Let $A=1989^{1990}-1988^{1990}, \quad B=1989^{1989}-1988^{1989}$. What is the greatest common divisor of $A$ and $B$?
It is known that the greatest common divisor $d=(A, B)$ of the numbers $A$ and $B$ also divides the number $A-n \cdot B$ for any integer $n$.

If $n=1988$, then

$$
\begin{aligned}
a-1988 B= & 1989^{1990}-1988^{1990}-1988 \cdot 1989^{1989}+1988^{1990}= \\
& =1989^{1989}(1989-1988)=1989^{1989}
\end{aligned}
$$

Thus, the greatest common divisor of $A$ and $B$ is a multiple of the first term of the difference that produces $B$, $1989^{1989}$, and hence $d$ is also a divisor of the second term, $1988^{1989}$.

Since 1988 and 1989 are consecutive integers, their greatest common divisor is 1, so they have no common prime factors. According to the fundamental theorem of arithmetic, their powers also cannot have common prime factors. Therefore, $1988^{1989}$ and $1989^{1989}$ are relatively prime, and thus $A$ and $B$ are also relatively prime.

Remarks. 1. We arrive at a similar result if we perform the first division step of the Euclidean algorithm that finds the greatest common divisor of the two numbers:

$$
\begin{gathered}
\left(1989^{1990}-1988^{1990}\right):\left(1989^{1989}-1988^{1989}\right)=1989 \\
\frac{1989^{1990}-1989 \cdot 1988^{1989}}{1989 \cdot 1988^{1989}-1988^{1990}}
\end{gathered}
$$

We obtained that $A=1989 B+1988^{1989}$, so $(A, B)=\left(B, 1988^{1989}\right)$, from which we can obtain the final result as described in the solution.

2. By similar reasoning, it can be generally proven that $\left(a^{n}-b^{n}, a^{m}-b^{m}\right)=a^{(n, m)}-b^{(n, m)}$, if $a$ and $b$ are relatively prime. In the problem, $a=1989, b=1988, n=1990, m=1989$.