Task B-4.3. For which natural numbers $n$ will the value of the expression $\left(\frac{1}{\sqrt{5-2 \sqrt{6}}}-\frac{1}{\sqrt{5+2 \sqrt{6}}}\right)^{n}$ be a rational number?
## Solution.

Determine the value of the number $x=\frac{1}{\sqrt{5-2 \sqrt{6}}}-\frac{1}{\sqrt{5+2 \sqrt{6}}}$,

$$
\begin{aligned}
x & =\frac{\sqrt{5+2 \sqrt{6}}-\sqrt{5-2 \sqrt{6}}}{\sqrt{5-2 \sqrt{6}} \cdot \sqrt{5+2 \sqrt{6}}}= \\
& =\frac{\sqrt{5+2 \sqrt{6}}-\sqrt{5-2 \sqrt{6}}}{\sqrt{25-24}}= \\
& =\sqrt{5+2 \sqrt{6}}-\sqrt{5-2 \sqrt{6}}
\end{aligned}
$$

Notice that $x>0$. Then,

$$
\begin{aligned}
x^{2} & =(\sqrt{5+2 \sqrt{6}}-\sqrt{5-2 \sqrt{6}})^{2}= \\
& =5+2 \sqrt{6}-2 \sqrt{5+2 \sqrt{6}} \sqrt{5-2 \sqrt{6}}+5-2 \sqrt{6}= \\
& =8
\end{aligned}
$$

Since $x>0$, it follows that $x=\sqrt{8}=2^{\frac{3}{2}}$.

Then the number $x^{n}=\left(2^{\frac{3}{2}}\right)^{n}$ is rational if and only if $n$ is an even number.