<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] Since
$$
\overbrace{n^2-n+\sqrt{n^2-n}}^{\text{$n^2-n$ is too small}}\lt n^2-\frac12\iff\overbrace{\sqrt{n^2-n}\lt n-\frac12}^{n^2-n\,\lt\,n^2-n+\frac14}
$$
and
$$
\overbrace{n^2-n+1+\sqrt{n^2-n+1}}^{\text{$n^2-n+1$ is too big}}\gt n^2+\frac12\iff\overbrace{\sqrt{n^2-n+1}\gt n-\frac12}^{n^2-n+1\,\gt\,n^2-n+\frac14}
$$
there can be no integer $x$ so that $x+\sqrt{x}$ rounds to a square.
</s>