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

Question

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

Answer 1

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.

Answer 2

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.

Answer 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 $x=9$.

Answer 4

Side comment: $\frac{\sqrt{m-1} + \sqrt{m+1}}2 \ge \sqrt{\sqrt{m^2 - 1}}>\sqrt m$ by $AM-GM$ theorem.

So $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} \approx 3\sqrt{x+1}$ but slightly larger.

$x \le \sqrt{x} + \sqrt{x+1} + \sqrt{x+2}< 3\sqrt{x+1} < x+1$ implies $x^2 < 9x + 9 < x^2 + 2x +1$ (the first inequality is definite, the second is aproximate).

$x^2 <9x+9$ (which is definitely true) implies $x^2 - 9x -9 = (x -\frac {9+\sqrt{81 + 4*9}}2)(x -\frac {9-\sqrt{81 + 4*9}}2) < 0$ which implies $x < \frac {9+\sqrt{81 + 4*9}}2= \frac {9+3\sqrt{13}}2\approx 9.9$ but is *definitely less than $10$. $9$ is a definite upper limit of $x$

$9x + 9 < x^2 + 2x + 1$ (which is only approximate) implies $x^2 - 7x -8 = (x-8)(x+1) > 0$ implies $x > 8$ (or $x < 1$ which would not be possible.)

So $9$ is a solution and $8$ is possible if $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} < 9 \le 3\sqrt{x+1}$ ... which, ... actually is the case for $x= 8$. (As $3\sqrt{8+1} = 9$ exactly.)

So solutions are $8$ and $9$.

Hmmm.... I haven't actually shown it fails for $x < 8$ but for that to hold the difference between $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$ and $3\sqrt{x+1}$ must be larger than $1$ and that's .... well, clearly not true for $x \approx 8$. We can just check that $7$ fails.

I'm also assuming the rate of which $x$ increases will "outstrip" $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2}$. Calculus verifies this.

Answer 5

It is clear that for some $x$ you have $LHS\lt x$ and for other $x$ you have $LHS\gt x$ and besides that you could have the equality for small enough $x$.

A straightforward calculation gives you $LHS\lt x$ for $x\lt8$ and $LHS\gt x$ for $\gt9$.

The solutions are $x=8$ and $x=9$ because $\sqrt8+\sqrt9+\sqrt{10}=8.9907...$ and $\sqrt9+\sqrt{10}+\sqrt{11}=9.4789...$

Answer 6

Here's a way to get bounds on $x$ for a generalized version of this.

We want to solve $x =\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor $.

Upper bound:

If $a = \max(a_k)$, then

$\begin{array}\\ x &=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\ &\le\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\ &=\lfloor n \sqrt{x+a} \rfloor\\ &\le n \sqrt{x+a} \\ \text{so}\\ x^2 &\le n^2(x+a)\\ \end{array} $

Therefore $x^2-n^2x+n^4/4 \le n^2(a+n^2/4) $ or $|x-n^2/2| \le n\sqrt{a+n^2/4} = (n/2)\sqrt{4a+n^2} $ so $x \le (n/2)(n+\sqrt{4a+n^2}) $.

For this case, $n=3, a=2$, so $x \le (3/2)(3+\sqrt{17}) \lt 10.7 $ so $x \le 10$.

Lower bound:

If $a = \min(a_k)$, then

$\begin{array}\\ x &=\lfloor \sum_{k=1}^n \sqrt{x+a_k} \rfloor\\ &\ge\lfloor \sum_{k=1}^n \sqrt{x+a} \rfloor\\ &=\lfloor n \sqrt{x+a} \rfloor\\ &\gt n \sqrt{x+a}-1 \\ \text{so}\\ (x+1)^2 &\gt n^2(x+a)\\ &= n^2(x+1+a-1)\\ \end{array} $

Therefore $(x+1)^2-n^2(x+1)+n^4/4 \gt n^2(a-1+n^2/4) $ or $|x+1-n^2/2| \gt n\sqrt{a-1+n^2/4} = (n/2)\sqrt{4a-4+n^2} $ so $x+1-n^2/2 \gt (n/2)\sqrt{4a-4+n^2} $ or $x+1-n^2/2 \lt -(n/2)\sqrt{4a-4+n^2} $.

Therefore $x \gt n^2/2-1+(n/2)\sqrt{4a-4+n^2} $ or $x \lt n^2/2-1-(n/2)\sqrt{4a-4+n^2} $.

For this case, $n=3, a=0$, so $(n/2)\sqrt{4a-4+n^2} =\frac32\sqrt{-4+9} =\frac32\sqrt{5} \approx 3.35 $ and $n^2/2-1 =3.5 $ so $x > 6.85$ or $x < 0$.

Therefore $x \ge 7$, so that, combining the bounds, $7 \le x \le 10$.

Answer 7

Just narrowed the possible solutions using inequality of integer at the both side

$\left[\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}+\sqrt{{x}+\mathrm{2}}\right]={x} \\ $ $\mathrm{3}\sqrt{{x}}\leqslant{x}\leqslant\mathrm{3}\sqrt{{x}+\mathrm{2}} \\ $ $\mathrm{9}{x}\leqslant{x}^{\mathrm{2}} \leqslant\mathrm{9}{x}+\mathrm{18} \\ $ ${x}\in{N}\Rightarrow{x}\in\left\{\mathrm{9},\mathrm{10}\right\} \\ $