A divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\sqrt{n}<d<2\sqrt{n}$. Does there exist a positive integer with exactly $2020$ close divisors?
Problem
Source: BxMO 2020, Problem 4
Tags: number theory, BxMO, Benelux
03.05.2020 21:57
03.05.2020 22:26
Perhaps a more interesting question: Quote: A divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\sqrt{n}<d<1.99\sqrt{n}$. Does there exist a positive integer with exactly $2020$ close divisors?
30.06.2020 14:04
ThE-dArK-lOrD wrote: Perhaps a more interesting question: Quote: A divisor $d$ of a positive integer $n$ is said to be a close divisor of $n$ if $\sqrt{n}<d<1.99\sqrt{n}$. Does there exist a positive integer with exactly $2020$ close divisors? Actually, let $d$ be close if it is in $(\sqrt{n},(1+\epsilon)\sqrt{n})$ for some real $\epsilon >0$, and it is still possible for $n$ to have any number of close divisors, but my proof uses some pretty "wild" results First note that there is a prime in $(\sqrt{n},(1+\epsilon)\sqrt{n})$ if $n$ is large enough, say $n>N$ (a kind of strong Bertrands postulate) Proof: One can show by the prime number theorem that $\lim_{n\rightarrow \infty}\frac{\pi((1+\epsilon)n)-\pi(n)}{n/\log n} = \epsilon$ so $\pi((1+\epsilon)n)-\pi(n)\geq 1$ for $n>N$ where $N$ is some adequately big number Now knowing $f(n)$, take some prime $p\in (\sqrt{n},(1+\epsilon)\sqrt{n})$ and look at $f(np^2)$, we split this into cases. Take any positive divisor $d$ of $n$: $1)$ $d \in (p\sqrt{n},(1+\epsilon)p\sqrt{n})$ is not possible because $d\leq n$ but $p\sqrt{n}>n$ $2)$ $dp \in (p\sqrt{n},(1+\epsilon)p\sqrt{n}) \iff d \in (\sqrt{n},(1+\epsilon)\sqrt{n})$, i.e. $dp$ is a close divisor of $np^2$ iff $d$ is a close divisor of $n$ $3)$ $dp^2 \in (p\sqrt{n},(1+\epsilon)p\sqrt{n}) \iff d \in (\frac{\sqrt{n}}{p},(1+\epsilon)\frac{\sqrt{n}}{p})$, but $\frac{\sqrt{n}}{p}<1$ and $1<(1+\epsilon)\frac{\sqrt{n}}{p}<2$, so this is only possible if $d=1$ This tells us $f(np^2)=f(n)+1$ for $n>N$ and $p\in(\sqrt{n},(1+\epsilon)\sqrt{n})$, which is the main idea. But if $n>N$ then $np^2>N$ so we can continue doing this by the first claim. To finish, we only need to find some $n>N$ with $f(n)=1$ A famous recent result tells us that the gap between two consecutive primes is less then $70000000$ infinitely often, so the is some gap that happens infinitely often, lets say $k$. Then take $n=p(p+k)$ for some really large $p$ where $p+k$ is also prime. Rewrite $p+k=p(1+\delta)$. Then obviously $p<\sqrt{n}$, so $p(1+\delta)$ is the only candidate for a close divisor. For having $p(1+\delta)\in(\sqrt{n},(1+\epsilon)\sqrt{n})$ it is sufficient to have $p(1+\delta)<(1+\epsilon)p\sqrt{1+\delta} \iff \sqrt{1+\delta}<1+\epsilon$. This is the case for all $p$ sufficiently large, because $\delta=\frac{k}{p}$ gets arbitrarily small, which proves the claim, and thus finishes the proof $\blacksquare$