Misiurewicz polynomials and dynamical units, part II

Abstract

Fix an integer $d\ge 2$. The parameters $c_0\in \overline{\mathbb {Q}}$ for which the unicritical polynomial $f_{d,c}(z)=z^d+c\in \mathbb {C}[z]$ has finite postcritical orbit, also known as Misiurewicz parameters, play a significant role in complex dynamics. Recent work of Buff, Epstein, and Koch proved the first known cases of a long-standing dynamical conjecture of Milnor using their arithmetic properties, about which relatively little is otherwise known. Continuing our work from a companion paper, we address further arithmetic properties of Misiurewicz parameters, especially the nature of the algebraic integers obtained by evaluating the polynomial defining one such parameter at a different Misiurewicz parameter. In the most challenging such combinations, we describe a connection between such algebraic integers and the multipliers of associated periodic points. As part of our considerations, we also introduce a new class of polynomials we call p-special, which may be of independent number theoretic interest.

1 Introduction

Let $f\in \mathbb {C}(z)$ be a rational function. For each integer $n\ge 0$, we write $f^n$ for the n-th iterate of f under composition, i.e., $f^0(z):=z$, and $f^n:=f\circ f^{n-1}$ for each $n\ge 1$.

A point $x\in \mathbb {P}^1(\mathbb {C})$ is periodic (of period n) if there is an integer $n\ge 1$ such that $f^n(x)=x$. In that case, the smallest such integer n is the exact period of x, and if $x\ne \infty$, then the multiplier $\lambda \in \mathbb {C}$ of x is

$$\begin{aligned} \lambda :=\big (f^n\big )'(x) = \prod _{i=0}^{n-1} f'\big (f^i(x)\big ) . \end{aligned}$$

(1)

(One can also define the multiplier of a periodic point at $\infty$ via coordinate change.)

More generally, we say $x\in \mathbb {P}^1(\mathbb {C})$ is preperiodic if there is some $m\ge 0$ such that $f^m(x)$ is periodic. That is, x is preperiodic if and only if its (forward) orbit

$$\begin{aligned} {{\,\textrm{Orb}\,}}_f^+(x) := \{f^n(x) : n\ge 0\} \end{aligned}$$

is finite. In that case, the tail length of x is the smallest integer $m\ge 0$ such that $f^m(x)$ is periodic. We say x is preperiodic of type (m, n) if x is preperiodic with tail length m, and n is the exact period of $f^m(x)$. Equivalently, x is preperiodic of type (m, n) if $m\ge 0$ is the minimal nonnegative integer and $n\ge 1$ is the minimal positive integer such that $f^{m+n}(x)=f^m(x)$.

We call $f\in \mathbb {C}(z)$ postcritically finite (or PCF) if all of its critical points are preperiodic.

In this paper we consider polynomials in the unicritical family $f_{d,c}:=z^d+c\in \mathbb {C}[z]$, whose only critical points are $\infty$ and 0. Since $\infty$ is a fixed point, the polynomial $f_{d,c}$ is postcritically finite if and only if the forward orbit

$$\begin{aligned} \{0,c,c^d+c,(c^d+c)^d+c,\ldots \} \end{aligned}$$

of the critical point 0 is finite. We set the notation $a_1=c$, and $a_{i+1}=a_i^d+c$ for $i\ge 1$, where we consider each $a_i$ as an element of the polynomial ring $\mathbb {Z}[c]$.

Any parameter $c_0$ for which $f_{d,c_0}$ is postcritically finite is an algebraic integer. Indeed, as we described in Sect. 1 of [3], if 0 is preperiodic but not periodic under $f_{d,c_0}$ , then $c_0$ is a root of a monic polynomial $G_{d,m,n}^{\zeta }\in \mathbb {Z}[\zeta ][c]$ for some $m\ge 2, n\ge 1$, where $\zeta \ne 1$ is a d-th root of unity. Here, the pair (m, n) is the preperiodic type of the parameter $c_0$, as defined above, and $\zeta \ne 1$ specifies the relation $\zeta a_{m-1}(c_0) = a_{m+n-1}(c_0)$. The polynomials $G_{d,m,n}^{\zeta }$ are called (m, n)-Misiurewicz polynomials, and they are defined by

$$\begin{aligned} G_{d,m,n}^{\zeta }(c):= \prod _{k|n} \big ( a_{m+k-1} - \zeta a_{m-1}\big )^{\mu (n/k)} \cdot {\left\{ \begin{array}{ll} \prod _{k|n} \big ( a_{k}\big )^{-\mu (n/k)} &{} \text { if } n|m-1, \\ 1 &{} \text { if } n \not \mid m-1. \end{array}\right. } \end{aligned}$$

(2)

For $d=2$, the d-th root of unity $\zeta \ne 1$ is necessarily $-1$. For this reason, we sometimes write simply $G_{m,n}$ instead of $G_{d,m,n}^{\zeta }$ in this case.

Misiurewicz parameters have been extensively studied in complex dynamics, especially in the quadratic case. Here are just a few examples of known results. Douady and Hubbard [8, Chapter 8] proved that Misiurewicz parameters are dense in the boundary of the Mandelbrot set. Poirer [22] used Hubbard trees to give a classification of the dynamics of PCF polynomials (see Theorems A and B in [22]). Eberlein [9] showed that the periodic cycle in the post-critical orbit of $f_{d,c_0}$ for any Misiurewicz parameter $c_0$ is repelling. Favre and Gauthier [11, Theorem 1] strengthened Douady and Hubbard’s density result by proving that Misiurewicz parameters are equidistributed in the boundary of the Mandelbrot set; see also [13].

Given their arithmetic nature, Misiurewicz polynomials have drawn number theoretic interest as well. For example, in Theorems 1.1 and 1.3 of [12], Fakhruddin used the simplicity of roots of Misiurewicz polynomials to prove dynamical analogues of the Mordell–Lang and Manin–Mumford conjectures for generic endomorphisms of $\mathbb {P}^n$. In [1], Baker and DeMarco presented a dynamical analogue of the André–Oort Conjecture by considering PCF parameters (such as Misiurewicz parameters) in dynamical moduli spaces in a role corresponding to that of André–Oort special points on Shimura varieties. That is, PCF parameters in dynamical moduli spaces are analogous to CM points on modular curves, and they are expected to have correspondingly analogous arithmetic properties. See, for example, [13] for more on the dynamical André–Oort conjecture.

One key open question from both the complex dynamical and number theoretic perspectives is the following variant of a conjecture of Milnor from [21].

Conjecture 1.1

Let $d,m\ge 2, n\ge 1$. Suppose that $\zeta \ne 1$ is a d-th root of unity. Then $G_{d,m,n}^{\zeta }$ is irreducible over $\mathbb {Q}(\zeta )$.

Besides some limited computational evidence and partial results, very little is known about Conjecture 1.1. The irreducibility is known, for instance, in the cases that $d=2$ and $n\le 3$, that $d=3$ and $n=2$, and that d is a prime power and $n=1$. See [4, 14, 15] for these results and more. In particular, in Theorems 1 and 4 of [4], Buff, Epstein, and Koch used some special cases of Conjecture 1.1 proven by the second author in [14, Corollary 1.1] to prove the first known cases of a different conjecture of Milnor [19, 20] on the irreducibility of certain curves arising as dynamical moduli spaces.

Although Conjecture 1.1 appears to be currently out of reach, it is but one piece of the broader question that always accompanies arithmetically interesting families of polynomials: what are the properties of the number fields generated by their roots? Inspired by parallels with cyclotomic and elliptic units, in [3], we posed the following question:

Question 1.2

Fix $d,m\ge 2$, $n\ge 1$, and $\zeta \ne 1$ a d-th root of unity. Let $c_0$ be a root of $G_{d,m,n}^\zeta$, and let $K:=\mathbb {Q}(c_0)$. For which integers $j\ge 2$ and $\ell \ge 1$ is $G_{d,j,\ell }^{\zeta }(c_0)$ an algebraic unit in $\mathcal {O}_K$?

Similar questions arose in [4, 14] surrounding Gleason polynomials, which are analogues of Misiurewicz polynomials for the case that the critical point is periodic rather than strictly preperiodic. In particular, in Lemma 3.1 of [14], the second author proved that evaluating one Gleason polynomial at the root of another yields an algebraic unit. On the other hand, Buff, Epstein, and Koch [4, Lemma 26] considered resultants of Misiurewicz polynomials with Gleason polynomials, and they proved that evaluating a Misiurewicz polynomial at a Gleason parameter gives an algebraic unit unless the periods of these polynomials match. They then leveraged this result to prove Misiurewicz irreducibility results towards Conjecture 1.1.

In [3], when $d=p^e$ is a prime power, we gave a complete answer to Question 1.2 if $j\ne m$. However, the much harder case seems to be when $j=m$, for which we posed the following conjecture, based on Magma computations for small values of m and n.

Conjecture 1.3

Let $d=p^e$, where p is a prime and $e\ge 1$. Let $c_0$ be a root of $G_{d,m,n}^\zeta$ for some $m\ge 2$, $n\ge 1$, and $\zeta \ne 1$ a d-th root of unity. Set $K:=\mathbb {Q}(c_0)$. Suppose that $1\le \ell \le n$. Then

$$\begin{aligned} G_{d,m,\ell }^{\zeta }(c_0) \text { is a unit in } \mathcal {O}_K \quad \text { if and only if }\quad \ell \not \mid n. \end{aligned}$$

When $\ell =n$, we have $G_{d,m,\ell }^{\zeta }(c_0)=0$, which is not a unit, and hence Conjecture 1.3 holds trivially. Thus, the conjecture is immediate for $n=1$, and hence we will often assume $n\ge 2$. In addition, when considering the forward implication of Conjecture 1.3, we can restrict our attention to the case that $\ell$ is a proper divisor of n.

The main results of the current paper are as follows. We prove the reverse implication of Conjecture 1.3 in Proposition 2.1 of Sect. 2. We also describe a connection between the forward implication and certain arithmetic properties of the multiplier of the periodic cycle in the postcritical orbit ${{\,\textrm{Orb}\,}}_f^{+}(c_0)$. Namely, for $m\ge 2, n\ge 1$, we define the multiplier polynomial $P_{d,m,n}^{\zeta }\in \mathbb {Z}[\zeta ][x]$ to be the monic polynomial whose roots are the multipliers of the periodic cycle portions of the critical preperiodic orbits that correspond to Misiurewicz parameters of type (m, n). (See Definition 5.1 for details.) In the case $d=2$, assuming the irreducibility of $G_{m,n}$ over $\mathbb {Q}$, we reduce Conjecture 1.3 to the following conjecture, which relates our question to classical cyclotomic polynomials. We denote by $\Phi _\ell$ the $\ell$-th cyclotomic polynomial.

Conjecture 1.4

Let $m\ge 2, n,\ell \ge 1$. Then $|{{\,\textrm{Res}\,}}(P_{m,n}, \Phi _{\ell })|>1$.

Magma computations suggest that the resultants of Conjecture 1.4 are huge integers, with no readily apparent patterns in their prime factorizations. In particular, Table 1 in Sect. 8 gives values of $\log |{{\,\textrm{Res}\,}}(P_{m,n}, \Phi _{\ell })|$ for $m+n\le 7$ and $\ell \le 8$, illustrating that these values are very large. Nevertheless, the relatively modest claim of Conjecture 1.4, merely that these resultants are greater than 1, appears to be quite difficult. For fixed m and n, one can prove such a result for sufficiently large $\ell$ using Diophantine approximation methods (see, for example, [17]), but we need the statement for all $\ell \ge 1$. In addition, the prime factorizations of the resultants (see Table 2 for all such factorizations for $m+n\le 5$ and $\ell \le 5$) suggest no patterns that might be leveraged to predict primes dividing them.

Another obstacle one faces in studying Conjecture 1.4 is that we do not yet have a closed formula for the multiplier polynomials $P_{m,n}$ , as we do for Misiurewicz and Gleason polynomials. To overcome this difficulty, we have introduced a broader class of polynomials that we conjecture the multiplier polynomials $P_{m,n}$ may belong to, and which may be of independent number theoretic interest.

