Let $n$ be a positive integer. We will call a divisor $d(<\sqrt{n})$ of $n$ special if there exists no perfect squares between $d$ and $\frac{n}{d}$. Prove that $n$ can have at-most one special divisor.
My progress: I boiled down the problem to the following: Suppose $k^2\le a,b,c,d\le (k+1)^2$, then $ab=cd\implies \{a,b\}=\{c,d\}$. But I can't seem to prove this.
Arriving here isn't difficult so I am omitting any further details(one more reason being I am not sure if I am on the correct path).
This follows from a result I have asked about a few years ago, namely:
For any $n\in\Bbb N$ there is at most one divisor of $n$ in the interval $[\sqrt{n},\sqrt{n}+\sqrt[4]{n}]$.
I claim that if $d<\sqrt{n}$ is special, then $n/d\in[\sqrt{n},\sqrt{n}+\sqrt[4]{n}]$. Indeed, if this were not the case, we would have $n/d>\sqrt{n}+\sqrt[4]{n}$ and $$d<\frac{n}{\sqrt{n}+\sqrt[4]{n}}=\frac{n-\sqrt{n}}{\sqrt{n}+\sqrt[4]{n}}+\frac{\sqrt{n}}{\sqrt{n}+\sqrt[4]{n}}<\sqrt{n}-\sqrt[4]{n}+1.$$ Letting $x=\sqrt[4]{n}$, it remains to show that for any $x>2$ there is a perfect square between $x^2-x+1$ and $x^2+x$. By monotonicity, it is enough to show that if $x^2-x+1=k^2$ for some $k\in\mathbb N$, then $x^2+x\geq (k+1)^2$. The first equality resolves to $x=\frac{1}{2}(\sqrt{4k^2-3}+1)$, and we have $$x^2+x=x^2-x+1+2x-1=k^2+\sqrt{4k^2-3}<k^2+2k<(k+1)^2.$$ I'm sure the last part of the argument can be carried out more cleanly, but it checks out.
Here is an elementary approach. We want to rule out finding distinct $a,b,c,d$ in short interval (specifically $[n^2+1,n^2+2n]$) with $ad=bc.$ We may assume that $a$ is the smallest and that $b<c$. Then $a<b<c<d.$ I will show that $a+2\sqrt{a}+1\leq d$ so that if $n^2 \leq a$ then $(n+1)^2 \leq d.$
Claim: There are integers $u<v$ and $x<y$ with $$a,b,c,d=ux,uy,vx,vy.$$
Proof: Let $u=\gcd(a,b)$ so $a=ux$ and $b=uy$ with $\gcd(x,y)=1.$ Then $uxd=uyc$ so $xd=yc$ and thus (since $x$ and $y$ are co-prime) there is $v$ with $c=xv$ and $d=yv.$
ASIDE We might as well use $v=u+1$ and $y=x+1$ since $$a,b',c',d'=ux,u(x+1),(u+1)x,(u+1)(x+1)$$ give $ad'=b'c'$ with $a <b',c',d'\leq d.$ I'll comment a bit more about this at the end.
So we want to show
if $ad=bc$ with $n^2<a<b<c<d$ then $d>(n+1)^2.$
From the claim above, $n^2<a=ux$ and $d\geq (u+1)(x+1)=ux+u+x+1.$ But given $ux=a>n^2,$ we know $u+x\geq 2\sqrt{a}>2n.$ Thus $ux+u+x+1>n^2+2n+1$ as desired.
Consider this problem: Given $a$, find $a<b<c<d$ with $d$ minimal such that $ad=bc.$ The work above shows that the solution is to have $$a,b,c,d=ux,u(x+1),(u+1)x,(u+1)(x+1)$$ with $|u-x|$ minimal and that $d>a+2\sqrt{a}+1.$ If we allow $b=c$ then $$n^2\cdot (n+1)^2=(n^2+n)\cdot (n^2+n).$$ If we want $b < c$ then $(n^2-n)\cdot(n^2+n)=(n^2-1)\cdot(n^2).$ With $d-a\approx 2\sqrt{a}.$
If $ad=bc$ then $abcd$ is a perfect square. However this property is weaker and there are solutions such as $$a,b,c,d=2\cdot 120^2,3\cdot 98^2,30\cdot 31^2,5\cdot 76^2=$$ $$28800, 28812, 28830, 28880$$ with $d<a+\frac12\sqrt{a}$ and all four factors between $28561=169^2$ and $28900=170^2.$
Only a partial answer for now. Let $d_{-1}$ the largest divisor of $n$ below $\sqrt{n}$, $d_{-(k+1)}$ the largest divisor of $n$ below $d_{-k}$ and $d_{k}:=n/d_{-k}$. Let $r_{k}:=d_{k}-\sqrt{n}$ and $l_{k}:=\sqrt{n}-d_{-k}$.
Define the $k$-th "square root divisor span" of $n$ as $s_{k}(n):=\sqrt{d_{k}}-\sqrt{d_{-k}}$. The sequence $(s_{k}(n))_k$ is strictly increasing, and its general term equals $\sqrt{\sqrt{n}+r_{k}}-\sqrt{\sqrt{n}-l_{k}}$ which is greater or equal than $\sqrt{\sqrt{n}+k}-\sqrt{\sqrt{n}-k}$.
I think the condition in my comment, namely $s_{k}(n)<1$, is fulfilled only when $\max(l_{k},r_{k})<n^{1/4}$, so for $m(n)=O(1)$ values of $k$, with $\displaystyle{\lim_{n\to\infty}m(n)=0}$.
