= Solution
Write $n=km$, with positive integers $k,m$, and put $Q=P^m$. <Polynomial> substitution respects congruence, so $Q(z)\equiv z\pmod{Q(z)-z}$ implies inductively $Q^j(z)\equiv z$ for every $j\geq1$. Since $Q^k=P^n$,
$$
\boxed{P^m(z)-z\mid P^n(z)-z.}
$$
Equivalently one can factor each $Q^{j+1}(z)-Q^j(z)$ by $Q(z)-z$ and telescope. If $Q(z)-z$ is identically zero, then $Q$ is the identity and $P^n-z$ is also zero, so the assertion still holds in the sense that the latter is a <polynomial> multiple of the former.
For the <holomorphic cycle>, the <chain rule> gives
$$
(P^n)'(w_j)=P'(w_j)P'(w_{j+1})\cdots P'(w_{j+n-1}),
$$
with indices cyclically understood. These are the same product at every point. If its modulus is less than one at one point, it is less than one at every point. <Taylor expansion> of each return map gives a contracting neighborhood of its <fixed point>, exactly as in question 1. Hence \b[every point of the <holomorphic cycle> is attracting for $P^n$ whenever one is]; a zero factor makes every return <periodic-orbit multiplier> zero.
For the quadratic family, factor the second-iterate fixed-point equation:
$$
P^2(z)-z=(z^2-z+c)(z^2+z+c+1).
$$
The first factor gives <fixed points>. A root of the second is also fixed only if subtracting the two factors gives $2z+1=0$, so $z=-1/2$ and $c=-3/4$. If $c\ne-3/4$, the second factor has discriminant $-4c-3\ne0$ and has two distinct roots, neither fixed. They are exchanged by $P$, since each is fixed by $P^2$ but not by $P$, and form an exact two-cycle. Therefore the unique parameter with no exact two-cycle is
$$
\boxed{c=-\frac34.}
$$
At this parameter,
$$
\boxed{P^2(z)-z=(z-3/2)(z+1/2)^3.}
$$
Thus the four finite solutions counted with multiplicity are $3/2$ once and $-1/2$ three times: there are only two distinct finite solutions, both already fixed by $P$. The <periodic-orbit multiplier> at $3/2$ is three (nine for $P^2$), so it is repelling. At $-1/2$ the <periodic-orbit multiplier> for $P$ is minus one; for $P^2$ it is one. Indeed, with $u=z+1/2$, the local first-return expression for $P$ is $-u+u^2$, and its square is $u-2u^3+u^4$. This is a parabolic return with a triple fixed-point root. The two-cycle has merged into this <fixed point>; the triple root is not three distinct <periodic points>. In the <Mandelbrot set> picture this is the <period-two bulb>'s attachment to the <main cardioid>.
Back to article page
