10.1. Determine all natural numbers $n$ for which the number $\sqrt{n+\sqrt{n-2}}$ is rational.
Solution: Suppose for some value of the number $n, n \geq 2$ we have $m=\sqrt{n+\sqrt{n-2}} \in Q, m \in N^{*}$. The equality $m^{2}=n+\sqrt{n-2}$ holds. Then there exists $k \in N^{*}$, such that $k=\sqrt{n-2}$, i.e., $n=k^{2}+2$. We can write $m^{2}=k^{2}+k+2$.

The inequality $k^{2}<k^{2}+k+2<k^{2}+4 k+4=(k+2)^{2}$ holds, or $k^{2}<m^{2}<(k+2)^{2}$. We obtain that $m=k+1$, then $k=1$ and $n=3$.

Answer: $n=3$