At most one divisor in $[\sqrt{n},\sqrt{n}+\sqrt[4]{n}]$ In one math book I'm reading there was the following problem, given as an exercise:

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 was unable to solve this problem myself, and author didn't leave any hint for this problem (which, presumably, means that it's quite simple).
Friend of mine has found a proof which uses AM-GM inequality, but I'm quite certain there is a proof avoiding it, because the convention in this book appears to be that every exercise can be solved only using methods already covered in earlier chapters, and this exercise is at the end of a chapter about divisibility (gcd, prime numbers and so on).
Is anyone aware of, or can come up with, such a solution of this problem?
Thanks in advance.
Edit: I thought it might be a good idea to post the solution my friend came up with. So here it is:
Suppose we have two distinct divisors in the interval in question, and let $d$ be their GCD. Write these divisors as $da,db$, so that $gcd(a,b)=1$. Wlog, let $a<b$. Because $\sqrt{n}\leq da<db\leq\sqrt{n}+\sqrt[4]{n}$, we know that $d\leq\sqrt[4]{n}$.
Because $da\mid n,db\mid n$, we get $dab=lcm(da,db)\mid n$. Write $n=dabk$. Because $db>da\geq\sqrt{n}$ we get $dadb>n=dabk$, so $d>k$.
From $a<b$ we know that $a\leq b-1$, similarly we know from $k<d$ that $k\leq d$. So $n=dabk\leq d(d-1)b(b-1)<(d-1/2)^2(b-1/2)^2$ (first use of AM-GM). We can also find $\sqrt{(b-1/2)(d-1/2)}\leq(b-1/2+d-1/2)/2=b/2+d/2-1/2$ (second use of AM-GM).
From last two inequalities we can find $\sqrt{n}+\sqrt[4]{n}<(b-1/2)(d-1/2)+b/2+d/2-1/2=bd-1/4<bd$, contradicting assumption that $bd\leq\sqrt{n}+\sqrt[4]{n}$.
I know that, as we just use AM-GM for two numbers we can easily prove the inequalities other way, but I feel like it's a bit of cheating here.
 A: Assume that $x\in \mathbb{R}$ is positive and $\sqrt{n} - x$ is an integer divisor of $n$. Now, consider $$\frac{n}{\sqrt{n} - x -1} - \frac{n}{\sqrt{n} - x}.$$ I claim that this is a little bit more than $1$. Not much more for small $x$, but a little. And, as we decrease the first denominator in steps of $1$, the difference keeps increasing by just a little more than $1$ at each step. For ${\sqrt{n} - x -k}$ to be a divisor, we need these little bits to add to some integer. At the very least, we need them to add to $1$ or more. That is, it is necessary that $$f(x, y) = \frac{n}{\sqrt{n} - y} - \frac{n}{\sqrt{n} - x} -(y - x) \geq 1$$ for $\sqrt{n} - x$ and $\sqrt{n} - y$ to be integer divisors of $n$ with $0\leq x<y$. Some calculus will tell you that $f$ increases as $x$ goes to $0$ and as $y$ increases.
Let $y_0$ be such that $$f(0, y_0) = \frac{n}{\sqrt{n} - y_0} - \sqrt{n} - y_0 = 1.$$ Therefore$$y_0^2 +y_0 -\sqrt{n} = 0$$ and $$ y_0= -\frac{1}{2} + \sqrt{\frac{1}{4} + \sqrt{n}} > -\frac{1}{2} + \sqrt[4]{n}$$ and finally $$\frac{n}{\sqrt{n} - y_0} = 1 + \sqrt{n} + y_0 > \frac{1}{2} + \sqrt{n} + \sqrt[4]{n}.$$ We conclude that there isn't enough space in the interval $[\sqrt{n}, \sqrt{n} + \sqrt[4]{n}]$ to get the increase in $f$ that is a necessary condition for there being two divisors.
A: Here is an argument that needs some justification of the assumptions that you want the factors to be close and $b,c$ to be a big as possible, but is quite elementary.  
Let the factors be $ab, (a+1)b$, so $b$ is their GCD. Let $n=a(a+1)bc$.  The $a+1$ comes from the fact that you want the factors to be as close as possible.  Then we can't have $\sqrt n=ab$ because we would have $n=a^2b^2$ and $(a+1)b$ can't be a factor, so $\sqrt n \lt ab$  or $a^2bc+abc \lt a^2b^2.$  This forces $c \lt b$  From the difference of the factors ($b$) being less than the fourth root of $n$ we get $b^4 \lt a^2bc+abc \lt a^2b^2+ab^2$, which requires $b \lt a$  
A little playing shows we want $b,c$ to be as large as possible to get close, so let $b=a-1,c=a-2$ then $n=(a+1)a(a-1)(a-2)=a^4-2a^3-a^2+2a$.  The factors are $a(a-1)$ and $(a+1)(a-1)=a^2-1$  Then $\sqrt n \lt a(a-1)-1, \sqrt[4]n \lt a, \text { and } \sqrt n +\sqrt[4]n \lt a^2-1$, which was the second factor.
A: This argument uses the notation above, and still uses the AM-GM inequality in one step:
$k\le d-1 \text{ and } a\le b-1\implies bd-ka\ge bd-(d-1)(b-1)=b+d-1,$ so
$\displaystyle b+d-1\ge a+d\ge2\sqrt{ad}\ge2\sqrt[4]{n}\implies \frac{bd(b+d-1)}{2bd}\ge\sqrt[4]{n}\implies\frac{bd(bd-ka)}{2bd}\ge\sqrt[4]{n}\implies$
$\displaystyle bd-\sqrt{n}=\frac{b^2 d^2-n}{bd+\sqrt{n}}=\frac{bd(bd-ka)}{bd+\sqrt{n}}>\frac{bd(bd-ka)}{2bd}\ge\sqrt[4]{n}\implies bd>\sqrt{n}+\sqrt[4]{n}$,
which gives a contradiction.
A: Using the notation given by the OP, we have that
$(a^2-bd)^2+a^2(b^2-a^2)>0\implies a^4-2a^2bd+b^2d^2+a^2b^2-a^4>0\implies b^2[d^2+a^2]>2a^2bd$,
so $\;b^2[a^2+ad+d^2]>2a^2bd+b^2ad>2a^2bd+a^3d\implies$
$\;\;\;\;b^2[a^2+2ad+d^2]>2a^2bd+ab^2d+a^3d=ad[a^2+2ab+b^2]\implies$
$\;\;\;\;b^2(a+d)^2>ad(a+b)^2\implies b(a+d)>(a+b)\sqrt{ad}\implies \color{blue}{bd(a+d)>(bd+ad)\sqrt{ad}}$.
Then $d-1\ge k \text{ and } b-1\ge a\implies bd-ka\ge bd-(d-1)(b-1)=b+d-1\ge a+d$, 
so$\displaystyle\;\;\; \color{blue}{bd-\sqrt{n}=\frac{b^2d^2-n}{bd+\sqrt{n}}=\frac{bd(bd-ak)}{bd+\sqrt{n}}>\frac{bd(a+d)}{bd+ad}>\sqrt{ad}>\sqrt[4]{n}}$.
Therefore $bd>\sqrt{n}+\sqrt[4]{n}$, which gives a contradiction.
