For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$? For what $(m, n)$, does $1+x+x^2 +\dots+x^m | 1 + x^n + x^{2n}+\dots+x^{mn}$?
Well, 
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1}$$
and, 
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1}$$
Notice that $x^{m+1} - 1|(x^{m+1})^n - 1$, therefore, 
$$x^{n(m+1)} - 1 = q(x^{m+1} - 1)$$
Let 
$$\frac{x^{m+1} - 1}{x - 1} = \alpha$$
Therefore, 
$$\sum_{i = 0}^{m} x^i = \frac{x^{m+1} - 1}{x - 1} = \alpha$$
and, 
$$\sum_{i = 0}^m x^{in} = \frac{x^{n(m+1)} - 1}{x-1} = \frac{q(x^{m+1} - 1)}{x-1} = \alpha q$$
Clearly, $\alpha|\alpha q$. Therefore, all $(m, n)$ such that $m, n \in \mathbb{N}$ should work. But the correct answer, according to the text is that $(m+1)$ and $n$ have to be relatively prime. Where did I go wrong?

In your second formula you need $x^n$ in denominator.
