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Math-5975 · mathematics · Mathematics

Erdős Problem #896

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Statement

Fix $N\ge 1$. For subsets $A,B\subseteq\{1,2,\dots,N\}$ define, for each integer $m\ge 1$, \[ r_{A,B}(m):=\#\{(a,b)\in A\times B: ab=m\}. \] Define \[ F(A,B):=\#\{m\in\mathbb{N}: r_{A,B}(m)=1\}. \] Question: estimate \[ M(N):=\max_{A,B\subseteq\{1,\dots,N\}} F(A,B)\] as $N\to\infty$.

Restatement from Math-5975-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-5975-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5975

Erdős Problem 896: uniquely represented products

Statement and result.
For $A,B\subseteq[1,N]$, let $F(A,B)$ count products having exactly one ordered representation $ab$ with $(a,b)\in A\times B$. Write \[ \delta=1-\frac{1+\log\log2}{\log2},\qquad \Phi(x)=(\log x)^\delta(\log\log x)^{3/2}. \] We reconstruct the known order-of-magnitude answer \[ \max_{A,B\subseteq[1,N]}F(A,B)\asymp N^2/\Phi(N). \tag{1} \] The analytic inputs below are explicit external theorems; the collision-avoidance construction is proved in detail.
Ford's inputs and their applicable ranges.
Ford's multiplication-table theorem, Corollary 3 of The distribution of integers with a divisor in a given interval, gives \[ |\{ab:1\leq a,b\leq N\}|\asymp N^2/\Phi(N). \tag{2} \] This immediately bounds $F(A,B)$ from above. Let $H_1(x,y,z)$ count integers at most $x$ having exactly one divisor in $(y,z]$, and let $H$ count those having at least one. Ford's Corollary 2 and Theorem 4 imply, uniformly for sufficiently large $y\leq N^{1/3}$, \[ H_1(Ny,y,2y)\gg Ny/\Phi(y). \tag{3} \] Here Corollary 2 gives the estimate for $H$. For Theorem 4 choose its fixed parameter $c=1/4$. Its additional conditions are \[ y+1\leq2y\leq(Ny)^{5/8}, \qquad 2y^2\leq(Ny)^{3/4}, \] which hold uniformly in that range for large $N$. The ratio $H_1/H$ is then bounded below by a positive constant because $z/y=2$. These analytic theorems are not reproved here.
Prime labels and their scale.
Take the primes \[ {\cal Q}=\{p:N^{2/3}<p\leq N^{3/4}\}, \quad y_p=N/(2p),\quad R_p=(y_p,2y_p]\cap\mathbb N. \] Then $y_p$ lies between $N^{1/4}/2$ and $N^{1/3}/2$, so $\Phi(y_p)\asymp\Phi(N)$ uniformly, and $3$ applies. Every $r\in R_p$ satisfies $r<N^{1/3}<q$ for all $q\in{\cal Q}$, and $pr\leq N$. The label representation $pr$ is unique: if $pr=qs$ with distinct labels, then $p\mid s<p$, which is impossible. The standard prime reciprocal estimate gives \[ \sum_{p\in{\cal Q}}\frac1p =\log(9/8)+o(1). \tag{4} \] For example, $4$ follows from Mertens' theorem for primes.
Candidates with a unique divisor.
For each $p\in{\cal Q}$, consider the integers $v\leq Ny_p$ having exactly one divisor $r_p(v)\in R_p$. Set $b_p(v)=v/r_p(v)$; then $b_p(v)\leq N$. Discard those with $p\mid b_p(v)$. The discarded number is $O(y_p^2)$, because $b_p(v)=pt$ with $t\leq N/p=2y_p$, and there are only $O(y_p)$ possibilities for $r_p(v)$. By $3$, the retained family ${\cal V}_p$ satisfies \[ |{\cal V}_p|\gg Ny_p/\Phi(N). \tag{5} \] Indeed $y_p^2=o(Ny_p/\Phi(N))$ uniformly over the range.
Random thinning and uniqueness.
Keep each prime of ${\cal Q}$ independently with probability $1/2$, obtaining ${\cal P}$, and define \[ A=\{pr:p\in{\cal P},r\in R_p\},\qquad B=\{b\leq N:\text{no }p\in{\cal P}\text{ divides }b\}. \] Both sets lie in $[1,N]$. For a retained candidate $(p,v)$, the event $p\in{\cal P}$ and $b_p(v)\in B$ has probability at least $1/4$. In fact $b_p(v)\leq N$ has at most one prime factor from ${\cal Q}$, all of which exceed $\sqrt N$; if that prime exists it differs from $p$.