Definition 1.5

Let $P(x)=x^i+A_{i-1}x^{i-1}+\cdots +A_1x+A_0\in \mathbb {Z}[x]$ be a monic polynomial with integer coefficients, and let p be a prime number. We say that P(x) is p-special if it satisfies the following two properties:

• $v_p(A_{i-1})> v_p(2)$, and

• $v_p(A_j)>v_p(A_{i-1})$ for $j=0,1,\ldots ,i-2$.

In particular, we prove the following result about p-special polynomials, intended as a stepping stone towards Conjecture 1.4.

Theorem 1.6

Let p be a prime number, and let $\ell \ge 1$. For any p-special polynomial $f(x)\in \mathbb {Z}[x]$, we have $|{{\,\textrm{Res}\,}}(f,\Phi _{\ell })|>1$.

Thus, if the multiplier polynomials are 2-special, and if Misiurewicz polynomials are irreducible as in Conjecture 1.1, then Theorem 1.6 would imply Conjecture 1.3. In particular, we are able to establish 2-specialness in the cases $n=1,2$, yielding the following result.

Theorem 1.7

Let $d=2$ and $m,n\ge 2$. Assume that $G_{m,n}$ is irreducible over $\mathbb {Q}$. Let $c_0$ be a root of $G_{m,n}$ , and let $K:=\mathbb {Q}(c_0)$. Then:

• (1) $G_{m,1}(c_0)$ is not a unit in $\mathcal {O}_K$.

• (2) If n is even, then $G_{m,2}(c_0)$ is not a unit in $\mathcal {O}_K$.

Because $G_{m,n}$ is already known to be irreducible for $n\le 3$ (see [4, 15]), we have the following unconditional corollary.

Corollary 1.8

Let $d=2$ and $m\ge 2$. Conjecture 1.3 holds for $n\le 3$.

The structure of the paper is as follows. In Sect. 2, we prove the if direction of Conjecture 1.3. In Sect. 3, we prove some technical lemmas on the ideals generated by certain values of Misiurewicz polynomials. Using these lemmas, we relate those Misiurewicz values to multipliers of the associated periodic cycles in Sect. 4. Then, in Sect. 5, we define polynomials $P_{d,m,n}^{\zeta }$ whose roots are these multipliers, and we relate Conjecture 1.3 to the resultants of these multiplier polynomials with cyclotomic polynomials. In Sect. 6, we introduce the class of polynomials we dub p-special, and in Sect. 7, we conjecture that when $d=2$, our multiplier polynomials are 2-special. In Sect. 8, we prove that multiplier polynomials are 2-special for $d=2$ and $n=1,2$, and we use this fact to deduce Theorem 1.7. Finally, in Sect. 8, we provide some empirical data related to Conjecture 1.4.

2 The reverse implication

In this section, we prove the easier if direction of Conjecture 1.3.

Proposition 2.1

Let $d,m,n\ge 2$. Let $c_0$ be a root of $G_{d,m,n}^{\zeta }$ , where $\zeta \ne 1$ is a d-th root of unity. Set $K=\mathbb {Q}(c_0)$. For $1\le \ell \le n$, if $\ell \not \mid n$, then $G_{d,m,\ell }^{\zeta }(c_0)$ is an algebraic unit in $\mathcal {O}_K$.

To prove Proposition 2.1, we will need the following two lemmas.

Lemma 2.2

Let $d,m,n\ge 2$. Let $c_0$ be a root of $G_{d,m,n}^{\zeta }$ , where $\zeta \ne 1$ is a d-th root of unity. Set $K=\mathbb {Q}(c_0)$. For any positive integers $u,v \ge 1$ with $\gcd (u,n)=\gcd (v,n)$, we have

$$\begin{aligned} \big \langle a_{m+u-1}(c_0)-\zeta a_{m-1}(c_0)\big \rangle =\big \langle a_{m+v-1}(c_0)-\zeta a_{m-1}(c_0)\big \rangle . \end{aligned}$$

as ideals in $\mathcal {O}_K$.

Proof

For every finite place $\mathfrak {p}$ of $\mathcal {O}_K$, we will establish the equality

$$\begin{aligned} v_{\mathfrak {p} }\big (a_{m+u-1}(c_0)-\zeta a_{m-1}(c_0)\big )=v_{\mathfrak {p} }\big (a_{m+\gcd (u,n)-1}(c_0)-\zeta a_{m-1}(c_0)\big ) \end{aligned}$$

(3)

for every positive integer $u \ge 1$, from which the lemma follows immediately.

Fix such an integer u, and consider an arbitrary finite place $\mathfrak {p}$ of $\mathcal {O}_K$.

Suppose first that $a_{m+u-1}(c_0)\equiv \zeta a_{m-1}(c_0) \pmod {\mathfrak {p} ^i}$ for some integer $i\ge 1$. Applying u iterations of $f:=f_{d,c_0}$ to this congruence, we obtain

$$\begin{aligned} a_{m+2u-1}(c_0)\equiv a_{m+u-1}(c_0)\equiv \zeta a_{m-1}(c_0)\pmod {\mathfrak {p} ^i}. \end{aligned}$$

Thus, proceeding inductively, we have $a_{m+k u-1}(c_0)\equiv \zeta a_{m-1}(c_0)\pmod {\mathfrak {p} ^i}$ for any integer $k\ge 1$. Since f has exact type (m, n), it follows that $a_{m+ku+\ell n-1}(c_0)\equiv \zeta a_{m-1}(c_0) \pmod {\mathfrak {p} ^i}$ for any integers $k,\ell \ge 1$. It is possible to choose positive integers $k,\ell , t\ge 1$ such that $k u+\ell n=\gcd (u,n)+nt$, and hence we have $a_{m+\gcd (u,n)-1}(c_0)\equiv \zeta a_{m-1}(c_0) \pmod {\mathfrak {p} ^i}$. Thus, we have proven the $\le$ direction of Eq. (3).

Conversely, suppose that $a_{m+\gcd (u,n)-1}(c_0)\equiv \zeta a_{m-1}(c_0)\pmod {\mathfrak {p} ^i}$ for some integer $i\ge 1$. By a similar inductive argument as above, we have $a_{m+t\gcd (u,n)-1}(c_0)\equiv \zeta a_{m-1}(c_0)\pmod {\mathfrak {p} ^i}$ for any integer $t\ge 1$. In particular, therefore, we have $a_{m+u-1}(c_0)\equiv \zeta a_{m-1}(c_0)\pmod {\mathfrak {p} ^i}$, proving the $\ge$ direction of Eq. (3). $\square$

Remark 2.3

If $\gcd (u,n)=\gcd (v,n)=n$, i.e., if n|u and n|v, then the ideal in the statement of Lemma 2.2 is the zero ideal. After all, in that case, we have

$$\begin{aligned} a_{m+u-1}(c_0) = a_{m+n-1}(c_0)=\zeta a_{m-1}(c_0), \end{aligned}$$

because $c_0$ is a root of $G_{d,m,n}^{\zeta }$.

On the other hand, if $n\not \mid u$, i.e., if $\ell :=\gcd (u,n)$ satisfies $1\le \ell < n$, then $a_{m+u-1}(c_0) \ne \zeta a_{m-1}(c_0)$, by Lemma 2.2 of [3], and hence the ideal in Lemma 2.2 above is nonzero.

Lemma 2.4

Let $\ell ,n\ge 1$ be positive integers with $\ell \not \mid n$. Then, for any positive integer t|n, we have

$$\begin{aligned} \sum _{\begin{array}{c} k|\ell \\ \gcd (k,n)=t \end{array}} \mu \bigg (\frac{\ell }{k}\bigg )=0. \end{aligned}$$

Proof

For each integer k in the sum, we may write $k=tk_1$ and $n=tn_1$, where $k_1,n_1\ge 1$ are relatively prime positive integers. Moreover, because $k|\ell$, we may write $\ell =t\ell _1$ for some positive integer $\ell _1\ge 1$ such that $\ell _1\not \mid n_1$. Thus,

$$\begin{aligned} \sum _{\begin{array}{c} k|\ell \\ \gcd (k,n)=t \end{array}} \mu \bigg (\frac{\ell }{k}\bigg )= \sum _{\begin{array}{c} k_1|\ell _1 \\ \gcd (k_1,n_1)=1 \end{array}} \mu \bigg (\frac{\ell _1}{k_1}\bigg ). \end{aligned}$$

Hence, it suffices to prove the lemma in the case that $t=1$, which we assume hereafter.

Define a function $v_n:\mathbb {N}\rightarrow \mathbb {N}$ by

$$\begin{aligned} v_n(\ell ) = {\left\{ \begin{array}{ll} 1 &{} \text { if } \text {gcd}(\ell ,n)=1\\ 0 &{} \text { if } \text {gcd}(\ell ,n)>1. \end{array}\right. } \end{aligned}$$

It is straightforward to check that $v_n$ is multiplicative. Therefore, being the Dirichlet convolution of two multiplicative functions, the function

$$\begin{aligned} F_n(\ell ):=\sum _{k|\ell }\mu \bigg (\frac{\ell }{k}\bigg )v_n(k) = \sum _{\begin{array}{c} k|\ell \\ \gcd (k,n)=1 \end{array}} \mu \bigg (\frac{\ell }{k}\bigg ) \end{aligned}$$

is also multiplicative. We wish to show that $F_n(\ell )=0$ for all $\ell \in \mathbb {N}$ with $\ell \not \mid n$. Since $F_n$ is multiplicative, it suffices to show that $F_n(p^e)=0$ for any prime p and any integer $e\ge 1$ such that $p^e\not \mid n$.

Given such p and e, we consider two cases. In the first case, suppose that p|n. Then because $p^e\not \mid n$, we must have $e\ge 2$. In addition, the only positive integer k such that $k|p^e$ and $\gcd (k,n)=1$ is $k=1$. Thus,

$$\begin{aligned} F_n(p^e) = \sum _{\begin{array}{c} k|p^e \\ \gcd (k,n)=1 \end{array}} \mu \bigg (\frac{p^e}{k}\bigg ) = \mu (p^e) = 0, \end{aligned}$$

where the final equality is because $e\ge 2$.

The only other case is that $p\not \mid n$. In this case, all divisors of $p^e$ are relatively prime to n, and hence

$$\begin{aligned} F_n(p^e) = \sum _{\begin{array}{c} k|p^e \\ \gcd (k,n)=1 \end{array}} \mu \bigg (\frac{p^e}{k}\bigg ) = \sum _{k|p^e} \mu \bigg (\frac{p^e}{k}\bigg ) = \mu (p)+\mu (1)=0, \end{aligned}$$

where the third equality is because $\mu (p^i)=0$ for $i\ge 2$. $\square$

Proof of Proposition 2.1

By the Möbius product definition of $G_{d,m,\ell }^{\zeta }$ , we have

$$\begin{aligned} \big \langle G_{d,m,\ell }^{\zeta }(c_0)\big \rangle \bigg | \prod _{k|\ell } \big \langle a_{m+k-1}(c_0)-\zeta a_{m-1}(c_0)\big \rangle ^{\mu (\ell /k)}. \end{aligned}$$

Therefore, it suffices to show that

$$\begin{aligned} \prod _{k|\ell } \big \langle a_{m+k-1}(c_0)-\zeta a_{m-1}(c_0)\big \rangle ^{\mu (\ell /k)}=\mathcal {O}_K. \end{aligned}$$

(4)

To this end, the product in Eq. (4) is

$$\begin{aligned}&\prod _{t|n} \prod _{\begin{array}{c} k|\ell \\ \gcd (k,n)=t \end{array}} \big \langle a_{m+k-1}(c_0)-\zeta a_{m-1}(c_0)\big \rangle ^{\mu (\ell /k)} = \prod _{t|n} \prod _{\begin{array}{c} k|\ell \\ \gcd (k,n)=t \end{array}}\big \langle a_{m+t-1}(c_0)-\zeta a_{m-1}(c_0)\big \rangle ^{\mu (\ell /k)} \\&\quad = \prod _{t|n} \big \langle a_{m+t-1}(c_0)-\zeta a_{m-1}(c_0)\big \rangle ^{E_t}, \quad \text {where}\quad E_t:=\sum _{\begin{array}{c} k|\ell \\ \gcd (k,n)=t \end{array}} \mu \bigg (\frac{\ell }{k}\bigg ) , \end{aligned}$$

and where we have applied Lemma 2.2 in the first equality above. However, we have $E_t=0$ for all t|n, by Lemma 2.4, so that the product is simply $\mathcal {O}_K$, as desired.

