## Task $3 / 66$

Solve the equation

$$
(\sqrt{2+\sqrt{3}})^{x}+(\sqrt{2-\sqrt{3}})^{x}=4
$$

If we set $y=(\sqrt{2+\sqrt{3}})^{x}$, then we get
$$
\frac{1}{y}=\frac{1}{(\sqrt{2+\sqrt{3}})^{x}}=\left(\frac{1}{\sqrt{2+\sqrt{3}}}\right)^{x}=\left(\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}} \sqrt{2-\sqrt{3}}}\right)^{x}=(\sqrt{2+\sqrt{3}})^{x}
$$

Thus, the given equation takes the form $y+\frac{1}{y}=4$. Since $y \neq 0$ always holds, it follows that $y^{2}-4 y+1=0$ and thus $y=2 \pm \sqrt{3}$.

1.) If $y=2+\sqrt{3}$, then $(\sqrt{2+\sqrt{3}})^{x}=2+\sqrt{3}$. The solution $x=2$ is obvious.

2.) If $y=2-\sqrt{3}$, then $(\sqrt{2+\sqrt{3}})^{x}=2-\sqrt{3}$. Furthermore, since

$$
\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3}
$$

it follows that $(\sqrt{2+\sqrt{3}})^{x}=\frac{1}{2+\sqrt{3}}$ with the solution $x=-2$.