veeltermen

Stel in $\mathbb{C}$ :
$P(S, m, x) \doteq \prod_{k \in S}^{ } \left (x - e^{\frac{k}{m}2\pi i} \right )$ met $ m \in \mathbb{N}_0$ en $S \subset S_m = \left \{ 1,2,\cdots , m \right \} \subset \mathbb{N}_0$
dan is
$x^m - 1
= \prod_{k=1}^{m} \left (x - e^{\frac{k}{m}2\pi i} \right )
= P \left (S_m, m, x \right ) $
en er geldt ook
$ P \left (S_m, m, x \right ) = P \left (n S_m, mn, x \right ) \forall n \in \mathbb{N}_0$ met $n S_m = \left \{ n,2n,\cdots , mn \right \}$.
Stel
$F \left ( m, n, x \right ) \doteq \frac{x^{mn} - 1}{x^m - 1}
= \frac{P( S_{mn} , mn,x )}{P( S_{m} , m,x )}
= \frac{P( S_{mn} , mn,x )}{P( n S_{m} , mn,x )}
= P( S_{mn} \setminus n S_m, mn,x )$
Verder:
$x^a + x^{a-1} + \cdots + x + 1
= \frac {x^{a+1} - 1}{x - 1}
= F \left ( 1, a+1, x \right )
= P(S_{a+1} \setminus (a+1) S_{1}, a+1,x)$
$= P(b S_{a+1} \setminus b (a+1) S_{1}, b (a+1),x)
$
en
$x^{ba} + x^{b\left (a-1 \right )} + \cdots + x^b + 1
= \frac {x^{b \left (a+1 \right )} - 1}{x^b - 1}
= F \left ( b, a+1, x \right )
= P(S_{b(a+1)} \setminus (a+1) S_{b}, b(a+1),x)$.
Dat het eerste het tweede deelt, is nu equivalent met
$F \left ( 1, a+1, x \right ) \mathop \backslash F \left ( b, a+1, x \right )$
$\Leftrightarrow
b S_{a+1} \setminus b (a+1) S_{1} \subset S_{b(a+1)} \setminus (a+1) S_{b}$
$\Leftrightarrow
\left \{ b, 2b, \cdots, (a+1)b \right \} \setminus \left \{ (a+1)b \right \} \subset
\left \{ 1, 2, 3, \cdots, (a+1)b \right \} \setminus \left \{ (a+1); 2(a+1), \cdots, b(a+1) \right \}$
$\Leftrightarrow
\left \{ b, 2b, \cdots, (a+1)b \right \} \cap \left \{ (a+1); 2(a+1), \cdots, b(a+1) \right \} = \left \{ b(a+1) \right \}$
$\Leftrightarrow
lcm{\left \{a+1, b \right \}} = (a+1)b$
$\Leftrightarrow
\gcd{\left \{a+1, b \right \}} = 1$