Solve $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$

$$\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor=x$$ I tried to solve this equation. First thing is $\lfloor \sqrt x +\sqrt{x+1}+\sqrt{x+2}\rfloor \in \mathbb{Z} $ so $x \in \mathbb{Z}$
second $$\sqrt x +\sqrt{x+1}+\sqrt{x+2} \geq \sqrt 0 +\sqrt{0+1}+\sqrt{0+2} \\\to x \in \mathbb{N}$$ so we can check $x=1,2,3,4,5,6,7,8,9,\ldots$ by a MATLAB program. I checked the natural numbers to find solution. I found $x=8,9$ worked here.
Now my question is about somehow an analytical solving of the equation, or another idea. Can any one help me? Thanks in advance.

Solution 1:

You can use inequalities to simplify your problem.

Since $\lfloor x \rfloor \le x$. Therefore we've

\begin{align} x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\ &\le \sqrt x+\sqrt {x+1}+\sqrt{x+2} \\ &\le 3\sqrt {x+2}\\ \end{align}

$$\implies x^2 \le 9(x+2) \; ; x \in \mathbb Z$$

This gives us the range $x \in [-1,10] \tag1$.

Also $\lfloor x \rfloor \ge x-1$. Therefore we've

\begin{align} x&= \lfloor \sqrt x+\sqrt {x+1}+\sqrt{x+2} \rfloor \\ &\ge \sqrt x+\sqrt {x+1}+\sqrt{x+2} \color{red}{-1}\\ &\ge 3\sqrt {x}-1\\ \end{align}

$$\implies x+1 \ge 3\sqrt x$$

$$\implies x^2-7x+1\ge 0 \, ; x \in \mathbb Z$$

This gives us $x \in (-\infty, 0]\cup [7,\infty)\tag2$

Taking intersection of $(1)$ and $(2)$, and taking care of domain, I.e. $x\ge 0$, we get

$$\color{blue}{x \in \{0,7,8,9,10\}}$$

Now you can check for $x=0,7,8,9,10$, which is quite easy now.

Solution 2:

This is a little plodding, but $x\in\mathbb{Z}$ and $x\ge0$ (required in order for $\sqrt x$ to be real) tells us

$$x\ge\lfloor\sqrt0+\sqrt1+\sqrt2\rfloor=\lfloor2.414\rfloor=2$$

which tells us

$$x\ge\lfloor\sqrt2+\sqrt3+\sqrt4\rfloor=\lfloor5.145\rfloor=5$$

which tells us

$$x\ge\lfloor\sqrt5+\sqrt6+\sqrt7\rfloor=\lfloor7.331\rfloor=7$$

which tells us

$$x\ge\lfloor\sqrt7+\sqrt8+\sqrt9\rfloor=\lfloor8.474\rfloor=8$$

Finally, we see that if $x\ge10$, then the function $f(x)=\sqrt x+\sqrt{x+1}+\sqrt{x+2}-x$ is negative, since $f(10)=-0.056997...$ and

$$f'(x)={1\over2\sqrt x}+{1\over2\sqrt{x+1}}+{1\over2\sqrt{x+2}}-1\lt{3\over2\sqrt 9}-1=-{1\over2}\lt0$$

so we cannot have $x\le\sqrt x+\sqrt{x+1}+\sqrt{x+2}$, which is required in order to have $x=\lfloor\sqrt x+\sqrt{x+1}+\sqrt{x+2}\rfloor$, for $x\ge10$. This leaves the two possibilities $x=8$ and $9$, which do solve the equation.

Solution 3:

First solve the real inequalities $$ x \le \sqrt{x}+\sqrt{x+1}+\sqrt{x+2} < x+1 \tag{1}$$ Solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}$ is numerically $9.8956$ and solution of $x = \sqrt{x}+\sqrt{x+1}+\sqrt{x+2}=x+1$ is numerically $7.9813$. So solution of (1) is $$ 7.9813 < x \le 9.8956 $$ Finally, assume $x$ is an integer. We get $x=8$ or $x=9$.