On this event the product \[ pv=(p r_p(v))b_p(v) \] has a unique representation in $A\times B$. For if $pv=qsc$, with $q\in{\cal P},s\in R_q,c\in B$, then $q\ne p$ would force $p\mid c$, because $s<p$. This contradicts the definition of $B$. Thus $q=p$; cancelling it makes $s$ the unique divisor of $v$ in $R_p$, determining both factors. This also proves that successful pairs $(p,v)$ yield distinct products: equality would give the same unique factorization and hence the same label and the same $v$.

Counting the successful products.
Taking expectations and using $4$--$5$ gives \[ \mathbb E F(A,B) \geq\frac14\sum_{p\in{\cal Q}}|{\cal V}_p| \gg\frac{N}{\Phi(N)}\sum_{p\in{\cal Q}}y_p =\frac{N^2}{2\Phi(N)} \sum_{p\in{\cal Q}}\frac1p \gg N^2/\Phi(N). \] Some choice realizes this bound, proving the lower half of $1$. Together with $2$ it proves the full order of magnitude.
Attribution and assessment.
The prime-label construction follows the GPT-5.5 Pro write-up prompted by Chojecki, as attributed on https://www.erdosproblems.com/896: https://www.ulam.ai/research/erdos896.pdf. The narrower prime range above makes the reciprocal sum immediate. Ford's primary paper is https://annals.math.princeton.edu/2008/168-2/p01. All sources were checked 24 September 2026. This is a complete reconstruction relative to the stated analytic inputs. A leading asymptotic constant is neither claimed nor needed for the known resolution.
Completion estimate.
COMPLETION: 100%

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Math-5975-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-5975

1) FORMAL RESTATEMENT

Fix $N\ge 1$. For subsets $A,B\subseteq\{1,2,\dots,N\}$ define, for each integer $m\ge 1$, \[ r_{A,B}(m):=\#\{(a,b)\in A\times B: ab=m\}. \] Define \[ F(A,B):=\#\{m\in\mathbb{N}: r_{A,B}(m)=1\}. \] Question: estimate \[ M(N):=\max_{A,B\subseteq\{1,\dots,N\}} F(A,B)\] as $N\to\infty$.

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement records bounds (attributed there to van Doorn): \[ (1+o(1))\frac{N^2}{\log N}\le M(N)\ll \frac{N^2}{(\log N)^\delta(\log\log N)^{3/2}} \quad\text{with }\delta\approx 0.086. \]

3) ATTACK PLAN

  • Give an explicit construction of $A,B$ with many uniquely represented products and prove uniqueness directly.
  • Give simple general identities/inequalities involving $F(A,B)$ (e.g. characterizing when $F=|A||B|$).
  • Do a brute-force reality check for small $N$.

4) WORK

PHASE 1 — FAST REALITY CHECK (brute force for small $N$).

I brute-forced $M(N)$ for $2\le N\le 10$ by enumerating all pairs of subsets $(A,B)$ (using bitmasks). The exact maxima found were: \[ \begin{array}{c|ccccccccc} N&2&3&4&5&6&7&8&9&10\\\hline M(N)&2&4&7&10&13&18&22&29&32 \end{array} \] (Examples of maximizers exist; one example for $N=10$ is $A=\{1,2,3,4,5,6,7,8\}$ and $B=\{1,7,8,9,10\}$, which yields $F(A,B)=32$.)

Problem-specific lemmas (injectivity and a clean lower-bound construction).

