq:1904:divisor).

For each positive product $v$, at most $d(v)$ ordered pairs of positive integers have that product. Thus \[ m^2\leq |A_0A_0|\,C_\delta(9D^6)^\delta \leq |A_0A_0|\,C_\delta9^\delta n^{6C\delta}. \] Taking $\delta=\epsilon/(6C)$ and using $|AA|\geq|A_0A_0|$ proves the lemma. The case $D=0$ has at most one element and is harmless. $\square$

This proves a substantial restricted form without using $|A+A|$ at all. It includes sets of polynomial diameter, but also progressions with arbitrarily large step and translation. The argument is supplied as a self-contained deduction, without asserting that the result is new in the literature.

Attempt to remove the progression hypothesis.

Small additive doubling often suggests modeling a set by a generalized arithmetic progression. But neither an arbitrary additive model nor a translation is a multiplicative homomorphism. The simple divisibility identity (eq:1904:divisibility) involves only one step; with several steps, independent linear and quadratic expressions in those steps enter, and no one-dimensional coprimality argument eliminates all of them. The precise missing statement for this route is a uniform product-growth or sum-growth dichotomy for arbitrary integer sets of large normalized progression span, not a height estimate with an element-dependent constant.

There is also no automatic descent of the 2026 real construction to integers. For example, if $\alpha$ is an irrational algebraic integer with irreducible polynomial $f$, there is no unital ring homomorphism ${\mathbb{Z}}[\alpha]\to{\mathbb{Z}}$: the image of $\alpha$ would be an integer root of $f$. A finite encoding preserving the additions and multiplications expressing $f(\alpha)=0$ and fixing integers has the same obstruction. Substituting a large integer for $\alpha$ keeps some additive information but breaks that exact multiplicative relation. This does not rule out every possible finite combinatorial transfer; it rules out the naive ring-specialization step. The growing number-field degree in the manuscript is therefore an actual target distinction, not just a cosmetic normalization. [E2]

Stress test and exact computations.

The companion script checks the sharp boundary collisions and the absence of collisions above the threshold for $3\leq D\leq30$. It also tests the gcd dichotomy for many progressions, including steps larger than $2D$. These checks test the formulas; the proofs, not the finite search, establish the lemmas for every permitted parameter.

The set $A=\{1,2,4,\ldots,2^{n-1}\}$ has $|AA|=2n-1$, so no universal near-quadratic product-only lemma is possible. Its unordered sums, however, are all distinct by binary expansion, including the doubled terms, and $|A+A|=n(n+1)/2$. It is a test against the proposed intermediate assertion, not against the actual max-of-two-sets conjecture. It also has exponential span and lies outside the hypotheses just proved.

Outcome.

Outcome: Exploratory approach with unresolved gap.

Two exact lemmas were obtained: a sharp product-injectivity threshold for translated integer intervals, and near-quadratic product growth for sets in polynomial-length arithmetic progressions, uniformly in the progression's height and step. The remaining arbitrary-set reduction is unproved. No integer counterexample, full sum-product proof, or improvement to the unrestricted exponent is claimed.

Sources

• S1 Noga Alon, Omer Angel, Itai Benjamini, and Eyal Lubetzky, Sums and products along sparse graphs, Israel Journal of Mathematics 188 $2012$, 353-384. https://www.math.tau.ac.il/~nogaa/PDFS/sumprod2.pdf. Catalogue formulation source.
• E1 Adam Cushman, A Note on the Sum-Product Problem and the Convex Sumset Problem, arXiv:2512.13849v1 (15 December 2025), Theorem 1.3. https://arxiv.org/html/2512.13849v1.
• E2 Thomas F. Bloom, Will Sawin, Carl Schildkraut and Dmitrii Zhelezov, The sum-product conjecture is false for real numbers, arXiv:2605.28781v1 (27 May 2026), Theorem 1.1 and Sections 1--2. The real/algebraic-number target is not the integer target of this entry. https://arxiv.org/html/2605.28781v1.

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.

OBJECTIVE AND SOURCE REVIEW

Erdős #52; independent attempt, 2026-09-23. The two-fold sum-product target is $\max(|A+A|,|AA|)\gg_\varepsilon |A|^{2-\varepsilon}$ for every finite integer set $A$. I read the entire supplied 52.tex. Cushman's Theorem 1.3 does state the quoted $4/3+10/4407-\varepsilon$ exponent. Requoting it is not an independent improvement. Instead I isolate a regime where an elementary divisor argument reaches the conjectured exponent.

WORK: SETS OF POLYNOMIAL HEIGHT

Fix $C>0$ and $\varepsilon>0$. If $|A|=n\ge2$ and every $a\in A$ satisfies $|a|\le n^C$, then \[ |AA|\gg_{C,\varepsilon}n^{2-\varepsilon}. \] First, for every $\delta>0$ there is $K_\delta$ such that $d(m)\le K_\delta m^\delta$ for all positive integers $m$. To see this, factor $m=\prod p^{e_p}$. For $p^\delta\ge2$, $(e+1)\le2^e\le p^{\delta e}$. For each of the finitely many smaller primes the supremum of $(e+1)/p^{\delta e}$ over $e\ge0$ is finite. Multiplying these finitely many constants proves the assertion.

Remove zero from $A$ and write $B=A\setminus\{0\}$. For nonzero $t$, the number of ordered pairs $(a,b)\in B^2$ with $ab=t$ is at most $2d(|t|)$, because the sign and the absolute value of the first factor determine the pair. Since $|t|\le n^{2C}$, choosing $\delta=\varepsilon/(2C)$ gives \[ |B|^2=\sum_{t\in BB}r(t)\le |BB|\,2K_\delta n^\varepsilon. \] Now $|B|\ge n/2$ yields the result with constant $1/(8K_\delta)$.

ADVERSARIAL VERIFICATION AND REMAINING GAP

This proof includes negative elements and removes the exceptional zero product before counting divisors. Constants depend on the fixed height exponent $C$; taking $C=C(A)$ destroys the uniform assertion needed in the problem. Additive translation can reduce the diameter but does not preserve product-set cardinality, so one cannot normalize arbitrary sets into this regime. The proof is a standard divisor-bound special case, not a new unrestricted sum-product theorem. It leaves sets with superpolynomial integer height uncontrolled. The multi-fold variants mentioned in the official entry are also not settled here.

SOURCES CHECKED

Complete local source: gpt_pro_5.4/52.tex. Cushman, Theorem 1.3, https://arxiv.org/html/2512.13849v2; official indexed entry https://www.erdosproblems.com/52, checked 2026-09-23. The displayed special-case proof is self-contained.

FINAL

UNRESOLVED. Polynomial-height sets satisfy the desired exponent, but no uniform bound for all integer sets is proved.

COMPLETION ESTIMATE

Conservative subjective progress toward the unrestricted problem. COMPLETION: 15%