<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] $n+\sqrt{n}$ never rounds to a square
By way of introduction, note first that $N^{2}+N$ can never be a
perfect square (when $N$ is a whole number) -- this is because perfect squares
of whole numbers must differ by (at least) $2N+1$.  If we let $n=N^{2}$ then
$N^{2}+N=n+\sqrt{n}$ and this can never be a perfect square; however, we can
ask if $n+\sqrt{n}$ can ever round to a perfect square -- i.e. is
there a perfect square which is within $\frac{1}{2}$ of it?
Let $G(n)=$ the nearest integer to $n+\sqrt{n}$, so that $G(n)$ lies within
$\frac{1}{2}$ of $n+\sqrt{n}$.$.$
Let $K$ be the nearest integer to $\sqrt{n}$, so
\begin{align}
K-\frac{1}{2}  & <\sqrt{n}<K+\frac{1}{2}\text{ and}\tag{A}\\
\left(  K-\frac{1}{2}\right)  ^{2}  & <n<\left(  K+\frac{1}{2}\right)
^{2}\text{ and}\nonumber\\
K^{2}-K+\frac{1}{4}  & <n<K^{2}+K+\frac{1}{4}\text{. Adding:}\tag{B}\\
K^{2}-\frac{1}{4}  & <n+\sqrt{n}<K^{2}+2K+\frac{3}{4}\text{ or}\nonumber
\end{align}
\begin{equation}
K^{2}-\frac{1}{4}<n+\sqrt{n}<\left(  K+1\right)  ^{2}-\frac{1}{4}\tag{C}%
\end{equation}
Thus: $n+\sqrt{n}$ lies between $K^{2}-\frac{1}{4}$ and $\left(  K+1\right)
^{2}-\frac{1}{4}$. It follows that $G(n)=K^{2}$ or $G(n)=\left(  K+1\right)
^{2}$ if $G(n)$ is to be a perfect square. We eliminate each case separately.
CASE 1: Suppose $G(n)=K^{2}$ Then $n+\sqrt{n}<K^{2}+\frac{1}{2}$ so (from line C):
\begin{equation}
K^{2}-\frac{1}{4}<n+\sqrt{n}<K^{2}+\frac{1}{2}\text{.}\tag{D}%
\end{equation}
Multiplying line (A) above by $-1$ and adding:
\begin{align*}
-K-\frac{1}{2}  & <-\sqrt{n}<-K+\frac{1}{2}\text{; adding to line (D):}\\
\left(  K^{2}-K\right)  -\frac{3}{4}  & <n<\left(  K^{2}-K\right)  +1\text{.}%
\end{align*}
Since $n$ and $K$ are whole numbers, we must have $n=K^{2}-K$. But from line
(B) above this gives $\ n+\frac{1}{4}<n$ or $\frac{1}{4}<0$: a contradiction.
CASE 2: Suppose $G(n)=\left(  K+1\right)  ^{2}$. Then
\begin{align*}
\left(  K+1\right)  ^{2}-\frac{1}{2}  & <n+\sqrt{n}<\left(  K+1\right)
^{2}+\frac{1}{2}\text{ and from line (A):}\\
-K-\frac{1}{2}  & <-\sqrt{n}<-K+\frac{1}{2}\text{. Adding:}\\
\left(  K+1\right)  ^{2}-K-1  & <n<\left(  K+1\right)  ^{2}-K+1\text{ or:}\\
\left(  K+1\right)  ^{2}-(K+1)  & <n<\left(  K+1\right)  ^{2}-\left(
K+1\right)  +2\text{.}%
\end{align*}
Since $n$ and $K$ are whole numbers and these are strict inequalities, we must
have $n=\left(  K+1\right)  ^{2}-(K+1)+1=(K+1)^{2}-K$. But from line (B) above:
\begin{align*}
\left(  K+1\right)  ^{2}-K  & <K^{2}+K+\frac{1}{4}\text{ or:}\\
K^{2}+2K+1-K  & <K^{2}+K+\frac{1}{4}\text{ or}\\
1  & <\frac{1}{4}\text{.}%
\end{align*}
This contradiction shows that $G(n)$ can never be a perfect (integer) square.
</s>