3 Technical Lemmas

The rest of this paper is devoted to the only if direction of Conjecture 1.3, which is much more involved than the if direction. To that end, we present two auxiliary lemmas in this section.

Lemma 3.1

Let $d,m,n\ge 2$. Let $c_0$ be a root of $G_{d,m,n}^{\zeta }$ , where $\zeta \ne 1$ is a d-th root of unity, and let $h\in \mathbb {Z}[\zeta ][c]$ be the minimal polynomial of $c_0$ over $\mathbb {Q}(\zeta )$. Let $1\le \ell <n$, and let $\alpha _0$ be a root of $G_{d,m,\ell }^{\zeta }$. Set $K=\mathbb {Q}(c_0)$ and $L=\mathbb {Q}(\alpha _0)$. If $h(\alpha _0)$ is not a unit in $\mathcal {O}_L$, then $G_{d,m,\ell }^{\zeta }(c_0)$ is not a unit in $\mathcal {O}_K$.

Proof

Let $g\in \mathbb {Z}[\zeta ][c]$ be the minimal polynomial of $G_{d,m,\ell }^{\zeta }(c_0)$ over $\mathbb {Q}(\zeta )$. We need to show that the norm $\text {N}_{K/\mathbb {Q}(\zeta )}(G_{d,m,\ell }^{\zeta }(c_0))$ is not a unit in $\mathbb {Z}[\zeta ]$, or equivalently, that g(0) is not a unit in $\mathbb {Z}[\zeta ]$.

By definition of g, we have $(g\circ G_{d,m,\ell }^{\zeta })(c_0)=0$. Since h is the minimal polynomial of $c_0$ over $\mathbb {Q}(\zeta )$, there is a polynomial $u\in \mathbb {Z}[\zeta ][c]$ such that $g\circ G_{d,m,\ell }^{\zeta }=h\cdot u$. Evaluating at $\alpha _0$ and recalling that $G_{d,m,\ell }^{\zeta }(\alpha _0)=0$, we obtain

$$\begin{aligned} g(0) = h(\alpha _0)u(\alpha _0). \end{aligned}$$

By hypothesis, $h(\alpha _0)$ is not a unit in $\mathcal {O}_L$. Since $u(\alpha _0)$ is also an algebraic integer, it follows that $g(0)=h(\alpha _0)u(\alpha _0)$ is also not a unit in $\mathcal {O}_L$. Therefore g(0) is not a unit in $\mathcal {O}_L\cap \mathbb {Q}(\zeta ) = \mathbb {Z}[\zeta ]$, as desired. $\square$

Remark 3.2

With notation as above, if $G_{d,m,n}^{\zeta }$ is irreducible over $\mathbb {Q}(\zeta )$, then Lemma 3.1 says that if $G_{d,m,n}^{\zeta }(\alpha _0)$ is not a unit in $\mathcal {O}_L$, then $G_{d,m,\ell }^{\zeta }(c_0)$ is not a unit in $\mathcal {O}_K$.

Motivated by Remark 3.2, we turn our attention to the algebraic integers $G_{d,m,n}^{\zeta }(\alpha _0)$, where $\alpha _0$ is a root of $G_{d,m,\ell }^{\zeta }$ for some proper divisor $\ell$ of n. To prove the only if direction of Conjecture 1.3, it suffices to show that these algebraic integers are not algebraic units.

Fix integers $d,m,n\ge 2$, and let $\zeta \ne 1$ be a d-th root of unity. Let $c_0$ be a root of $G_{d,m,n}^{\zeta }$ , fix a positive integer $\ell$ that is a proper divisor of n, and let $\alpha _0$ be a root of $G_{d,m,\ell }^{\zeta }$. Define sequences $\{B_j\}_{j\ge 1}$ and $\{b_j\}_{j\ge 1}$ in the polynomial ring $\mathbb {Z}[\zeta ,c]$ by

$$\begin{aligned} B_j:= a_{m+\ell j-1} - \zeta a_{m-1} \quad \text {and}\quad b_j := \frac{B_j}{B_1} = \frac{a_{m+\ell j-1} - \zeta a_{m-1}}{a_{m+\ell -1} - \zeta a_{m-1}}. \end{aligned}$$

(5)

To see that $b_j$ is indeed a polynomial in $\mathbb {Z}[\zeta ,c]$, both its numerator $a_{m+\ell j-1} - \zeta a_{m-1}$ and denominator $a_{m+\ell -1} - \zeta a_{m-1}$ are monic polynomials in $\mathbb {Z}[\zeta ,c]$, and as shown in the proof of Theorem A.1 of [10], they both have only simple roots. Moreover, because $\ell | \ell j$, Lemma 2.2 of [3] shows that every root of the denominator is also a root of the numerator. Thus, as claimed, the quotient $b_j$ is indeed a monic polynomial in $\mathbb {Z}[\zeta ,c]$.

Lemma 3.3

Let $d,m,n\ge 2$. Let $\ell$ be a proper divisor of n, and let $\alpha _0$ be a root of $G_{d,m,\ell }^{\zeta }$ , where $\zeta \ne 1$ is a d-th root of unity. Set $L=\mathbb {Q}(\alpha _0)$. Then

$$\begin{aligned} \big \langle G_{d,m,n}^{\zeta }(\alpha _0)\big \rangle = \prod _{j| (n/\ell )} \big \langle b_j(\alpha _0)\big \rangle ^{\mu (n/ (j \ell ))} \end{aligned}$$

as ideals in $\mathcal {O}_L$.

Proof

Suppose first that $n\not \mid m-1$. By the definition of $G_{d,m,n}^{\zeta }$ , we have

$$\begin{aligned} G_{d,m,n}^{\zeta } = \prod _{k|n} \big (a_{m+k-1}-\zeta a_{m-1}\big )^{\mu (n/k)} = H_1 \cdot H_2, \end{aligned}$$

where

$$\begin{aligned} H_1 := \prod _{\begin{array}{c} t|\ell \\ t\ne \ell \end{array}} \prod _{\begin{array}{c} k|n\\ \gcd (k,\ell )=t \end{array}} \big (a_{m+k-1}-\zeta a_{m-1}\big )^{\mu (n/k)} \quad \text {and}\quad H_2 := \prod _{\begin{array}{c} k|n \\ \ell | k \end{array}} \big (a_{m+k-1}-\zeta a_{m-1}\big )^{\mu (n/k)}. \end{aligned}$$

By Remark 2.3, none of the polynomials $a_{m+k-1}-\zeta a_{m-1}$ appearing in the product defining $H_1$ has a zero at $\alpha _0$, whereas all of the terms in the product defining $H_2$ are zero at $\alpha _0$.

For any proper divisor t of $\ell$, and for any integer $k\ge 1$ with $\gcd (k,\ell )=t$, Lemma 2.2 and the fact that $f_{d,\alpha _0}$ has exact type $(m,\ell )$ yield

$$\begin{aligned} \big \langle a_{m+k-1}(\alpha _0)-\zeta a_{m-1}(\alpha _0)\big \rangle = \big \langle a_{m+t-1}(\alpha _0)-\zeta a_{m-1}(\alpha _0)\big \rangle \end{aligned}$$

as ideals in $\mathcal {O}_L$. It follows that for any proper divisor t of $\ell$, we have

$$\begin{aligned} \prod _{\begin{array}{c} k|n\\ \text {gcd}(k,\ell )=t \end{array}} \big \langle a_{m+k-1}(\alpha _0)-\zeta a_{m-1}(\alpha _0)\big \rangle ^{\mu (n/k)} = \big \langle a_{m+t-1}(\alpha _0)-\zeta a_{m-1}(\alpha _0)\big \rangle ^{E_t} , \end{aligned}$$

where

$$\begin{aligned} E_t := \sum _{\begin{array}{c} k|n \\ \gcd (k,\ell )=t \end{array}} \mu \bigg (\frac{n}{k}\bigg ) . \end{aligned}$$

However, according to Lemma 2.4, we have $E_t=0$. Taking the product over all such t, it follows that the ideal $\big \langle H_1(\alpha _0) \big \rangle$ is simply the identity ideal $\mathcal {O}_L$.

By writing $k=j\ell$, we may rewrite the product defining $H_2$ as

$$\begin{aligned} H_2 = \prod _{j|(n/\ell )} B_j^{\mu (n/(j\ell ))} = \prod _{j|(n/\ell )} \bigg (\frac{B_j}{B_1}\bigg )^{\mu (n/(j\ell ))} = \prod _{j|(n/\ell )} b_j^{\mu (n/(j\ell ))}, \end{aligned}$$

by the definition of $B_j$ and $b_j$ from Eq. (5), where we have used the fact that $\ell$ is a proper divisor of n, and hence that $\sum _{j|(n/\ell )} \mu ( n/(j\ell )) = 0$. Thus, we have

$$\begin{aligned} \big \langle G_{d,m,n}^{\zeta }(\alpha _0)\big \rangle = \big \langle H_1(\alpha _0) \big \rangle \big \langle H_2(\alpha _0) \big \rangle = \prod _{j| (n/\ell )} \big \langle b_j(\alpha _0)\big \rangle ^{\mu (n/ (j \ell ))}, \end{aligned}$$

completing the proof in the case that $n\not \mid m-1$.

Next, we suppose that $n \, | \, m-1$. With $H_1$ and $H_2$ as before, we have

$$\begin{aligned} \langle G_{d,m,n}^{\zeta }(\alpha _0)\rangle = \langle H_1(\alpha _0)\rangle \cdot \langle H_2(\alpha _0)\rangle \cdot \prod _{k|n} \langle a_k(\alpha _0)\rangle ^{-\mu (n/k)} . \end{aligned}$$

Hence, to finish the proof, it suffices to show that

$$\begin{aligned} \prod _{k|n} \big \langle a_k(\alpha _0)\big \rangle ^{\mu (n/k)}=\mathcal {O}_L. \end{aligned}$$

Since $\alpha _0$ is a root of $G_{d,m,\ell }^{\zeta }$ , Proposition 3.1 of [3] shows that if $\ell | k$, then $\langle a_k(\alpha _0)\rangle =\langle a_{\ell }(\alpha _0)\rangle$ as ideals in $\mathcal {O}_L$; and if $\ell \not \mid k$, then $a_k(\alpha _0)$ is a unit in $\mathcal {O}_L$, so that $\langle a_k(\alpha _0)\rangle =\mathcal {O}_L$. Thus,

$$\begin{aligned} \prod _{k|n} \big \langle a_k(\alpha _0)\big \rangle ^{\mu (n/k)} = \prod _{j|(n/\ell )} \big \langle a_{\ell }(\alpha _0)\big \rangle ^{\mu (n/(j\ell ))} = \mathcal {O}_L, \end{aligned}$$

where the last equality is because the nonzero ideal $\langle a_{\ell }(\alpha _0)\rangle$ is being raised to the power $\sum _{j|(n/\ell )}\mu (\frac{n}{j\ell })=0$, since $n / \ell > 1$. $\square$

4 Results on multipliers

For a Misiurewicz parameter $\alpha _0$ with type $(m,\ell )$, the algebraic integers $b_j(\alpha )$ defined by Eq. (5) of Sect. 3 turn out to be related to the multiplier of the periodic cycle in the associated critical orbit, as we now discuss.

