Baltic Way 1991 · Problem 2

Solution: Factorizing, we get

$$102^{1991}+103^{1991}=(102+103)\left(102^{1990}-102^{1989} \cdot 103+102^{1988} \cdot 103^{2}-\cdots+103^{1990}\right),$$

where $102+103=205=5 \cdot 41$. It suffices to show that the other factor is not divisible by $5$. Let $a_{k}=102^{k} \cdot 103^{1990-k}$, then $a_{k} \equiv 4\ (\bmod\ 5)$ if $k$ is even and $a_{k} \equiv -4\ (\bmod\ 5)$ if $k$ is odd. Thus the whole second factor is congruent to $4 \cdot 1991 \equiv 4\ (\bmod\ 5)$.