1.79 Determine all pairs of positive integers $(m, n)$ such that
$$
1+x^{n}+x^{2 n}+\cdots+x^{m n}
$$

can be divided by $1+x+x^{2}+\cdots+x^{m}$.
(6th United States of America Mathematical Olympiad, 1977)
[Solution] Since $1+x^{n}+x^{2 n}+\cdots+x^{m n}=\frac{x^{(m+1) n}-1}{x^{n}-1}$,
$$
1+x+x^{2}+\cdots+x^{m}=\frac{x^{m+1}-1}{x-1} .
$$

The polynomial $1+x^{n}+x^{2 n}+\cdots+x^{m n}$ can be divided by $1+x+x^{2}+\cdots+x^{m}$ if and only if
$$
\frac{x^{(m+1) n}-1}{x^{m+1}-1} \cdot \frac{x-1}{x^{n}-1}
$$

is an integer-coefficient polynomial.
For all positive integers $m$ and $n$, we have
$$
\begin{array}{r}
x^{m+1}-1 \mid x^{(m+1) n}-1, \\
x^{n}-1 \mid x^{(m+1) n}-1 .
\end{array}
$$

Therefore, for (1) to be an integer-coefficient polynomial, it is sufficient that $x^{m+1}-1$ and $x^{n}-1$ have no common factors other than $x-1$. In fact, if $x^{m+1}-1$ and $x^{n}-1$ have common factors other than $x-1$, then $x^{m+1}-1=0$ and $x^{n}-1=0$ have common roots other than 1, but $x^{(m+1) n}-1=0$ has no repeated roots, so $\frac{\left(x^{(m+1) n}-1\right)(x-1)}{\left(x^{n}-1\right)\left(x^{m+1}-1\right)}$ is not an integer-coefficient polynomial. Therefore, the necessary and sufficient condition for the problem is that $m+1$ and $n$ are coprime.