 Solution

Let $p$ be prime and defineThe recursion $P_{n+1}=P_n^2+c$ shows that $P_p$ has degree $2^{p-1}$. Moreover $P_p(c)=c+O(c^2)$ at zero, so $c=0$ is a simple root. Since the degree is greater than one, $P_p$ has a nonzero root $c_0$. At $c_0$, the critical point zero is periodic with period dividing $p$. It is not fixed because $c_0\ne 0$, so primality makes its exact period $p$. Thus $c_0$ is the center of a hyperbolic component of exact period $p$.

The multiplier map on this component covers the unit disc. Move to its boundary along parameters whose attracting-cycle multiplier tends to $-1$. Compactness of the Mandelbrot set gives a limiting parameter $c_{\ast }$. The periodic cycle persists with exact period $p$: at multiplier $-1$, every point is a simple root of $f_{c_{\ast }}^p(z)-z$, so no collision to a lower-period orbit occurs. Its multiplier is the root of unity $-1$, and hence it is a parabolic cycle after squaring the return map. We have produced a parabolic cycle of exact period $p$ for every prime $p$. Therefore the parabolic periods in the quadratic family form an infinite set.