245*. Can the number $5^{n}-1$ be divisible by the number $4^{n}-1$ for the same natural $n$?
$\triangle$ The difference $4^{n}-1$ for any natural $n$ is divisible by 3, hence the difference $5^{n}-1$ is also divisible by 3. However, the latter is true only for even $n$ (check this!): $n=2k (n \in N)$. But for even $n$,

$$
4^{n}-1=4^{2k}-1
$$

is also divisible by 5, while the difference $5^{n}-1$ is never divisible by 5 for any natural $n$; therefore, the number $5^{n}-1$ is never divisible by $4^{n}-1$.

Answer: cannot.