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I was trying to solve the following problem, but realized that I probably reached a dead end:

Prove that the $n$th positive nonsquare integer is $n+\{ \sqrt{n} \}$ where $\{x\}$ is denotes the integer closest to $x$.

Nevertheless, I have a question related to the above about the closest integer function: $\{x\}$. On close inspection it is not obvious to me that there is a nice limititing behavior for this function.

One argument I was working on relied on showing that

$\{\sqrt{M+2}\}>\{\sqrt{M}\}$, or $\{\sqrt{M+2}\}=\{\sqrt{M}\}+1$ for finitely many $M$, which I believe is false but I am not sure if my idea is correct.

I think one way to disprove this is to show that given any $N \in \mathbb{N}$ we can find an integer $M$ such that $N \leq M$ and $\{\sqrt{M+2}\}=\{\sqrt{M}\}+1$ , (which I think is equivalent to the following) meaning we can find an integer $k$

such that $\sqrt{M}\ < k+\frac{1}{2}$, and $\sqrt{M+2}>k+\frac{1}{2} \implies $

$M < k^{2}+k+\frac{1}{4}$, and $M>k^{2}+k-\frac{3}{4} \implies$

$k^{2}+k-\frac{3}{4}<M<k^{2}+k+\frac{1}{4}$.

It looks like this is trivial choosing some very large $k > N$ where the steps can be reversed. Is this thinking correct?

I am also interested if the generalization for

$\{\sqrt{M+n}\}=\{\sqrt{M}\}+1$ where $n\in \mathbb{N}$ follows as easily.

Derek Luna
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You just want to have $$k^2\leq M<(k+1)^2\leq M+n<(k+2)^2$$ so essentially $(k+1)^2-n\leq M<(k+1)^2$ for $k>n$ so you have plenty solutions for large enough $k$