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. 
 A: Assume $x\in\Bbb Z^+$.
If $x=m^2$ is a pefect square, then
$$m^2<x+\sqrt x=m^2+m<m^2+2m+1=(m+1)^2$$
an so $x+\sqrt x=R(x+\sqrt x)$ cannot be a perfect square.
Thus we need only consider the case that $x$ is not a perfect square, which makes $\lfloor x+\sqrt x\rfloor <\lceil x+\sqrt x\rceil$.
Let $n\in\Bbb Z^+$ be maximal with $n(n+1)<x$. Then $x=n^2+n+d$ with $1\le d\le (n+1)(n+2)-n(n+1)=2n+2$.
This makes
$$\lfloor x+\sqrt x\rfloor = n^2+n+d+n=(n+1)^2+d-1.$$
This is $\ge (n+1)^2$ and $\le n^2+4n+2<(n+2)^2 $. 
Hence $ k^2=\lfloor x+\sqrt x\rfloor$
implies $k=n+1$, $x=n^2+n+1$. But $$(n+\tfrac12)^2=n^2+n+\tfrac14<x$$ implies that $x+\sqrt x$ should round up, not down.
Similarly, $k^2=\lceil x+\sqrt x\rceil = \lfloor x+\sqrt x\rfloor+1$ implies  $k=n+2$, $d=(n+2)^2+1-(n+1)^2=2n+2$, $x=n^2+3n+2$. 
But $$ (n+\tfrac32)^2=n^2+3n+\tfrac 94>x$$
implies that $x+\sqrt x$ should be rounded down, not up.
We conclude that $R(x+\sqrt x)$ is never a perfect square for $x\in\Bbb Z^+$.
A: $$\sqrt{n^2+n+1}= \sqrt{\left(n+\frac12\right)^2+\frac 34} > n+\frac 12$$
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 $\rm{fractional part}{(n^2+n+1)}>0.5$
A: 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.
A: $n+\sqrt{n}$ 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.