Fix integers $d,m\ge 2$ and $n \ge 1$. Let $\zeta \ne 1$ be a d-th root of unity. Consider a root $\alpha _0$ of $G_{d,m,n}^\zeta$. Since $f:=f_{d,\alpha _0}$ has $f'(z)=dz^{d-1}$, Eq. (1) shows that the multiplier of the periodic cycle $\{a_m(\alpha _0),\ldots , a_{m+n-1}(\alpha _0)\}$ is

$$\begin{aligned} \lambda _{d,m,n}^{\zeta }(\alpha _0) = d^n \Bigg (\prod _{i=0}^{n-1}a_{m+i}(\alpha _0)\Bigg )^{d-1}. \end{aligned}$$

(6)

The following result shows that the sequence $\{b_j(\alpha _0)\}_{j\ge 1}$ satisfies a linear recurrence, which will immediately allow us to write a simple and explicit formula for $b_j(\alpha _0)$.

Theorem 4.1

Fix integers $d,m,n\ge 2$, a proper divisor $\ell \ge 1$ of n, and a d-th root of unity $\zeta \ne 1$. Define

$$\begin{aligned} C_{d,m,\ell }:= d^{\ell } \bigg ( \prod _{i=0}^{\ell -1} a_{m-1+i} \bigg )^{d-1} \in \mathbb {Z}[c] . \end{aligned}$$

Then for every integer $j\ge 1$, we have

$$\begin{aligned} b_{j+1} \equiv \zeta ^{d-1}C_{d,m,\ell } b_j + 1 \pmod {I}, \end{aligned}$$

where $I:=\langle B_1 \rangle$ is the principal ideal of $\mathbb {Z}[\zeta ,c]$ generated by $B_1=a_{m+\ell - 1}-\zeta a_{m-1}$. In particular, for any root $\alpha _0$ of $B_1$, we have

$$\begin{aligned} b_{j+1}(\alpha _0) = \zeta ^{d-1} C_{d,m,\ell }(\alpha _0) b_j(\alpha _0) + 1. \end{aligned}$$

Proof

Clearly, we have $a_{m+\ell -1} \equiv \zeta a_{m-1} \pmod {I}$, and hence, repeatedly raising each side to the power d, we have $a_k\equiv a_{k+\ell } \pmod {I}$ for every $k\ge m$. Therefore, for every $j\ge 1$ and $i\ge 0$, we have

$$\begin{aligned} a_{m+\ell j-1} \equiv \zeta a_{m-1} \pmod {I} \quad \text {and}\quad a_{m+\ell j+i} \equiv a_{m+i} \pmod {I} . \end{aligned}$$

(7)

Also observe that for any $P, Q\in \mathbb {Z}[\zeta ,c]$, if we let $f(z):=z^d+c$, we have

$$\begin{aligned} f(P)-f(Q) = P^d - Q^d = (P-Q) \sum _{t=0}^{d-1} P^t Q^{d-1-t} . \end{aligned}$$

Thus, for any $j\ge 1$, applying the first congruence of (7) gives

$$\begin{aligned} a_{m+\ell j} - a_{m}&= (a_{m+\ell j-1} - \zeta a_{m-1} ) \sum _{t=0}^{d-1} a_{m+\ell j-1}^t (\zeta a_{m-1})^{d-1-t} \nonumber \\&\equiv d (\zeta a_{m-1})^{d-1} (a_{m+\ell j-1} - \zeta a_{m-1} ) \pmod {I^2} , \end{aligned}$$

(8)

where we have used the fact that $a_{m+\ell j-1} - \zeta a_{m-1}\in I$ to obtain the congruence modulo $I^2$. Similarly, for every $j\ge 1$ and every $i=1,\ldots ,\ell -1$, the second congruence of (7) yields

$$\begin{aligned} a_{m+\ell j+i} - a_{m+i}&= (a_{m+\ell j+i-1} - a_{m+i-1} ) \sum _{t=0}^{d-1} a_{m+\ell j+i-1}^t (a_{m+i-1})^{d-1-t} \\&\equiv d (a_{m+i-1})^{d-1} (a_{m+\ell j+i-1} - a_{m+i-1} ) \pmod {I^2} . \nonumber \end{aligned}$$

(9)

Thus, for any $j\ge 1$, working in the ring $\mathbb {Z}[\zeta ,c]$ modulo $I^2$, we have

$$\begin{aligned} \zeta ^{d-1} C_{d,m,\ell } B_j&= \Bigg [ d^{\ell -1} \Bigg ( \prod _{i=1}^{\ell -1} a_{m-1+i} \Bigg )^{d-1}\Bigg ] \cdot d (\zeta a_{m-1})^{d-1} (a_{m+\ell j-1} - \zeta a_{m-1} ) \\&\equiv d^{\ell -1} \Bigg ( \prod _{i=1}^{\ell -1} a_{m-1+i} \Bigg )^{d-1} (a_{m+\ell j} - a_m) \\&= \Bigg [ d^{\ell -2} \Bigg ( \prod _{i=2}^{\ell -1} a_{m-1+i} \Bigg )^{d-1} \Bigg ] \cdot d (a_{m})^{d-1} (a_{m+\ell j} - a_{m} ) \\&\equiv d^{\ell -2} \Bigg ( \prod _{i=2}^{\ell -1} a_{m-1+i} \Bigg )^{d-1} (a_{m+\ell j+1} - a_{m+1}) \\&\cdots \\&\equiv a_{m+\ell j+\ell -1} - a_{m+\ell -1} = B_{j+1} - B_1, \end{aligned}$$

where we have used identity (8) in the first congruence, and identity (9) in all the subsequent congruences. Since these are congruences modulo $I^2$, where $I=\langle B_1 \rangle$, and since $B_k\in I$ for every k (by the first congruence of (7)), we may divide both sides by $B_1$ to obtain

$$\begin{aligned} \zeta ^{d-1} C_{d,m,\ell } b_j \equiv b_{j+1} - 1 \pmod {I} \quad \text {for every } j\ge 1, \end{aligned}$$

yielding the first desired conclusion. Evaluating both sides at $c=\alpha _0$ and using the fact that $F(\alpha _0)=0$ for every $F\in I$, the second conclusion follows immediately. $\square$

Corollary 4.2

With notation as in Theorem 4.1, for every $j\ge 1$ we have

$$\begin{aligned} b_j(\alpha _0) = 1+C+\cdots + C^{j-1} , \end{aligned}$$

where $C=\zeta ^{d-1} C_{d,m,\ell }(\alpha _0)$.

Proof

We have $b_1(\alpha _0)=1$ by definition, and $b_{j+1}(\alpha _0)=Cb_j(\alpha _0)+1$ for every $j\ge 1$ by Theorem 4.1. The conclusion is immediate by induction. $\square$

Remark 4.3

With the notation in Corollary 4.2, we have $C=\lambda _{d,m,\ell }^{\zeta }(\alpha _0)$. Indeed, substituting $a_{m+\ell -1}(\alpha _0) = \zeta a_{m-1}(\alpha _0)$ into $C=\zeta ^{d-1}d^{\ell } \prod _{i=0}^{\ell -1} a_{m-1+i}^{d-1}(\alpha _0)$ immediately yields $C= d^{\ell } \prod _{i=0}^{\ell -1} a_{m+i}^{d-1}(\alpha _0) = \lambda _{d,m,\ell }^{\zeta }(\alpha _0)$.

The following result reduces Conjecture 1.3 to a question about multipliers and cyclotomic polynomials $\Phi _j$.

Proposition 4.4

Fix integers $d,m,n\ge 2$, a proper divisor $\ell \ge 1$ of n, and a d-th root of unity $\zeta \ne 1$. Let $\alpha _0$ be a root of $G_{d,m,\ell }^{\zeta }$ , and set $L=\mathbb {Q}(\alpha _0)$. Suppose that $G_{d,m,n}^{\zeta }$ is irreducible over $\mathbb {Q}(\zeta )$, and let $c_0$ be one of its roots. Set $K=\mathbb {Q}(c_0)$. If $\Phi _{n/\ell }(\lambda _{d,m,\ell }^{\zeta }(\alpha _0))$ is not a unit in $\mathcal {O}_L$, then $G_{d,m,\ell }^{\zeta }(c_0)$ is not a unit in $\mathcal {O}_K$.

Proof

Define $C=\zeta ^{d-1} C_{d,m,\ell }(\alpha _0)$, which is the multiplier $\lambda _{d,m,\ell }^{\zeta }(\alpha _0)$ of the periodic cycle of $f:=f_{d,\alpha _0}$ , by Remark 4.3. If C is a root of unity (in particular, if $C=1$), then f has a parabolic periodic point, and hence by basic complex dynamics (see, for example, [7, Theorem 2.3] or [18, Corollary 14.5]), some critical point of f must be wandering, contradicting the fact that the only two critical points of f (at $z=0$ and $z=\infty$) are preperiodic. Thus, $C\ne 1$, and hence Corollary 4.2 yields

$$\begin{aligned} b_j(\alpha _0) = \frac{C^j - 1}{C-1} \quad \text {for every integer } j\ge 1 . \end{aligned}$$

Therefore, by Lemma 3.3, there is a unit $u\in \mathcal {O}_L$ such that

$$\begin{aligned} G_{d,m,n}^{\zeta }(\alpha _0) = u \prod _{j| (n/\ell )} \bigg (\frac{C^j -1}{C-1}\bigg )^{\mu (n/ (j\ell ))} = u \prod _{j| (n/\ell )} \big (C^j -1\big )^{\mu (n/ (j\ell ))}, \end{aligned}$$

where the second equality is because $C-1\ne 0$ is being raised to the power $\sum _{j|(n/\ell )}\mu (\frac{n}{j\ell })=0$. Hence, by the definition of the cyclotomic polynomial $\Phi _{n/\ell }$ , we obtain

$$\begin{aligned} G_{d,m,n}^{\zeta }(\alpha _0) = u \Phi _{n/\ell }(C) = u \Phi _{n/\ell }\big (\lambda _{d,m,\ell }^{\zeta }(\alpha _0)\big ). \end{aligned}$$

The result now immediately follows from Remark 3.2. $\square$

5 Multiplier polynomials

In light of Proposition 4.4, we state the following definition and conjecture.

Definition 5.1

Let $d,m\ge 2$ and $n\ge 1$ be integers, and let $\zeta \ne 1$ be a d-th root of unity. Let $c_1,\ldots ,c_k$ be all the roots of $G_{d,m,n}^{\zeta }$. The multiplier polynomial $P_{d,m,n}^{\zeta }$ associated with $G_{d,m,n}^{\zeta }$ is

$$\begin{aligned} P_{d,m,n}^{\zeta }(x)= \prod _{j=1}^{k} (x-\lambda _{d,m,n}^{\zeta }(c_j)) \in \mathbb {Z}[\zeta ][x], \end{aligned}$$

where $\lambda _{d,m,n}^\zeta$ is defined as in Eq. (6).

Conjecture 5.2

Let $d,m\ge 2$ and $n\ge 1$ be integers, and let $\zeta \ne 1$ be a d-th root of unity. Then for every integer $i\ge 1$, the resultant ${{\,\textrm{Res}\,}}(P_{d,m,n}^{\zeta },\Phi _{i})$ is not a unit in $\mathbb {Z}[\zeta ]$, where $\Phi _i$ is the i-th cyclotomic polynomial.

The following result shows that if every $G_{d,m,n}^{\zeta }$ is irreducible over $\mathbb {Q}(\zeta )$, then Conjecture 5.2 implies Conjecture 1.3.

Proposition 5.3

Let $d,m,n\ge 2$ be integers. Let $c_0$ be a root of $G_{d,m,n}^{\zeta }$ , where $\zeta \ne 1$ is a d-th root of unity, and set $K=\mathbb {Q}(c_0)$. Let $\ell \ge 1$ be a proper divisor of n, and suppose that ${{\,\textrm{Res}\,}}(P_{d,m,\ell }^{\zeta }, \Phi _{n/\ell })$ is not a unit in $\mathbb {Z}[\zeta ]$. If $G_{d,m,n}^{\zeta }$ and $G_{d,m,\ell }^{\zeta }$ are irreducible over $\mathbb {Q}(\zeta )$, then $G_{d,m,\ell }^{\zeta }(c_0)$ is not a unit in $\mathcal {O}_K$.

