72nd Czech and Slovak Mathematical Olympiad

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 $x<y$ is eliminated similarly. The equality of $x=y$ is thus proved.

Let us therefore deal with the (only possible) case $x=y$ next. The original system of two equations is then obviously reduced to a single equation $$\sqrt{\sqrt{x}+2}=x-2.(1)$$ After substituting $s=\sqrt{x}$, when $x=s^2$, the equation (1) becomes $\sqrt{s+2}=s^2-2$, while obviously $s>\sqrt{2}$. For each such $s$, we square the equality to obtain $s+2=(s^2-2{)}^2$, which we rewrite in the form $s^4-4s^2-s+2=0$. Note that this equation has a root $s=2$. This is confirmed by the decomposition $$s^4-4s^2-s+2=s^2(s^2-4)-(s-2)=s^2(s-2)(s+2)-(s-2)=(s-2)(s^3+2s^2-1),$$ 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 $s>\sqrt{2}$, $s^3+2s^2-1>0$ (this is even true for $s\ge 1$). 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.

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

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Alternative solution.

We again use the observation that both numbers $x$ and $y$ are greater than 2, and introduce the function $f:(2,\infty )\to (2,\infty )$, $f(t)=\sqrt{t}+2$ for every $t>2$. The equations from the problem, rewritten in the form $$\begin{array}{rl}\sqrt{\sqrt{x}+2}+2& =y,\\ \sqrt{\sqrt{y}+2}+2& =x\end{array}$$ can then be written as a system of equations using the function $f$ $$\begin{array}{rl}f(f(x))& =y,\\ f(f(y))& =x.\end{array}$$ We see that its solutions are just pairs of the form $(x,y)=(x,f(f(x)))$, where the number $x$ satisfies the relation $f(f(f(x)))=x$. 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 a quadruple of inequalities $$f\left(f\left(f\left(f(x)\right)\right)\right)<f\left(f\left(f(x)\right)\right)<f(f(x))<f(x)<x$$ where the last inequality is obvious and every previous inequality is the consequence of the immediately following inequality and the fact that $f$ is increasing. Similarly in the case of $f(x)>x$ we have $$f\left(f\left(f(f(x))\right)\right)>f\left(f(f(x))\right)>f(f(x))>f(x)>x.$$ Thus, we have proved equivalence of $f\left(f\left(f(f(x))\right)\right)=x$ and $f(x)=x$.

It remains to solve the equation $f(x)=x$ with unknown $x>2$, which is easy: $$f(x)=x\iff \sqrt{x}+2=x\iff 0=(\sqrt{x}-2)(\sqrt{x}+1)\iff \sqrt{x}=2\iff x=4.$$ We arrive at the same conclusion as in the first solution.