# Perfect squares between certain divisors of a number

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 MO website is for questions of math research. It's not clear to me that this question qualifies. How did you come across it? (Also, you are overloading the symbol $n$.) – Gerry Myerson Mar 24 at 11:38
• @Gerry Myerson My math professor says that this problem came up while he was trying to solve some open problem. I didn't bother asking him the details though. I would be very grateful if you can help me solve this problem. Or even any good idea would do. – user154024 Mar 24 at 12:07
• Simulposted to m.se, math.stackexchange.com/questions/3592817/… without notice to either site. That's an abuse. – Gerry Myerson Mar 24 at 12:12
• Doesn't $d$ being special imply $\sqrt{n/d}-\sqrt{d}<1$? – Sylvain JULIEN Mar 24 at 12:13

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} I'm sure the last part of the argument can be carried out more cleanly, but it checks out.

• Thanks for answering this question, so that I can stop racking my brain on it! – Sylvain JULIEN Mar 24 at 14:23

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 $$b. Then $$a 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 $$u and $$x 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 I'll comment a bit more about this at the end.

So we want to show

if $$ad=bc$$ with $$n^2 then $$d>(n+1)^2.$$

From the claim above, $$n^2 and $$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 $$a with $$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 $$d and all four factors between $$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 $$\max(l_{k},r_{k}), so for $$m(n)=O(1)$$ values of $$k$$, with $$\displaystyle{\lim_{n\to\infty}m(n)=0}$$.