Proof

Fix a proper divisor $\ell$ of n. Enumerate the roots of $G_{d,m,\ell }^{\zeta }$ as $\alpha _1,\ldots ,\alpha _k$. Let $r\ge 1$ be the number of indices $j\in \{1,\ldots ,k\}$ for which $\lambda _{d,m,\ell }^{\zeta }(\alpha _1) = \lambda _{d,m,\ell }^{\zeta }(\alpha _j)$. Because $\lambda _{d,m,\ell }^{\zeta }\in \mathbb {Z}[\zeta ][c]$ is a polynomial over $\mathbb {Q}(\zeta )$, and because we have assumed $G_{d,m,\ell }^{\zeta }$ is irreducible over $\mathbb {Q}(\zeta )$, we know that

$$\begin{aligned} \lambda _{d,m,\ell }^{\zeta }(\alpha _1),\ldots ,\lambda _{d,m,\ell }^{\zeta }(\alpha _k) \end{aligned}$$

are Galois conjugates over $\mathbb {Q}(\zeta )$. It follows that for any $1\le i\le k$, there are exactly r distinct indices $j\in \{1,\ldots ,k\}$ such that $\lambda _{d,m,\ell }^{\zeta }(\alpha _i) = \lambda _{d,m,\ell }^{\zeta }(\alpha _j)$. Therefore, the minimal polynomial $h_{d,m,\ell }^{\zeta }$ of $\lambda _{d,m,\ell }^{\zeta }(\alpha _i)$ over $\mathbb {Q}(\zeta )$ for $i=1,\ldots ,k$ satisfies

$$\begin{aligned} (h_{d,m,\ell }^{\zeta })^r = P_{d,m,\ell }^{\zeta }. \end{aligned}$$

(10)

Set $L_i=\mathbb {Q}(\alpha _i)$ for $1\le i\le k$. By the definition of resultant, we have

$$\begin{aligned} {{\,\textrm{Res}\,}}\big (h_{d,m,\ell }^{\zeta }, \Phi _{n/\ell }\big ) = \text {N}_{L_i/\mathbb {Q}(\zeta )}\Big (\Phi _{n/\ell }\big (\lambda _{d,m,\ell }^{\zeta }(\alpha _i)\big )\Big ) \quad \text {for } i=1,\ldots , k, \end{aligned}$$

where $N_{L_i/\mathbb {Q}(\zeta )}$ denotes the norm from $L_i$ to $\mathbb {Q}(\zeta )$. Therefore, if ${{\,\textrm{Res}\,}}(h_{d,m,\ell }^{\zeta }, \Phi _{n/\ell })$ is not a unit in $\mathbb {Z}[\zeta ]$, then $G_{d,m,\ell }^{\zeta }(c_0)$ is not a unit in $\mathcal {O}_K$ by Proposition 4.4. (Recall that we have assumed $G_{d,m,n}^{\zeta }$ is irreducible over $\mathbb {Q}(\zeta )$.) By Eq. (10) and the multiplicativity of the resultant, we have

$$\begin{aligned} {{\,\textrm{Res}\,}}(h_{d,m,\ell }^{\zeta }, \Phi _{n/\ell })\text { is not a unit in }\mathbb {Z}[\zeta ] \iff {{\,\textrm{Res}\,}}(P_{d,m,\ell }^{\zeta }, \Phi _{n/\ell })\text { is not a unit in }\mathbb {Z}[\zeta ] \end{aligned}$$

and hence the result follows from our hypothesis that ${{\,\textrm{Res}\,}}(P_{d,m,\ell }^{\zeta }, \Phi _{n/\ell })$ is not a unit. $\square$

6 p-special polynomials

In order to study the resultants ${{\,\textrm{Res}\,}}(P_{d,m,\ell }^{\zeta }, \Phi _{n/\ell })$ arising in Sect. 5, we wish to show that the multiplier polynomials $P_{d,m,\ell }^{\zeta }$ belong to a special class, which we now define.

Definition 6.1

Let $P(x)=x^i+A_{i-1}x^{i-1}+\cdots +A_1x+A_0\in \mathbb {Z}[x]$ be a monic polynomial with integer coefficients, and let p be a prime number. We say that P(x) is p-special if it satisfies the following two properties:

• $v_p(A_{i-1})> v_p(2)$, and

• $v_p(A_j)>v_p(A_{i-1})$ for $j=0,1,\ldots ,i-2$.

Note that if $p=2$, the first bullet point in Definition 6.1 says that $v_p(A_{i-1})\ge 2$, whereas for $p\ge 3$, it says that $v_p(A_{i-1})\ge 1$. Our main result about p-special polynomials is the following.

Theorem 6.2

Let $P(x)\in \mathbb {Z}[x]$ be a p-special polynomial for some prime p. Then for every integer $\ell \ge 1$, we have $|{{\,\textrm{Res}\,}}(P,\Phi _{\ell })|>1$.

To prove the theorem, we will need a lemma.

Lemma 6.3

Let $P(x)\in \mathbb {Z}[x]$ be a p-special polynomial for some prime p. For any $n\ge 1$, $(P(x))^n$ is also a p-special polynomial.

Proof

We first claim that $(P(x))^p$ is a p-special polynomial. Indeed, write

$$\begin{aligned} P(x)=x^i+ \big (A_{i-1} x^{i-1} + \cdots + A_0\big ). \end{aligned}$$

(11)

Expanding $(P(x))^p$ using the binomial theorem, the leading term is $x^{ip}$, and the second term is $pA_{i-1}x^{pi-1}$, which satisfies $v_p(pA_{i-1})> v_p(A_{i-1}) > v_p(2)$, verifying the first bullet point of Definition 6.1. Moreover, since $p \, | \, \left( {\begin{array}{c}p\\ j\end{array}}\right)$ for $1\le j\le p-1$, every other coefficient must be a sum of integers, each of which is divisible either by $pA_j$ for some $j\in \{0,1\ldots ,i-2\}$, or by $A_j^p$ for some $j\in \{0,1,\ldots ,i-1\}$. For $0\le j\le i-2$, we have

$$\begin{aligned} v_p(pA_j) > v_p(pA_{i-1}), \end{aligned}$$

and for $0\le j\le i-1$, we have

$$\begin{aligned} v_p(A_j^p) =pv_p(A_j) > 1 + v_p(A_j) = v_p(pA_{j}) \ge v_p(pA_{i-1}), \end{aligned}$$

using the fact that P is p-special for both inequalities. These bounds verify the second bullet point for $P^p$, proving the claim.

Second, we claim that for any integer $m\ge 1$ with $p\not \mid m$, the polynomial $P^m$ is also p-special. Again writing P as in Eq. (11) and expanding $P^m$, the lead term is $x^{im}$, and the second term is $mA_{i-1} x^{im-1}$. We have $v_p(mA_{i-1}) = v_p(A_{i-1}) > v_p(2)$, verifying the first bullet point of Definition 6.1. All other coefficients are sums of integers divisible either by $A_j A_k$ for some not necessarily distinct $j,k\in \{0,1,\ldots ,i-1\}$, or else divisible by $A_j$ for some $j\in \{0,1,\ldots ,i-2\}$. For $j,k\in \{0,1,\ldots ,i-1\}$, we have

$$\begin{aligned} v_p(A_jA_k)> v_p(A_j) > v_p(A_{i-1}) = v_p(mA_{i-1}), \end{aligned}$$

and for $0\le j\le i-2$, we have

$$\begin{aligned} v_p(A_j) > v_p(A_{i-1}) = v_p(mA_{i-1}), \end{aligned}$$

using the facts that $v_p(m)=0$ and that P is p-special. These bounds verify the second bullet point for $P^m$, proving our second claim.

Inductively applying the two above claims, it follows that $P^n$ is p-special for any positive integer n. $\square$

Proof of Theorem 6.2

It suffices to show that for any primitive $\ell$-th root of unity $\zeta$, the value $P(\zeta )$ is not a unit in the ring $\mathbb {Z}[\zeta ]$. For the sake of contradiction, we suppose that

$$\begin{aligned} P(\zeta )\in \mathbb {Z}[\zeta ]^{\times }. \end{aligned}$$

(12)

Case 1. Assume $\ell =1.$ Write $P(x)=x^i+\sum \limits _{t=0}^{i-1} A_tx^t$. Since $\zeta =1$, assumption (12) yields

$$\begin{aligned} 1+\sum _{t=0}^{i-1} A_t=\pm 1. \end{aligned}$$

(13)

If the right-hand side of Eq. (13) is 1, we have

$$\begin{aligned} A_{i-1} = -\sum _{t=0}^{i-2}A_t, \end{aligned}$$

(14)

which is impossible because all the terms on the right have valuation strictly greater than the term on the left. Similarly, if the right-hand side of Eq. (13) is $-1$, we have

$$\begin{aligned} -2 = \sum _{t=0}^{i-1}A_t, \end{aligned}$$

(15)

which again is impossible, because $v_p(A_t) > v_p(2)=v_p(-2)$ for all $0\le t\le i-1$. Either way, we have our desired contradiction.

Case 2. Assume $\ell =2$, so that $\zeta =-1$. With notation as in Case 1, we have

$$\begin{aligned} (-1)^i+\sum _{t=0}^{i-1} (-1)^{t}A_{t}=\pm 1, \quad \text {i.e.,}\quad \sum _{t=0}^{i-1} (-1)^{t}A_{t} \in \{-2,0,2\} . \end{aligned}$$

If the sum above is zero, we reach a contradiction in the same way as in Eq. (14); or if it is $\pm 2$, we reach a contradiction in the same way as in Eq. (15).

Case 3. Assume for the rest of the proof that $\ell \ge 3$. Note that the subgroup $\langle \zeta \rangle \mathbb {Z}[\zeta +\zeta ^{-1}]^{\times }$ of the unit group $\mathbb {Z}[\zeta ]^{\times }$ has index at most two inside of $\mathbb {Z}[\zeta ]^{\times }$, by [23, Theorem 4.12]. Define $Q(x)\in \mathbb {Z}[x]$ by $Q:=P$ if this index is 1, or $Q:=P^2$ if this index is 2. Then Q is p-special by Lemma 6.3, and $Q(\zeta )\in \langle \zeta \rangle \mathbb {Z}[\zeta +\zeta ^{-1}]^{\times }$. Assumption (12) therefore implies that

$$\begin{aligned} Q(\zeta ) = \zeta ^iF(\zeta +\zeta ^{-1}) \end{aligned}$$

(16)

for some polynomial $F(x)\in \mathbb {Z}[x]$ and some integer i. Setting $R:=Q^{\ell }$, which has degree $k\ell$ for some positive integer k, we have that R is again p-special by Lemma 6.3. Taking the $\ell$-th power of both sides of Eq. (16) yields

$$\begin{aligned} R(\zeta ) = G(\zeta +\zeta ^{-1}) \end{aligned}$$

(17)

for some polynomial $G(x)\in \mathbb {Z}[x]$ with $\deg (G)<\phi (\ell )/2$, where $\phi$ is the Euler totient function, since $[\mathbb {Q}(\zeta +\zeta ^{-1}):\mathbb {Q}]=\phi (\ell )/2$.

