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


  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.
Assume $x\in\Bbb Z^+$.
If $x=m^2$ is a pefect square, then
$$m^2&lt;x+\sqrt x=m^2+m&lt;m^2+2m+1=(m+1)^2$$
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 $\lfloor x+\sqrt x\rfloor &lt;\lceil x+\sqrt x\rceil$.

Let $n\in\Bbb Z^+$ be maximal with $n(n+1)&lt;x$. 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&lt;(n+2)^2 $. 

Hence $ k^2=\lfloor x+\sqrt x\rfloor$
implies $k=n+1$, $x=n^2+n+1$. But $$(n+\tfrac12)^2=n^2+n+\tfrac14&lt;x$$ 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 $$ (n+\tfrac32)^2=n^2+3n+\tfrac 94&gt;x$$
implies that $x+\sqrt x$ should be rounded down, not up.

We conclude that $R(x+\sqrt x)$ is never a perfect square for $x\in\Bbb Z^+$.
$n+\sqrt{n}$ 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}  &amp; &lt;\sqrt{n}&lt;K+\frac{1}{2}\text{ and}\tag{A}\\
\left(  K-\frac{1}{2}\right)  ^{2}  &amp; &lt;n&lt;\left(  K+\frac{1}{2}\right)
^{2}\text{ and}\nonumber\\
K^{2}-K+\frac{1}{4}  &amp; &lt;n&lt;K^{2}+K+\frac{1}{4}\text{. Adding:}\tag{B}\\
K^{2}-\frac{1}{4}  &amp; &lt;n+\sqrt{n}&lt;K^{2}+2K+\frac{3}{4}\text{ or}\nonumber
\end{align}

\begin{equation}
K^{2}-\frac{1}{4}&lt;n+\sqrt{n}&lt;\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 $n+\sqrt{n}&lt;K^{2}+\frac{1}{2}$ so (from line C):

\begin{equation}
K^{2}-\frac{1}{4}&lt;n+\sqrt{n}&lt;K^{2}+\frac{1}{2}\text{.}\tag{D}%
\end{equation}

Multiplying line (A) above by $-1$ and adding:

\begin{align*}
-K-\frac{1}{2}  &amp; &lt;-\sqrt{n}&lt;-K+\frac{1}{2}\text{; adding to line (D):}\\
\left(  K^{2}-K\right)  -\frac{3}{4}  &amp; &lt;n&lt;\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 $\ n+\frac{1}{4}&lt;n$ or $\frac{1}{4}&lt;0$: a contradiction.

CASE 2: Suppose $G(n)=\left(  K+1\right)  ^{2}$. Then

\begin{align*}
\left(  K+1\right)  ^{2}-\frac{1}{2}  &amp; &lt;n+\sqrt{n}&lt;\left(  K+1\right)
^{2}+\frac{1}{2}\text{ and from line (A):}\\
-K-\frac{1}{2}  &amp; &lt;-\sqrt{n}&lt;-K+\frac{1}{2}\text{. Adding:}\\
\left(  K+1\right)  ^{2}-K-1  &amp; &lt;n&lt;\left(  K+1\right)  ^{2}-K+1\text{ or:}\\
\left(  K+1\right)  ^{2}-(K+1)  &amp; &lt;n&lt;\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  &amp; &lt;K^{2}+K+\frac{1}{4}\text{ or:}\\
K^{2}+2K+1-K  &amp; &lt;K^{2}+K+\frac{1}{4}\text{ or}\\
1  &amp; &lt;\frac{1}{4}\text{.}%
\end{align*}

This contradiction shows that $G(n)$ can never be a perfect (integer) square.