Most of the solutions presented here are wrong, because they didn't provide an adequate explanation of why the only solutions must occur at $x=y$ . This was either assumed, or incorrectly shown.

Comments posted by Peter B., Abhishek S. and C L. provides solutions based on calculus techniques like mean value theorem, contraction mapping, Rolle's theorem.

I would encourage you to read through the long thread of discussions, and learn from each other.

Peter presents the following solution which doesn't use Calculus.

Let $f(x) = \sqrt{ x + 1000} - \sqrt{x + 100}$ . We are given that $f(x) = y$ and $f(y) = x$ . Observe that

$\begin{aligned} f(y) - f(x) & = \sqrt{y+1000} - \sqrt{x+1000} +\sqrt{x+100} - \sqrt{y+100} \\ &= \frac{y-x}{\sqrt{y+1000} + \sqrt{x+1000}} -\frac{y-x}{\sqrt{y+100} + \sqrt{x+100}} \\ & =(y-x)\cdot G(x,y) \end{aligned}$

where $G(x,y) = \frac{1}{\sqrt{y+1000} + \sqrt{x+1000}} -\frac{1}{\sqrt{y+100} + \sqrt{x+100}}$

But since $-\frac{1}{20} \le G(x,y) \le 0$ for any $x,y\ge0$ , in order for $y-x = f(y) - f(x)$ , we must have $y = x$ . We then proceed to solve the equation, and show that only 1 solution exists.

The following is my solution.

Multiplying by the algebraic conjugate, the system is equivalent to

$\begin{cases} x( \sqrt{y+1000} + \sqrt{y+100}) = 900 \\ y ( \sqrt{x+ 1000 } + \sqrt{ x + 100}) = 900 . \\ \end{cases}$

As such, $x$ and $y$ must be positive. If $x > y$ , then we have

$\begin{aligned} 900 & = \sqrt{x^2 y + 1000 x^2 } + \sqrt{ x^2 y + 100x^2 } \\ & > \sqrt{xy^2 + 1000y^2 } + \sqrt{ xy^2 + 100 y^2 } \\ & = 900 , \end{aligned}$

which is a contradiction. A similar inequality holds for $x < y$ , which means we must have $x = y$ .

Thus, $x( \sqrt{x+1000} + \sqrt{x+100}) = 900$ . Observe that the LHS is an increasing function on positive values of $x$ , and is equal to 0 when $x= 0$ and increases without bound. Hence, exactly 1 solution exists.

Note: This easily generalizes to other (positive) values. When $N$ is negative, we can't just substitute $x=0$ in. Instead, the smallest starting value is $x = -N$ , which gives us a value of $|N| \sqrt{M+|N|}$ .

Let $f(t) = \sqrt{t + 1000} - \sqrt{t + 100}.$ We can write

$\begin{aligned} f(t) &= \sqrt{t + 1000} - \sqrt{t + 100} \\ &= \frac{(\sqrt{t + 1000} - \sqrt{t + 100})(\sqrt{t + 1000} + \sqrt{t + 100})}{\sqrt{t + 1000} + \sqrt{t + 100}} \\ &= \frac{(t + 1000) - (t + 100)}{\sqrt{t + 1000} + \sqrt{t + 100}} \\ &= \frac{900}{\sqrt{t + 1000} + \sqrt{t + 100}}. \end{aligned}$

In this form, we can see that the function $f$ is decreasing.

We want to find $x$ and $y$ such that $x = f(y)$ and $y = f(x)$ .

If $x < y$ , then $f(x) > f(y)$ . But $x = f(y)$ and $y = f(x)$ , so $f(y) > f(x)$ , contradiction. Similarly, $x > y$ also leads to a contradiction. Therefore, if $x = f(y)$ and $y = f(x)$ , then $x = y$ . In other words, $f(x) = x$ .

Consider the continuous function $g(x) = f(x) - x$ . We see that $g(0) = f(0) = \sqrt{1000} - \sqrt{100} > 0,$ and

$\begin{aligned} g(24) &= f(24) - 24 \\ &= \sqrt{1024} - \sqrt{124} - 24 \\ &= 8 - \sqrt{124} < 0, \end{aligned}$

so by the Intermediate Value Theorem, there exists an $x_0 \in [0,24]$ such that $g(x_0) = 0$ , or $f(x_0) = x_0$ . Furthermore, the function $g(x) = f(x) - x$ is decreasing, so $x = x_0$ is the unique solution of $f(x) = x$ .

Hence, the system $x = f(y)$ and $y = f(x)$ has exactly one solution, namely $x = y = x_0$ .

I solved this equation graphically

It was evident that due to symmetry, the solution would lie on $x=y$ and it is evident that $x >= -100$ and $y >= -100$

So I wrote a MATLAB code as follows :

$x1 = [-100:200];$

$y1 = sqrt(x1+1000) - sqrt(x1 + 100);$

$y2 = x1 = [-100:200];$

$x2 = sqrt(y2+1000) - sqrt(y2 + 100);$

%Time to plot

$plot(x1,y1);$

$hold all;$

$plot(x2,y2);$

voila! As expected, the point lies on $x=y$ and there exist only $1$ such point

P.S. I myself dont support graphical solutions much, so I apologize if I offend any one of you. After Spending much time on the question, out of curiosity, I chose the Graphical Method ^_^

It can be seen that no matter how, it is impossible for x and y to take on distinct values. Hence, x=y and we have x= \sqrt{x+1000} - \sqrt{x+100}. Solving this, we get a quartic equation with four roots, x= -37.4, -22.5, 21.0 and 44.3. However, three of them are invalid. The proof is as follows: One must observe that since the square root of any number is positive, \sqrt{x+1000} > \sqrt{x+100} for all x. Hence, x must be positive and we reject the two negative roots. Manipulating the equation gives us x+ \sqrt{x+100} = \sqrt{x+1000} Squaring it gives us x^2 + 2x\sqrt{x+100} =900 then 2x\sqrt{x+100} = 900 -x^2. Since 2x\sqrt{x+100} >0, 900- x^2 >0 and we have 0<x<30. Hence we reject x= 44.3 leaving us with only one root and hence one ordered pair.

$\sqrt{x+100}$ is a curve (which looks like half of a parabola, like $\sqrt{x}$ ) lying in the first quadrant starting from the point $\{0,\sqrt{100}\}$ . Similarly $\sqrt{x+1000}$ has the same structure but starts from $\{0,\sqrt{1000}\}$ . Since both functions are monotonic increasing, their difference is also monotonic increasing, and hence $y=\sqrt{x+1000}-\sqrt{x+100}$ is a curve of similar shape but passes between the two individual curves and starts from $\{0,\sqrt{1000}-\sqrt{100}\}$ . Now for the curve $x=\sqrt{y+1000}-\sqrt{y+100}$ , it is simply a rotated version (also lying in the first quadrant and starting from the point $\{\sqrt{1000}-\sqrt{100},0\}$ and heading towards $y=$ infinity. Clearly, both the curves meet at a single point. And hence the solution is $1$

We can easily see that (just by looking at one of the two roots in each equation) we have two identical parabola (half of them) with only one point of intersection on the line y=x Then we only take the same quantity from each of them, so the intersection will remain one

This is not rigours solution, but it's how I guess the answer

$y=\sqrt{x+1000}-\sqrt{x+100}, f'(x)<0$ where $-100 \leq x \leq \infty$ , Stricly Decreasing.

$x=\sqrt{y+1000}-\sqrt{y+100}, f'(y)<0$ where $-100 \leq y \leq \infty$ , this graphic same with rotation fitst graphic irotation $180^{\circ}$ counterclockwise.