Write

$$\begin{aligned} R(x)=x^{k\ell }+\sum _{t=0}^{k\ell -1}B_{t}x^{t} \quad \text {and}\quad G(x)=\sum _{t=0}^{m}D_tx^t . \end{aligned}$$

Substituting in (17), we obtain

$$\begin{aligned} 1+\sum _{t=0}^{k\ell -1}B_{t}\zeta ^t=\sum \limits _{t=0}^{m} D_t(\zeta +\zeta ^{-1})^t. \end{aligned}$$

Multiplying by $\zeta ^m$ and moving one term to the other side yields

$$\begin{aligned} \sum _{t=0}^{k\ell -1} B_{t}\zeta ^{m+t}= \Bigg (\sum _{t=1}^{m} D_t(\zeta ^2+1)^t\zeta ^{m-t}\Bigg )+(D_0-1)\zeta ^m. \end{aligned}$$

(18)

Let $r:=v_p(B_{k\ell -1}) > v_p(2)$. Since R is a p-special polynomial, the left-hand side in (18) lies in $p^r\mathbb {Z}[\zeta ]$. We claim that $D_t\in p^r\mathbb {Z}$ for each $t=1,2,\ldots ,m$.

To prove the claim, suppose it were false, and consider the largest index $i\in \{1,2,\ldots ,m\}$ for which $D_i\not \equiv 0 \pmod {p^r}$. Note that the term with the largest power of $\zeta$ in $D_i(\zeta ^2+1)^i\zeta ^{m-i}$ is $D_i\zeta ^{m+i}$, and for any $t<i$ with $D_t\not \equiv 0 \pmod {p^r}$, the term $D_t(\zeta ^2+1)^t\zeta ^{m-t}$ only involves powers $\zeta ^e$ with $0\le e <m+i$. Hence, recalling that $0\le m<\phi (\ell )/2$, and thus $m+i<\phi (\ell )$, it follows immediately that the right-hand side of Eq. (18), when written as a $\mathbb {Z}$-linear combination of the integral basis $\{1,\zeta ,\zeta ^2,\ldots ,\zeta ^{\phi (\ell )-1}\}$, has $\zeta ^{m+i}$-coefficient not congruent to 0 modulo $p^r$. Thus, the right-hand side of Eq. (18) cannot lie in $p^r\mathbb {Z}[\zeta ]$, which is a contradiction, proving our claim.

By the claim, it follows that both the left-hand side and the sum in parentheses in Eq. (18) must lie in $p^r\mathbb {Z}[\zeta ]$. Therefore, we must also have $(D_0-1)\zeta ^m\in p^r \mathbb {Z}[\zeta ]$. Define $B_t' : =p^{-r} B_t\in \mathbb {Z}$ for each $0\le t\le k\ell -1$, and $D_t' : =p^{-r} D_t\in \mathbb {Z}$ for each $1\le t\le m$, and $D_0' : =p^{-r} (D_0-1)\in \mathbb {Z}$. Then Eq. (18) becomes

$$\begin{aligned} \sum _{t=0}^{k\ell -1}B'_{t}\zeta ^{m+t}=\sum _{t=0}^{m}D'_t(\zeta ^2+1)^t\zeta ^{m-t}. \end{aligned}$$

(19)

Note that $B'_{k\ell -1}\not \equiv 0 \pmod {p}$, and $B'_t\equiv 0\pmod {p}$ for $t=0,\ldots k\ell -2$, because R is p-special. In addition, we have $\zeta ^{m+k\ell -1}=\zeta ^{m-1}$. Thus, reducing Eq. (19) modulo $p\mathbb {Z}[\zeta ]$ yields

$$\begin{aligned} \Bigg (\sum _{t=0}^{m}D'_t(\zeta ^2+1)^t\zeta ^{m-t}\Bigg )-B'_{k\ell -1}\zeta ^{m-1}\equiv 0\pmod {p\mathbb {Z}[\zeta ]}. \end{aligned}$$

(20)

Because we assumed $\ell \ge 3$, we have $\zeta ^m\ne \pm \zeta ^{m-1}$. If $D'_1,D'_2,\ldots ,D'_m$ were all divisible by p, we would obtain $D'_0\zeta ^m-B_{k\ell -1}'\zeta ^{m-1}\equiv 0 \pmod {p\mathbb {Z}[\zeta ]}$, which is impossible. Thus, we may define i to be the largest index $1\le i\le m$ for which $p\not \mid D'_i$.

Note that the term with the largest power of $\zeta$ in $D_i'(\zeta ^2+1)^{i}\zeta ^{m-i}$ is $D_i'\zeta ^{m+i}$. As above, for any positive integer $t<i$ with $p\not \mid D'_i$, the term $D'_t(\zeta ^2+1)^t\zeta ^{m-t}$ only involves powers $\zeta ^a$ with $a<m+i$. Hence, when the left side of Eq. (20) is fully expanded, the coefficient of $\zeta ^{m+i}$ is relatively prime to p. This contradicts Eq. (20) itself, completing the proof of Theorem 6.2.

7 The case $d=2$: Theorem 1.7 via 2-special polynomials

For the remainder of the paper, we consider Conjecture 5.2 in the case $d=2$, i.e., $f_{d,c}(z)=z^2+c$. For ease of notation, we will hereafter write simply $G_{m,n}$ , $P_{m,n}$ , and $\lambda _{m,n}$ instead of $G_{2,m,n}^{\zeta }$ , $P_{2,m,n}^{\zeta }$ , and $\lambda _{2,m,n}^{\zeta }$ , respectively.

With this notation, we propose the following conjecture, which we have verified using Magma for all pairs (m, n) with $m+n\le 10$. By Theorem 6.2, it clearly implies Conjecture 5.2 in the case $d=2$.

Conjecture 7.1

Let $m\ge 2$ and $n\ge 1$. Then $P_{m,n}$ is a 2-special polynomial.

Most of the rest of this paper is devoted to the proof of the following theorem, which confirms Conjecture 7.1 for $n=1$ and $n=2$.

Theorem 7.2

Let $m\ge 2$ be an integer. Then both $P_{m,1}$ and $P_{m,2}$ are 2-special.

Assuming Theorem 7.2, we can now prove Theorem 1.7.

Proof of Theorem 1.7

By [14, Corollary 1.1], both $G_{m,1}$ and $G_{m,2}$ are irreducible over $\mathbb {Q}$. By Theorem 7.2, both $P_{m,1}$ and $P_{m,2}$ are 2-special, and therefore, by Theorem 6.2, we have

$$\begin{aligned} {{\,\textrm{Res}\,}}(P_{m,1},\Phi _i) \text { and } {{\,\textrm{Res}\,}}(P_{m,2},\Phi _i) \text { are not units in } \mathbb {Z}, \quad \text {for all } i\ge 1. \end{aligned}$$

(21)

With $i=n$ in Eq. (21), then choosing $\ell =1$ in Proposition 5.3 yields that $G_{m,1}(c_0)$ is not a unit in $\mathcal {O}_K$. Similarly, if n is even, choosing $i=n/2$ and $\ell =2$ yields that $G_{m,2}(c_0)$ is not a unit in $\mathcal {O}_K$. $\square$

It remains only to prove Theorem 7.2, for which we will need the following explicit formulas for $P_{m,1}$ and $P_{m,2}$.

Lemma 7.3

Let $m\ge 2$. Then:

• (a) $\displaystyle P_{2,1}(x) = x-4$,

• (b) $\displaystyle P_{m,1}(x) = 2^{2^{m-1}} x^{-1}a_{m-1}\bigg (\frac{-x^2+2x}{4}\bigg ) +2^{(2^{m-1}-1)}$ if $m\ge 3$, and

• (c) $\displaystyle P_{m,2}(x)= 4^k G_{m,2}\bigg (\frac{x-4}{4}\bigg )$, where $k:=\deg (G_{m,2})$.

Proof

(a) and (b). Enumerate the roots of $G_{m,1}$ as $c_1,c_2,\ldots ,c_k$, where $k:=\deg (G_{m,1})=2^{m-1}-1$. For each $i=1,\ldots ,k$, define

$$\begin{aligned} \gamma _i = -a_{m-1}(c_i) = a_m(c_i), \end{aligned}$$

so that the multiplier of the cycle in the periodic orbit is $\lambda _i:=2\gamma _i$. Observe that

$$\begin{aligned} \gamma _i^2+c_i = f_{c_i}(-a_{m-1}(c_i))=a_m(c_i)=\gamma _i, \end{aligned}$$

and hence $c_i = -\gamma _i^2+\gamma _i$. Thus,

$$\begin{aligned} a_{m-1}(-\gamma _i^2+\gamma _i)=a_{m-1}(c_i)=-\gamma _i. \end{aligned}$$

Adding $\gamma _i$ to both sides and dividing by $\gamma _i$ — while remembering that $a_{m-1}$ has no constant term — we obtain

$$\begin{aligned} Q_1(\gamma _i)=0, \quad \text {where}\quad Q_1(x):=\frac{1}{x}a_{m-1}(-x^2+x)+1. \end{aligned}$$

If $m=2$, then $a_{m-1}(x)=x$, and hence $Q_1(x)=2-x$. On the other hand, if $m\ge 3$, then $a_{m-1}$ is monic of (even) degree $2^{m-2}$, so that $Q_1$ is also monic, but of degree $2^{m-1}-1$. Thus, define ${\tilde{Q}}_1:=-Q_1$ if $m=2$, or ${\tilde{Q}}_1=Q_1$ if $m\ge 3$. Then ${\tilde{Q}}_1$ is monic of degree $2^{m-1}-1$, which is the same as the degree k of $G_{m,1}$. Because each root $c_i$ of $G_{m,1}$ yields a root of ${\tilde{Q}}_1$ via $\gamma _i = -a_{m-1}(c_i)$, and conversely via $c_i = -\gamma _i^2+\gamma _i$, the roots of ${\tilde{Q}}_1$ are precisely $\gamma _1,\ldots ,\gamma _k$. Therefore, since ${\tilde{Q}}_1$ is monic, we have

$$\begin{aligned} {\tilde{Q}}_1(x) = \prod _{i=1}^k (x-\gamma _i) . \end{aligned}$$

The corresponding multipliers satisfy $\lambda _i = 2\gamma _i$, and hence

$$\begin{aligned} P_{m,1}(x) = \prod _{i=1}^k (x-\lambda _i) = 2^{k}\prod _{i=1}^k \bigg (\frac{x}{2}-\gamma _i\bigg ) = 2^{(2^{m-1}-1)}{\tilde{Q}}_1\bigg (\frac{x}{2}\bigg ). \end{aligned}$$

If $m=2$, and hence ${\tilde{Q}}_1(x)=x-2$, this expression is $2(\frac{x}{2}-2) = x-4$, proving part (a). Otherwise, it is

$$\begin{aligned} P_{m,1}(x) = 2^{(2^{m-1}-1)} \bigg [ \frac{2}{x} a_{m-1} \bigg ( \frac{-x^2}{4} + \frac{x}{2} \bigg ) + 1 \bigg ] = 2^{2^{m-1}} x^{-1} a_{m-1} \bigg ( \frac{-x^2+2x}{4} \bigg ) + 2^{(2^{m-1}-1)}, \end{aligned}$$

proving part (b).

Part (c). Enumerate the roots of $G_{m,2}$ as $c_1,\ldots ,c_k$. (This time, $k=2^{m-1}$ if m is even, or $k=2^{m-1}-1$ if m is odd, although we will not need those exact values in this part of the proof.) For each $i=1,\ldots ,k$, define

$$\begin{aligned} \gamma _i:= -a_{m-1}(c_i)a_m(c_i) = a_m(c_i)a_{m+1}(c_i), \end{aligned}$$

so that the multiplier of the cycle in the periodic orbit is $\lambda _i = 4\gamma _i$. Since $a_m(c_i)$ and $a_{m+1}(c_i)$ are both 2-periodic points of $f_{c_i}$ , they are the two roots of the polynomial $z^2+z+(c_i+1)$, and therefore their product $\gamma _i$ is the constant term of this polynomial. That is, $\gamma _i = c_i + 1$. Thus, defining $Q_2$ to be the monic polynomial $Q_2(x):= G_{m,2}(x-1)\in \mathbb {Z}[x]$, we have

$$\begin{aligned} Q_2(x) = \prod _{i=1}^k (x-\gamma _i) . \end{aligned}$$

The corresponding multipliers satisfy $\lambda _i = 4\gamma _i$, and hence

$$\begin{aligned} P_{m,2}(x) = \prod _{i=1}^k (x-\lambda _i) = 4^{k}\prod _{i=1}^k \bigg (\frac{x}{4}-\gamma _i\bigg ) = 4^k Q_2\bigg (\frac{x}{4}\bigg ) = 4^k G_{m,2}\bigg (\frac{x-4}{4}\bigg ) . \end{aligned}$$

$\square$ $\square$

To prove Theorem 7.2, we will also need the following elementary lemma.

