Which are those positive integers $n$ for which $n^{3}+1$ and $n^{2}-1$ are both divisible by 101?
Solution. Let's write both expressions in product form:

$$
\begin{aligned}
& n^{2}-1=(n+1)(n-1) \\
& n^{3}+1=(n+1)\left(n^{2}-n+1\right)
\end{aligned}
$$

Since 101 is a prime number, for the product to be divisible by it, one of its factors must be a multiple of 101.

Let $n+1=k \cdot 101$, then $n=k \cdot 101-1$, where $k \in \mathbb{N}^{+}$. Since the factor $n+1$ appears in both products, choosing $n$ in this way ensures that both numbers will be divisible by 101.

If 101 were a divisor of $(n-1)$, then in the second case, $n^{2}-n+1=(n-1) n+1$ would mean that 101 could not be a divisor of the second expression.