**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 \( S = \{0, 2\} \)