Lemma 7.4

Let $\ell \ge 3$ and $0\le j\le 2^{\ell }-3$ be integers. Then

$$\begin{aligned} 2^{\ell } > j + v_2(j) + 1. \end{aligned}$$

Proof

Suppose first that $j=2^e$ for some $0\le e < \ell$. If $e\ge 2$, then $e+1<2^e$, and hence

$$\begin{aligned} j + v_2(j) + 1 = 2^e + e + 1 < 2^{e+1} \le 2^{\ell }, \end{aligned}$$

as desired. Otherwise, we have $e\le 1$, and hence $2^e+e+1\le 4 < 2^{\ell }$, since $\ell \ge 3$. Either way, we are done.

The only other possibility is that $2^{e-1}<j<2^e$ for some $2\le e\le \ell$. Note that we must have $v_2(j)\le e-2$ in this case. If $e\le \ell -1$, then since $e<2^e$, we have

$$\begin{aligned} j + v_2(j) + 1< 2^e + (e-2) + 1<2^{e+1} - 1 < 2^{\ell } . \end{aligned}$$

Otherwise, we have $e=\ell$. If $v_2(j)\le 1$, then because $j\le 2^{\ell }-3$, we have

$$\begin{aligned} j + v_2(j) + 1 \le 2^{\ell } -3 + 1 + 1 < 2^{\ell }. \end{aligned}$$

The only remaining case is that $e=\ell$ and $v_2(j)\ge 2$. We must have $j\le 2^{\ell }-2^{r}$, where $r:=v_2(j)\ge 2$, and hence

$$\begin{aligned} j + v_2(j) + 1 \le 2^{\ell } -2^r + r + 1 < 2^{\ell }. \end{aligned}$$

$\square$

8 The case $d=2$: Proving Theorem 7.2

Throughout this section, write

$$\begin{aligned} a_{\ell }(c) = \sum _{i=0}^{2^{\ell -1}} A_{\ell ,i} c^i \qquad \text {for any }\ell \ge 1. \end{aligned}$$

(22)

Using the fact that $a_{\ell +1}=(a_{\ell })^2 +c$, a simple induction on $\ell$ shows that for all $\ell \ge 2$, we have

$$\begin{aligned} A_{\ell ,2^{\ell -1}}=1, \quad A_{\ell ,2^{\ell -1}-1} = 2^{\ell -2}, \quad A_{\ell ,1}=1, \quad \text {and}\quad A_{\ell ,0}=0. \end{aligned}$$

(23)

We also have the following technical but more general bound.

Lemma 8.1

Let $\ell \ge 2$ and $0\le i\le 2^{\ell -1}-2$ be integers. Then

$$\begin{aligned} v_2(A_{\ell ,i}) > 2i + \ell - 2^\ell . \end{aligned}$$

(24)

Remark 8.2

For $i=2^{\ell -1}-1$, it follows immediately from (23) that $v_2(A_{\ell ,i}) = 2i+\ell -2^{\ell }$. We will need this fact in the proof of Lemma 8.1.

Proof of Lemma 8.1

We proceed by induction on $\ell \ge 2$. For $\ell =2$, we must have $i=0$, and hence the right side of inequality (24) is $0+2-4=-2<0$. Therefore, the desired bound holds trivially.

For the rest of the proof, fix some $\ell \ge 2$, and assume inequality (24) holds for that particular $\ell$; we must show it holds for $\ell +1$ as well. We have

$$\begin{aligned} a_{\ell +1}(c)&= a_{\ell }(c)^2+c =\Bigg (\sum _{i=1}^{2^{\ell -1}} A_{\ell ,i}c^i\Bigg )^2+c \\&= \sum _{i=1}^{2^{\ell -1}} A_{\ell ,i}^2 c^{2i} +2\Bigg (\sum _{1\le i<j\le 2^{\ell -1}} A_{\ell ,i}A_{\ell ,j}c^{i+j}\Bigg )+c. \end{aligned}$$

Therefore, the coefficients $A_{\ell +1,i}$ of $a_{\ell +1}$ are given by

$$\begin{aligned} A_{\ell +1,i} = {\left\{ \begin{array}{ll} \displaystyle A_{\ell ,i/2}^2+\sum _{\begin{array}{c} 0\le r< s\le 2^{\ell -1}\\ r+s=i \end{array}} 2A_{\ell ,r}A_{\ell ,s} &{} \text { if { i} is even}, \\ \displaystyle \sum _{\begin{array}{c} 0\le r< s\le 2^{\ell -1}\\ r+s=i \end{array}} 2A_{\ell ,r}A_{\ell ,s} &{} \text { if i>1 is odd}, \\ 1 &{} \text { if } i=1. \end{array}\right. } \end{aligned}$$

(25)

Since $A_{\ell +1,0}=0$, to finish the induction, we must show that

$$\begin{aligned} v_2(A_{\ell +1,i}) \overset{\text {?}}{>} \ell +1+2i-2^{\ell +1} \end{aligned}$$

(26)

for all $1\le i\le 2^{\ell }-2$.

Given such i, suppose first that i is even. We have

$$\begin{aligned} v_2\big (A^2_{\ell ,i/2}\big ) = 2v_2\big (A_{\ell ,i/2}\big ) \ge 2(\ell +i-2^{\ell }) > \ell +1 + 2i - 2^{\ell +1}, \end{aligned}$$

where the first inequality is by Remark 8.2 (needed for the case $i=2^{\ell }-2$) and our inductive hypothesis that (24) holds for $\ell$, and the second is because $\ell \ge 2$, and hence $2\ell > \ell +1$. Similarly,

$$\begin{aligned} v_2\big ( 2A_{\ell ,r}A_{\ell ,s} \big )&= 1+v_2(A_{\ell ,r})+v_2(A_{\ell ,s}) \ge 1+ v_2(A_{\ell ,r}) \nonumber \\&> 1+\ell +2r-2^{\ell } \ge \ell +1+2i-2^{\ell +1} \end{aligned}$$

(27)

where the first inequality is again by our inductive hypothesis, and the rest is because $r+s=i$ and $s\le 2^{\ell -1}$. Thus, according to Eq. (25), the desired bound (26) holds when i is even.

If $i>1$ is odd, inequality (26) is immediate from inequality (27) and Eq. (25).

Finally, if $i=1$, inequality (26) is $0>\ell +3-2^{\ell +1}$, which holds because $\ell \ge 2$. $\square$

At last, we are prepared to prove Theorem 7.2. We treat the cases $n=1$ and $n=2$ separately.

Proof of Theorem 7.2 for $n=1$

For $m=2,3,4$, direct computation with Magma gives:

$$\begin{aligned} P_{2,1} = x-4,\quad P_{3,1} = x^3 - 4x^2 + 16, \quad P_{4,1} = x^7 - 8x^6 + 16x^5 + 16x^4 - 64x^3 + 256, \end{aligned}$$

all of which are 2-special. Thus, we may assume hereafter that $m\ge 5$.

Part (b) of Lemma  therefore gives us

$$\begin{aligned} P_{m,1}(x)&= 2^{(2^{m-1}-1)} + 2^{2^{m-1}} \sum _{i=1}^{2^{m-2}} A_{m-1,i} x^{-1} \bigg ( \frac{2x -x^2}{4}\bigg )^i \\&=2^{(2^{m-1}-1)} + \sum _{i=1}^{2^{m-2}} 2^{(2^{m-1}-2i)} A_{m-1,i} x^{i-1} (2-x)^i \\&= 2^{(2^{m-1}-1)} + \sum _{i=1}^{2^{m-2}}\sum _{j=0}^{i}2^{(2^{m-1}-i-j)}A_{m-1,i }(-1)^j \left( {\begin{array}{c}i\\ j\end{array}}\right) x^{i+j-1} \\&= x^{(2^{m-1}-1)} + \sum _{t=0}^{2^{m-1}-2} B_t x^t, \end{aligned}$$

where $B_0= 2^{2^{m-1}}$ (using the fact that $A_{m-1,1}=1$), and

(28)

Formula (28) yields that $B_{2^{m-1}-2}=-2^{m-1}$. Recalling that $m\ge 5$, it follows that

$$\begin{aligned} v_2(B_{2^{m-1}-2})=m-1\ge 2, \quad \text {and}\quad v_2(B_0)=2^{m-1}>v_2(B_{2^{m-1}-2}), \end{aligned}$$

verifying the first bullet point of Definition 6.1, and a small part of the second. Therefore, to finish verifying that $P_{m,1}$ is 2-special, it remains to show that $v_2(B_t)>m-1$ for each integer $1\le t\le 2^{m-1}-3$. We achieve this goal in the following four steps.

Step 1 We claim that

$$\begin{aligned} v_2(B_{2^{m-1}-3})>m-1. \end{aligned}$$

Observe that there are only two pairs of indices $j\le i\le 2^{m-2}$ such that $i+j=2^{m-1}-2$, namely $i=j=2^{m-2}-1$, and $i=2^{m-2}$, $j=2^{m-2}-2$. Therefore, formula (28) gives

$$\begin{aligned} B_{2^{m-1}-3}&= -4A_{m-1,2^{m-2}-1}+4A_{m-1,2^{m-2}} \left( {\begin{array}{c}2^{m-2}\\ 2\end{array}}\right) \\&= -4(2^{m-3}) + 4(1) \big (2^{m-3}(2^{m-2}-1)\big ) = 2^{m-1}(2^{m-2}-2), \end{aligned}$$

where we used Eq. (23) with $\ell =m-1$ in the second equality. Thus, $v_2(B_{2^{m-1}-3})=m>m-1$, as desired.

Step 2 We claim that for $i=2^{m-2}$ and $0\le j\le 2^{m-2}-3$, we have

$$\begin{aligned} v_2\bigg ( 2^{2^{m-1}-i-j}A_{m-1,i}(-1)^{j} \left( {\begin{array}{c}i\\ j\end{array}}\right) \bigg ) > m-1. \end{aligned}$$

(29)

By Eq. (23), we have $A_{m-1,i}=1$. Therefore, by the identity $v_2(\left( {\begin{array}{c}2^{m-2}\\ j\end{array}}\right) )=m-2-v_2(j)$, inequality (29) becomes

$$\begin{aligned} \big (2^{m-1} - 2^{m-2} - j\big ) + \big ( m-2-v_2(j) \big ) > m-1, \end{aligned}$$

which holds by applying Lemma 7.4 with $\ell =m-2\ge 3$.

Step 3 We claim that inequality (29) also holds for $i=2^{m-2}-1$ and $0\le j<2^{m-2}-1$. That is, we are claiming that

$$\begin{aligned} 2^{m-2}+v_2\bigg (\left( {\begin{array}{c}2^{m-2}-1\\ j\end{array}}\right) \bigg ) \overset{? }{>} j+1 , \end{aligned}$$

which we obtained by substituting $i=2^{m-2}-1$ and $v_2(A_{m-1},i)=m-3$ from Eq. (23) into inequality (29), and simplifying. However, this desired inequality is immediate from our assumption that $j<2^{m-2}-1$.

Step 4 We claim that inequality (29) also holds for $0\le j \le i\le 2^{m-2}-2$. In that case, the left side of inequality (29) is

$$\begin{aligned} 2^{m-1}-i-j+v_2(A_{m-1,i}) + v_2\bigg (\left( {\begin{array}{c}i\\ j\end{array}}\right) \bigg ) \ge 2^{m-1} -2i + v_2(A_{m-1,i}) > m-1 \end{aligned}$$

as desired, where the second inequality is by Lemma 8.1 with $\ell =m-1\ge 4$.

