<s>[INST] Does $x+\sqrt{x}$ ever round to a perfect square, given $x\in \mathbb{N}$? I'll define rounding as $$R(x)=\begin{cases} \lfloor x \rfloor, & x-\lfloor x \rfloor <0.5 \\ \lceil x \rceil, & else\end{cases}$$

Does $x+\sqrt{x}$ ever round (to the nearest integer) to a perfect square, given $x\in \mathbb{N}$? 

For example, $7+\sqrt{7}=9.646...$ which rounds up, and $57+\sqrt{57}=64.549...$ which also round up. Also, $6+\sqrt{6}$ and $57+\sqrt{57}$ both round down. 
I think the positive integers $x$ such that $\lfloor x+\sqrt{x} \rfloor =k^2, k\in \mathbb{Z}$ are all of the form $n^2+n+1, n\in \mathbb{Z}^+$. The set of all $x$ begins as: $\{3, 7, 13, 21, 31, 43, 57, \dots \}$ and all those numbers are of the form $n^2+n+1$
I tried to find a pattern for whether the decimal part of $\sqrt{n^2+n+1}$ is less than $0.5$ or not, and I tried to modify $\sqrt{n^2+n+1}$ to $\sqrt{n^2+2n+1}=(n+1)^2$ but that didn't lead anywhere. 
Is there an algebraic proof/disproof of my above claim? 
Thanks. 
 [/INST] $$\sqrt{n^2+n+1}= \sqrt{\left(n+\frac12\right)^2+\frac 34} > n+\frac 12$$
And $$\sqrt{n^2+n+1} < \sqrt{n^2+2n+1} =n+1$$
$$\implies n+0.5 <\sqrt{n^2+n+1} <n+1$$
Thus $\rm{fractional part}{(n^2+n+1)}>0.5$
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