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

Question

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.

Answer 1

$$\sqrt{n^2+n+1}= \sqrt{\left(n+\frac12\right)^2+\frac 34} > n+\frac 12$$ 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 $\rm{fractional part}{(n^2+n+1)}>0.5$

Answer 2

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 $x\in\Bbb Z^+$.

Answer 3

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.

Answer 4

$n+\sqrt{n}$ 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.