The number of elements in {1,2,...,a}.{1,2,...,b}, where $ab=n^2$ Let $A_{a,b}$=$\{pq:p\leq a,q\leq b\}$, where $ab=n^2$ and $n^2$ is fixed.
How large is $A_{a,b}$? Does $A_{a,b}$ attain its lower value when $a=b=n$?
The case when $a=b=n$ is settled by Ford, and a stronger conjecture by Erdos is asking if $|M(G)|\geq |M(K_{n,n})|$ in general for all bipartite $G$ such that $e(G)=n^2$. Yet is there any development about the above weaker problem?
 A: Ford's methods provide lower bounds for this "asymmetric" multiplication table problem that match the lower bounds for the standard multiplication table problem.
Define $H(x,y,z) := \#\{n \le x : \exists d \in (y,z]  \text{ with } d \mid n\}$. Ford proved that $$H(x,y,cy) \asymp \frac{x}{(\log Y)^\delta (\log\log Y)^{2/3}}$$ whenever $c>1$ and $\frac{1}{c-1} \le y \le \frac{x}{c}$, where $Y := \min(y,x/y)+3$.
Now take $a \ge 1$ and consider $|[a]\cdot [n^2/a]|$, which is $|A_{a,n^2/a}|$ in your notation. If $a \le 4$ or $a \ge n^2/4$, then clearly $|[a]\cdot[n^2/a]| = \Omega(n^2)$, so we may assume otherwise. We claim $$\left|[a]\cdot[\frac{n^2}{a}]\right| \ge H(\frac{n^2}{4},\frac{a}{4},\frac{a}{2}).$$ Indeed, if $b \le n^2/4$ has some $d \in (a/4,a/2]$ with $d \mid b$, then $b/d \le (n^2/4)/(a/4) = n^2/a$, so $b = d\cdot \frac{b}{d} \in [a]\cdot[n^2/a]$. Therefore, using $x = n^2/4, y = a/4, c=2$, we obtain $$\left|[a]\cdot[\frac{n^2}{a}]\right| \gg \frac{n^2/4}{(\log Y)^\delta(\log\log Y)^{2/3}},$$ where $Y = \min(a/4,n^2/a)+3$. Since $Y \le n^2$, $\log Y \ge 2\log n$, so we obtain the desired result $$\left|[a]\cdot[\frac{n^2}{a}]\right| \ge c'\frac{n^2}{(\log n)^\delta (\log\log n)^{2/3}}$$ for some absolute $c' > 0$.
