5. Find all natural numbers $n \geq 2$ such that $20^{n}+19^{n}$ is divisible by $20^{n-2}+19^{n-2}$.
Solution.

Consider the expression

$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)
$$

By the condition, it is divisible by $20^{n-2}+19^{n-2}$. On the other hand,

$$
20^{n}+19^{n}-19^{2} \cdot\left(20^{n-2}+19^{n-2}\right)=20^{n-2}\left(20^{2}-19^{2}\right)=20^{n-2} \cdot 39
$$

Note that $20^{n-2}$ and $20^{n-2}+19^{n-2}$ are coprime, since no prime divisor of $20^{n-2}$ is a divisor of $19^{n-2}$, and the expressions $20^{n-2}$ and $20^{n-2}+19^{n-2}$ do not have
common prime divisors. Therefore, in the product $20^{n-2} \cdot 39$, only the second factor can be divisible by $20^{n-2}+19^{n-2}$. For $n-2>1$, the expression $20^{n-2}+19^{n-2}$ exceeds 39, so it cannot be its divisor. It remains to check by substitution $n=2,3$.

For $n=2$, $20^{0}+19^{0}=2$, the odd number $20^{2}+19^{2}=761$ is not divisible by 2. For $n=3$, $20^{1}+19^{1}=39,20^{3}+19^{3}=39 \cdot\left(20^{2}-20 \cdot 19+19^{2}\right)$ is divisible by $20^{1}+19^{1}=39$. Answer. $n=3$.