## Task 10/86

Let $a ; n \in N$ and $n \geq 2$. What is the remainder when $a^{n}$ is divided by $(a+1)$?
It is

$$
a^{n}=[(a+1)-1]^{n}=\sum_{k=0}^{n}\binom{n}{k}(a+1)^{n-k}(-1)^{k}
$$

All terms of the sum, except for the last one, contain the factor $(a+1)$. The remainder $R$ is thus determined by the last term $(-1)^{n}$: It is $R=1$ for even $n$ and $R=-1$ for odd $n$.