# Finding the Number of Positive integers such that $\lfloor{\sqrt{n}\rfloor} \mid n$

How does one find the no of positive integers such that find all possible numbers such that $$\lfloor{\sqrt{n}\rfloor} \mid n$$

What i did was to subsitute $n=t^{2}$ so that the equation becomes $\lfloor{t\rfloor} \mid t^{2}$ But this means that we want $t^{2} = k \lfloor{t\rfloor}$, where $k \in \mathbb{N}$. I don't really know what to do from here. By the way, this problem is in Apostol.

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There is a gap in the problem: If you look at $n=k^2$ (for $k\mathbb{N}$), then certainly $\sqrt{n}$ is an integer that divide $n$ - hence there are infinite number of such integers $n$. – AD. Oct 9 '10 at 13:13
@AD: Yes even i gave a thought on that! – anonymous Oct 9 '10 at 13:19
By the way, this problem is in Apostol. To be more specific, it is Exercise 21 in Chapter 3 in Apostol's Introduction to Analytic Number Theory. – Martin Sleziak Jan 14 '14 at 11:57

I suppose you want to find all possible numbers such that $\displaystyle [\sqrt{n}] \mid n$.

Assume $\displaystyle n$ is not a perfect square, then there must be some $\displaystyle k$ such that

$\displaystyle k^2 < n < (k+1)^2$. We have that $\displaystyle k = [\sqrt{n}]$.

The only numbers in the range $\displaystyle k^2 < n < (k+1)^2$ which are divisible by $k$ are $\displaystyle k^2+k$ and $\displaystyle k^2+2k$.

Thus the numbers $\displaystyle n$ such that $\displaystyle [\sqrt{n}] \mid n$ are of the form

$\displaystyle k^2, k^2+k, k^2+2k$

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Thanks a lot! – anonymous Oct 9 '10 at 15:32