The claims of Steps 1–4, together with Eq. (28), show that $v_2(B_t) > m-1$ for each $1\le t \le 2^{m-1} -3$, which as we noted completes our proof that $P_{m,1}$ is 2-special. $\square$

Proof of Theorem 7.2 for $n=2$

The main Möbius product in the definition of $G^{\zeta }_{d,m,n}$ in Eq. (2), in our case of $G_{m,2}$ , is

$$\begin{aligned} H_m(c) := \frac{a_{m+1}+a_{m-1}}{a_m+a_{m-1}} = \frac{a_{m+1}-a_m}{a_m+a_{m-1}} + \frac{a_m+a_{m-1}}{a_m+a_{m-1}} = a_m-a_{m-1}+1, \end{aligned}$$

since $a_{m+1}-a_m = (a^2_m +c) - (a^2_{m-1} +c) = a^2_m-a^2_{m-1}$. Note that $\text {deg}(H_m) = 2^{m-1}$. Inspired by part (c) of Lemma , we define

$$\begin{aligned} R_m(x):= 4^{2^{m-1}} H_m\bigg (\frac{x-4}{4}\bigg ) = 4^{2^{m-1}}\bigg (a_m\bigg (\frac{x-4}{4}\bigg )-a_{m-1}\bigg (\frac{x-4}{4}\bigg )+1\bigg ). \end{aligned}$$

(30)

If m is even, then Eq. (2) gives $G_{m,2}=H_m$ , and therefore part (c) of Lemma  gives $P_{m,2}=R_m$. On the other hand, if m is odd, then

$$\begin{aligned} G_{m,2} = H_m \prod _{d|2}a_d^{-\mu (2/d)} = H_m \cdot \frac{c}{c^2 + c} = (c+1)^{-1} H_m, \end{aligned}$$

which is a monic polynomial by [3, Theorem 2.1], and which has degree degree $2^{m-1}-1$. Thus, still assuming m is odd, part (c) of Lemma  gives

$$\begin{aligned} P_{m,2}(x) = 4^{(2^{m-1}-1)} G_{m,2}\bigg (\frac{x-4}{4}\bigg ) = \bigg ( \frac{1}{4} \bigg (\frac{x-4}{4} + 1\bigg )^{-1} \bigg ) R_m(x) = \frac{1}{x}R_m(x), \end{aligned}$$

which must be a polynomial (in spite of the 1/x) because $G_{m,2}$ is a polynomial. That is, for m odd, the polynomial $R_m$ of Eq. (30) must have constant term zero.

Thus, it suffices to show, for all $m\ge 2$, that $R_m$ is 2-special. Indeed, for m odd, if $R_m$ is 2-special, then since it also has constant term zero, it is immediate from Definition 6.1 that $P_{m,2}(x)=R_m(x)/x$ is 2-special as well; and for m even, we have $P_{m,2}=R_m$.

For $m=2,3,4$, direct computation of the above formulas with Magma gives:

$$\begin{aligned}{} & {} R_2 = x^2-8x+32, \quad R_3 = x^4 - 8x^3 + 128x, \quad \text {and} \\{} & {} R_4 = x^8 - 16x^7 + 96x^6 - 128x^5 - 1536x^4 + 8192x^3 - 8192x^2 - 32768x +131072, \end{aligned}$$

all of which are 2-special. Therefore, for the remainder of the proof, we assume $m\ge 5$, and we must prove that $R_m$ is 2-special.

Writing

$$\begin{aligned} a_m-a_{m-1}+1 = \sum _{i=0}^{2^{m-1}} D_{m,i}c^i \end{aligned}$$

(31)

and using this expansion in Eq. (30), we obtain

$$\begin{aligned} R_m(x)&= 4^{2^{m-1}}\sum _{i=0}^{2^{m-1}} D_{m,i}\bigg (\frac{x-4}{4}\bigg )^i = 4^{2^{m-1}} \sum _{i=0}^{2^{m-1}} D_{m,i}\sum _{j=0}^{i} (-1)^{i-j}\left( {\begin{array}{c}i\\ j\end{array}}\right) \bigg (\frac{x}{4}\bigg )^j\\&= \sum _{i=0}^{2^{m-1}} D_{m,i}\sum _{j=0}^{i} (-1)^{i-j} 4^{2^{m-1}-j}\left( {\begin{array}{c}i\\ j\end{array}}\right) x^j = x^{2^{m-1}} + \sum _{t=0}^{2^{m-1}-1}E_t x^t, \end{aligned}$$

where

$$\begin{aligned} E_t := 4^{2^{m-1}-t}\sum _{i=t}^{2^{m-1}} (-1)^{i-t}D_{m,i}\left( {\begin{array}{c}i\\ t\end{array}}\right) \quad \text {for } t=0,1,\ldots ,2^{m-1}-1. \end{aligned}$$

(32)

Observe that for $i>\deg (a_{m-1})$, the coefficients $D_{m,i}$ in Eq. (31) coincide with the coefficients $A_{m,i}$ from Eq. (22). That is, we have

$$\begin{aligned} D_{m,i} = A_{m,i} \quad \text {for } i=2^{m-2}+1,\ldots ,2^{m-1} . \end{aligned}$$

(33)

Recalling from Eq. (23) that $A_{m,2^{m-1}}=1$ and $A_{m,2^{m-1}-1}=2^{m-2}$, Eqs. (32) and (33) yield

$$\begin{aligned} E_{2^{m-1}-1}&= 4\sum _{i=2^{m-1}-1}^{2^{m-1}}(-1)^{i-2^{m-1}+1}D_{m,i}\left( {\begin{array}{c}i\\ 2^{m-1}-1\end{array}}\right) \\&= 4\sum _{i=2^{m-1}-1}^{2^{m-1}}(-1)^{i-2^{m-1}+1}A_{m,i}\left( {\begin{array}{c}i\\ 2^{m-1}-1\end{array}}\right) \\&= 4A_{m,2^{m-1}-1}-4A_{m,2^{m-1}}2^{m-1} = -2^m. \end{aligned}$$

Thus, this coefficient has valuation $v_2(-2^m)=m > v_2(2)$, verifying the first bullet point of Definition 6.1. Therefore, to finish the proof, we need to verify that

$$\begin{aligned} v_2(E_t)>m \quad \text {for } t=0,1,\ldots ,2^{m-1}-2 . \end{aligned}$$

(34)

For $0\le t\le 2^{m-2}$, Eq. (32) gives us

$$\begin{aligned} v_2(E_t)\ge v_2(4^{2^{m-1}-t})\ge 2^{m-1}>m, \end{aligned}$$

which establishes the desired inequality, since $m\ge 5$.

For the remainder of the proof of (34), then, we consider $t>2^{m-2}$. By Eqs. (32) and (33), it suffices to show

$$\begin{aligned} 2^m-2t+v_2(A_{m,i})+v_2\bigg (\left( {\begin{array}{c}i\\ t\end{array}}\right) \bigg ) \overset{? }{>} m \quad \text { for }2^{m-2} < t \le 2^{m-1}-2 \,\text { and }\, i\ge t . \end{aligned}$$

(35)

To prove this, we consider three cases.

Case 1 If $i=2^{m-1}$, then $A_{m,i}=1$ and $v_2(\left( {\begin{array}{c}i\\ t\end{array}}\right) )=m-1-v_2(t)$, and hence goal (35) becomes

$$\begin{aligned} 2^m-2t+m-1-v_2(t) \overset{? }{>} m \quad \text { for }2^{m-2} < t \le 2^{m-1}-2. \end{aligned}$$

For $t=2^{m-1}-2$, we have $v_2(t)=1$, and hence

$$\begin{aligned} 2^m-2t+m-1-v_2(t) =2^m - 2^m + 4 + m-2 = m+2 > m, \end{aligned}$$

as desired. Otherwise, we have $t\le 2^{m-1}-3$, in which case

$$\begin{aligned} 2^m-2t+m-1-v_2(t) \ge 2^{m-1} - t - v_2(t) +m+2> m+3 > m, \end{aligned}$$

where we used the fact that $t\le 2^{m-1}-3$ and, in the second inequality, Lemma 7.4 with $\ell =m-1 \ge 4$ and $j=t$.

Case 2 If $i=2^{m-1}-1$, then $A_{m,i}=2^{m-2}$, and hence

$$\begin{aligned} 2^m-2t+v_2(A_{m,i})+v_2\bigg (\left( {\begin{array}{c}i\\ t\end{array}}\right) \bigg ) \ge 2^m-2t+m-2 \ge m+2 > m, \end{aligned}$$

as desired, where we used the fact that $t\le 2^{m-1}-2$ in the second inequality.

Case 3 It remains to consider $t\le i\le 2^{m-1}-2$. In this case, we have

$$\begin{aligned} 2^m-2t+v_2(A_{m,i})+v_2\bigg (\left( {\begin{array}{c}i\\ t\end{array}}\right) \bigg ) \ge 2^m-2t+v_2(A_{m,i}) \ge 2^m-2i+v_2(A_{m,i}) > m, \end{aligned}$$

where the last inequality is by Lemma 8.1 with $\ell =m$. $\square$

Appendix

As noted in the introduction, the resultants ${{\,\textrm{Res}\,}}(P_{m,n},\Phi _\ell )$ of Conjecture 1.4 are in fact very large integers, according to computational evidence. Table 1 presents the Weil heights of these resultants, rounded down to the nearest integer, for $m+n\le 7$ and $\ell \le 8$. (Recall that the Weil height of an integer N is $\log \max \{|N|,1\}$.)

Table 1 Heights of the resultants ${{\,\textrm{Res}\,}}(P_{m,n},\Phi _\ell )$

The data in Table 1 lead us to propose the following conjecture. Recall that $a\asymp b$ means that the quantities a and b have the same growth rate, i.e., there are constants $c,C>0$ such that $ca\le b \le Ca$.

Conjecture 8.3

Let $m\ge 2$ and $n,\ell \ge 1$. Then

$$\begin{aligned} \log \big | {{\,\textrm{Res}\,}}\big (P_{m,n}, \Phi _{\ell }\big )\big | \asymp n\deg (P_{m,n}) \deg (\Phi _{\ell }) . \end{aligned}$$

Indeed, throughout Table 1, we have

$$\begin{aligned} 0.71 \le \frac{ \log \big | {{\,\textrm{Res}\,}}\big (P_{m,n}, \Phi _{\ell }\big ) \big | }{ n\deg (P_{m,n}) \deg (\Phi _{\ell }) } \le 1.44 \end{aligned}$$

and outside of the rows with $m+n\le 4$, the upper bound of 1.44 drops to 0.82.

This growth rate should be expected if there are no particular coincidences aligning the multiplier polynomial $P_{m,n}$ and the cyclotomic polynomial $\Phi _\ell$. After all, each root $\lambda$ of $P_{m,n}$ is of the form $2^n\alpha$ where $\alpha =a_n(c_0)\cdots a_{n+m-1}(c_0)$ is an algebraic integer whose p-adic valuation is 0 for all odd primes p and very small for $p=2$. (See Theorem 1.4 of [3].) Hence, the expected size of $\lambda -\zeta$ should be about $2^n$, where $\zeta$ is a root of unity. The resultant is the product of all such differences across all roots $\lambda$ of $P_{m,n}$ and all roots $\zeta$ of $\Phi _\ell$, suggesting that $\log | {{\,\textrm{Res}\,}}(P_{m,n}, \Phi _{\ell } ) |$ should be on the order of $n\deg (P_{m,n}) \deg (\Phi _{\ell })$. Thus, there is both empirical and theoretical evidence to support Conjecture 1.4, that $|{{\,\textrm{Res}\,}}(P_{m,n}, \Phi _{\ell } ) |> 1$.

In spite of this growth rate, there are no particularly apparent patterns in the primes that show up in the factorizations of these resultants. See Table 2 for the prime factorizations for $m+n\le 5$ and $\ell \le 5$.

Table 2 Prime factorizations of the resultants ${{\,\textrm{Res}\,}}(P_{m,n},\Phi _\ell )$