GPT 6 Astra Ultra attempt on Math-5975
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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 $v$.