Do iterated quadratic polynomials have the minimal number of irreducible factors?

29 Views Asked by At

Let $k\in\mathbb{Z}$, and $f\in \mathbb{Z}[X]$ be given by $f=X^2+k$. For all integers $n\ge 1$, let $f^{\circ n}$ denote $f$ iterated $n$ times. For example, $f^{\circ 2}=(X^2+k)^2+k$.

I'm interested in how $f^{\circ n}-X$ factors into irreducible polynomials over $\mathbb{Z}[X]$. There are two special cases, $k=0,-2$, in which this factorization is relatively easy to prove. In general, if $d\mid n$, then $f^{\circ d}-X\mid f^{\circ n}-X$.

Based on computational evidence ($1\le n\le 10$, $1\le k\le 100$), I conjecture that, given $k\neq 0,-2$, there exist polynomials $(g_m)_{m\ge 1}$ in $\mathbb{Z}[X]$ such that for all $n\ge 1$, $$f^{\circ n}-X=\prod_{d\mid n}g_d,$$ and such that $g_d$ is irreducible for all $d>2$. Can anyone prove or refute this conjecture?