Lemma 896.1 (Injectivity criterion). For any finite sets $A,B$ of positive integers, \[ F(A,B)=|A||B| \quad\Longleftrightarrow\quad \text{the map }A\times B\to\mathbb{N},\ (a,b)\mapsto ab\text{ is injective}. \]

Proof. We always have $F(A,B)\le |A||B|$ since each $m$ with $r_{A,B}(m)=1$ accounts for exactly one pair $(a,b)\in A\times B$.

If the map is injective, then every product $ab$ occurs for exactly one pair, so $r_{A,B}(m)\in\{0,1\}$ for all $m$ and $F(A,B)=|A||B|$.

Conversely, if $F(A,B)=|A||B|$, then every pair $(a,b)\in A\times B$ must produce an $m=ab$ with $r_{A,B}(m)=1$ (otherwise we would have strictly fewer than $|A||B|$ values with exactly one representation). Thus distinct pairs cannot have the same product, i.e. the map is injective. $\square$

Lemma 896.2 (Large-prime construction gives many unique products). Let \[A:=\{1,2,\dots,\lfloor N/2\rfloor\},\qquad B:=\{p\ \text{prime}: N/2<p\le N\}. \] Then every $m$ of the form $m=ap$ with $a\in A$ and $p\in B$ has exactly one representation $m=ab$ with $a\in A$, $b\in B$. In particular \[ F(A,B)=|A||B|=\lfloor N/2\rfloor\left(\pi(N)-\pi(N/2)\right). \]

Proof. Fix $a\in A$ and $p\in B$ and let $m=ap$. We show that if $m=a'p'$ with $a'\in A$ and $p'\in B$, then necessarily $p'=p$ and $a'=a$.

Since $p\mid m$ and $p'$ is prime, either $p'=p$ or $p\mid a'$. But $p> N/2$ and $a'\le \lfloor N/2\rfloor$, so $p\nmid a'$. Hence $p'=p$, and then $a'=m/p=a$. Therefore $r_{A,B}(m)=1$ for every $m$ in the image of $A\times B$. Moreover the same argument shows the map $(a,p)\mapsto ap$ is injective, so by Lemma 896.1 we have $F(A,B)=|A||B|$. $\square$

5) VERIFICATION

  • Lemma 896.1: the only subtle direction is that $F=|A||B|$ forces every pair $(a,b)$ to map to a uniquely represented product. This follows because the total number of pairs is $|A||B|$, and a value $m$ with $r(m)=1$ accounts for exactly one pair.
  • Lemma 896.2: the key inequality is $p>N/2\ge a'$, so $p\nmid a'$. This is correct.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. The explicit choice $A=\{1,\dots,\lfloor N/2\rfloor\}$ and $B=\{p\text{ prime}: N/2<p\le N\}$ yields $F(A,B)=\lfloor N/2\rfloor(\pi(N)-\pi(N/2))$ (Lemma 896.2), i.e. on the order of $N\cdot(\#\text{primes in }(N/2,N])$. Exact brute-force maxima for $N\le 10$ are $M(2)=2, M(3)=4,\dots,M(10)=32$.

(ii) First gap (crisp). Prove an asymptotic formula (or a tight order of magnitude) for $M(N)=\max_{A,B\subseteq[N]}F(A,B)$ as $N\to\infty$. In particular, determine whether $M(N)\sim cN^2/\log N$ for some constant $c>0$.

(iii) Top 3 next moves.

  1. Improve the lower bound construction by allowing $B$ to contain integers with a guaranteed “unique large prime factor” beyond $N/2$, while enlarging $A$.
  2. Develop an upper bound mechanism: bound the number of uniquely represented products by controlling collisions $ab=a'b'$ via divisor statistics or energy methods.
  3. Computational: for moderate $N$ (say $N\le 18$) use ILP/SAT to find exact $M(N)$ and examine the structure of extremizers, to guess the right asymptotic constant and shape.

(iv) Minimal counterexample structure (if the conjectured order is wrong). If $M(N)$ were substantially larger than $N^2/\log N$, extremizers would need to avoid the ubiquitous multiplicative collisions in $[1,N]^2$; this would likely force $A$ and $B$ to concentrate on numbers with a “signature” prime factor in a narrow range, so that products carry an essentially unique large prime.

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