
2. In the real numbers, solve the system of equations

$$
\begin{aligned}
& \sqrt{\sqrt{x}+2}=y-2 \\
& \sqrt{\sqrt{y}+2}=x-2
\end{aligned}
$$

(Radek Horenský)


Solution. Let $(x, y)$ be any solution of the given system. Since $\sqrt{\sqrt{x}+2}$ is obviously positive, we have $y>2$ by the first equation. Similarly, the second equation implies $x>2$.

Now we prove that the numbers $x$ and $y$ must be equal.* We will use the observation that the function square root is increasing. If $x>y$, then

$$
\sqrt{\sqrt{x}+2}>\sqrt{\sqrt{y}+2}
$$

i.e. $y-2>x-2, y>x$, and that is a contradiction. The case $x\sqrt{2}$. For each such $s$, we square the equality to obtain $s+2=\left(s^{2}-2\right)^{2}$, which we rewrite in the form $s^{4}-4 s^{2}-s+2=0$. Note that this equation has a root $s=2$. This is confirmed by the decomposition

$s^{4}-4 s^{2}-s+2=s^{2}\left(s^{2}-4\right)-(s-2)=s^{2}(s-2)(s+2)-(s-2)=(s-2)\left(s^{3}+2 s^{2}-1\right)$, by which we now show that $s=2$ is the only root of the derived equation, that satisfies our condition $s>\sqrt{2}$. Indeed, for every $s>\sqrt{2}, s^{3}+2 s^{2}-1>0$ (this is even true for $s \geq 1$ ). The only satisfactory value of $s=2$ corresponds to the only solution $x=s^{2}=4$ of the equation (1), and therefore to the only solution $x=y=4$ of the given problem.[^2]

Conclusion. The given system of equations has a unique solution $(x, y)=(4,4)$.

REMARK. Let us give a second possible derivation of $x=y$.

We begin the new proof of the equality $x=y$ by taking squares of both equations

$$
\begin{aligned}
& \sqrt{x}+2=(y-2)^{2}, \\
& \sqrt{y}+2=(x-2)^{2} .
\end{aligned}
$$

By subtracting the second equation from the first we get

$$
\begin{aligned}
\sqrt{x}-\sqrt{y} & =(y-2)^{2}-(x-2)^{2}=((y-2)-(x-2))((y-2)+(x-2))= \\
& =(y-x)(x+y-4)=(\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x})(x+y-4) .
\end{aligned}
$$

Assuming that $x \neq y$, after dividing the two outer expressions by $\sqrt{y}-\sqrt{x} \neq 0$, we get

$$
-1=(\sqrt{y}+\sqrt{x})(x+y-4) .
$$

However, as we know, both $x$ and $y$ are greater than 2 , so the right-hand side of the last equation is positive, which is a contradiction.

ANOTHER SOLUtion. We again use the observation that both numbers $x$ and $y$ are greater than 2, and introduce the function $f:(2, \infty) \rightarrow(2, \infty), f(t)=\sqrt{t}+2$ for every $t>2$. The equations from the problem, rewritten in the form

$$
\begin{aligned}
& \sqrt{\sqrt{x}+2}+2=y \\
& \sqrt{\sqrt{y}+2}+2=x
\end{aligned}
$$

can then be written as a system of equations using the function $f$

$$
\begin{aligned}
& f(f(x))=y, \\
& f(f(y))=x .
\end{aligned}
$$

We see that its solutions are just pairs of the form $(x, y)=(x, f(f(x)))$, where the number $x$ satisfies the relation $f(f(f(f(x))))=x$. This is certainly satisfied in the case where $f(x)=x$. We show that the equality holds only in this case.

If $f(x)x$ we have*

$$
f(f(f(f(x))))>f(f(f(x)))>f(f(x))>f(x)>x .
$$

Thus, we have proved equivalence of $f(f(f(f(x))))=x$ and $f(x)=x$.

It remains to solve the equation $f(x)=x$ with unknown $x>2$, which is easy:

$$
f(x)=x \Leftrightarrow \sqrt{x}+2=x \Leftrightarrow 0=(\sqrt{x}-2)(\sqrt{x}+1) \Leftrightarrow \sqrt{x}=2 \Leftrightarrow x=4 \text {. }
$$

We arrive at the same conclusion as in the first solution.[^3]

REMARK. The reasoning about the function $f$ from the second solution can also be used to solve equation (1) from the first solution without considering the fourth degree equation. Indeed, the equation (1) can be written as the equation $f(f(x))=x$, which is, of course, equivalent to the simpler equation $f(x)=x$, due to the implications

$$
f(x)x \Rightarrow f(f(x))>f(x)>x
$$

which is justified in the same way as in the second solution. There we also solved the simplified equation $f(x)=x$.
