Problem 4. Determine the natural numbers $n$ for which the number $\sqrt{\frac{20^{n}-18^{n}}{19}}$ is a rational number.
Solution. $n=0$ is a solution. ....................................................................... for $n \geq 1, \frac{20^{n}-18^{n}}{19}$ is a perfect square, $20^{n}-18^{n}=19 x^{2}, 2^{n}\left(10^{n}-9^{n}\right)=19 x^{2}, x$ is a non-zero natural number; $n$ is even, $n=2 m$, $m$ is a non-zero natural number 1 point. $2^{2 m}\left(10^{2 m}-9^{2 m}\right)=19 x^{2}, x$ is a non-zero natural number, $x=2^{m} y, y$ is odd, $10^{2 m}-9^{2 m}=19 y^{2}$, $\left(10^{m}-9^{m}\right) \cdot\left(10^{m}+9^{m}\right)=19 y^{2},\left(10^{m}-9^{m}, 10^{m}+9^{m}\right)=1 \ldots \ldots \ldots \ldots \ldots .1$ point. Case 1: There exist $a, b$ odd natural numbers, $(a, b)=1, a \cdot b=y$ with $10^{m}+9^{m}=19 a^{2}$ and $10^{m}-9^{m}=b^{2} ; 10^{m}=9^{m}+b^{2}=\mathcal{M} 4+1+\mathcal{M} 4+1=\mathcal{M} 4+2$, $m=1, b=1, a=1, y=1, x=2, n=2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots \ldots \ldots$ points.

Case 2: There exist $a, b$ odd natural numbers, $(a, b)=1, a \cdot b=y$ with $10^{m}+9^{m}=a^{2}$ and $10^{m}-9^{m}=19 b^{2} ; 10^{m}=9^{m}+19 b^{2}=\mathcal{M} 8+1+19(\mathcal{M} 8+1)=\mathcal{M} 8+4$, $m=2, b=1, a^{2}=181$, contradiction

$S=\{0,2\}$ 2 points.