Let $0 < \varepsilon < 1$. A natural number $x$ is called an $\varepsilon$-square if $x = ab$, $a, b \in \mathbb{N}$ and $(1 - \varepsilon)b \le a \le b$. Denote by $f(N)$ the number of $\varepsilon$-squares on the interval $[1, N]$. Is it true that $\lim\_{N\to\infty} \frac{f(N)}{N} = 0$?


There is at least $cN$, for some constant $c > 0$, pairs of natural numbers $(a, b)$ such that $(1 - \varepsilon)b \le a \le b$ and $ab \le N$. But the product of such pairs can define the same $\varepsilon$-squares. For example, we can take $x = \prod\_{i = 1}^{2m}p\_i$, where $p\_i$ are prime numbers close enough to each other. Then there are $\binom{2m}{m}$ representations $x = ab$.



Density of a set of natural numbers which are the product of close numbers
If $x\leq N$ is an $\varepsilon$-square, then $x=ab$ and $a\leq b\leq \sqrt{x/(1-\varepsilon)}=:K$. So, $x$ should appear in $K\times K$ multiplication table. It is known, however, that most numbers below $K^2$ are not in this multiplication table, see [Erdős–Tenenbaum–Ford constant](https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Tenenbaum%E2%80%93Ford_constant). If $M(K)$ is the number of distinct elements in multiplication table $K\times K$, then for $K\to +\infty$ we have
$$
M(K)=\frac{K^2}{(\ln K)^{\delta+o(1)}},
$$
where $\delta\approx 0.086$. In particular, for any $\varepsilon>0$ there are $o(K^2)=o(N)$ numbers below $N$, which are $\varepsilon$-squares.


