**Solution.** $n = 0$ is a solution For $n \geq 1$, $\frac{20^n - 18^n}{19}$ is a perfect square, $20^n - 18^n = 19x^2$, $2^n(10^n - 9^n) = 19x^2$, $x$ is a natural number; $n$ is even, $n = 2m$, $m$ is a natural non-zero number $2^{2m}(10^{2m} - 9^{2m}) = 19x^2$, $x$ is a natural non-zero number, $x = 2^my$, $y$ is odd, $10^{2m} - 9^{2m} = 19y^2$, $(10^m - 9^m) \cdot (10^m + 9^m) = 19y^2$, $(10^m - 9^m, 10^m + 9^m) = 1$ **Case 1:** There exist odd natural numbers $a, b$, $(a, b) = 1$, $a \cdot b = y$ with $10^m + 9^m = 19a^2$ and $10^m - 9^m = b^2$; $10^m = 9^m + b^2 = \mathbb{M}4 + 1 + \mathbb{M}4 + 1 = \mathbb{M}4 + 2$, $m = 1, b = 1, a = 1, y = 1, x = 2, n = 2$ **Case 2:** There exist odd natural numbers $a, b$, $(a, b) = 1$, $a \cdot b = y$ with $10^m + 9^m = a^2$ and $10^m - 9^m = 19b^2$; $10^m = 9^m + 19b^2 = \mathbb{M}8 + 1 + 19(\mathbb{M}8 + 1) = \mathbb{M}8 + 4$, $m = 2, b = 1, a^2 = 181$, contradiction $S = \{0, 2\}$