Reciprocal Monogenic Septinomials of Degree 2n3

Jones, Lenny

doi:10.2478/amsil-2024-0003

Full Article

1.

Introduction

Let f (x) ∈ ℤ[x]. When we say that f (x) is “irreducible” or “reducible”, without reference to a particular field, we mean that f (x) is “irreducible” or “reducible” over the rational numbers ℚ. We call f (x) reciprocal if f (x) = x^deg(f) f (1/x). We let Δ(f) and Δ(K) denote the discriminants over ℚ, respectively, of f (x) and a number field K. If f (x) is irreducible, with f (θ) = 0 and K = ℚ(θ), then we have the well-known equation [1] (1.1) $Δ (f) = {[ℤ_{K} : ℤ [θ]]}^{2} Δ (K),$ \Delta \left(f \right) = {[{{\mathbb Z}_K}:{\mathbb Z}\left[ \theta \right]]^2}\Delta \left(K \right), where ℤ_K is the ring of integers of K. We define f (x) to be monogenic if f (x) is irreducible and ℤ_K = ℤ[θ], or equivalently from (1.1), that Δ(f ) = Δ(K). When f (x) is monogenic, we have that {1, θ, θ², . . . , θ^{deg f−1}} is a basis for ℤ_K, commonly referred to as a power basis. The existence of a power basis makes computations in ℤ_K easier, as in the case of the cyclotomic polynomials Φ_n(x) [12]. We see from (1.1) that if Δ(f) is squarefree, then f (x) is monogenic. However, the converse is false in general, and when Δ(f ) is not squarefree, it can be quite difficult to determine whether f (x) is monogenic.

Recently [9], a procedure was presented to manufacture infinite families of reciprocal quintinomials of degree 2ⁿ that are monogenic for all n ≥ 2. In this article, we use similar methods to construct infinite families of reciprocal septinomials in ℤ[x] of degree 2ⁿ3 that are monogenic for all n ≥ 1. We should point out that, using different methods, infinite families of reciprocal monogenic septinomials of degree 6 were given in [8]. Our main results here are the following:

Theorem 1.1

Let n, A, B ∈ ℤ with n ≥ 1. Define the reciprocal polynomial (1.2) $ℱ_{n, A, B} (x) : = x^{2^{n} \cdot 3} + 9 A x^{2^{n - 1} \cdot 5} + (27 A^{2} + 3) x^{2^{n + 1}} + 3 B x^{2^{n - 1} \cdot 3} + (27 A^{2} + 3) x^{2^{n}} + 9 A x^{2^{n - 1}} + 1 .$ {{\cal F}_{n,A,B}}\left(x \right): = {x^{{2^n} \cdot 3}} + 9A{x^{{2^{n - 1}} \cdot 5}} + \left({27{A^2} + 3} \right){x^{{2^{n + 1}}}} + 3B{x^{{2^{n - 1}} \cdot 3}} + \left({27{A^2} + 3} \right){x^{{2^n}}} + 9A{x^{{2^{n - 1}}}} + 1. If B ≢ 0 (mod 3), (Â, B̂) ∈ Ψ := {(1, 1), (3, 3)}, where $\hat{*} \in {0, 1, 2, 3}$ \hat * \in \left\{{0,1,2,3} \right\} is the reduction modulo 4 of ∗, and $D : = (3 B + 54 A^{2} + 18 A + 8) (3 B - 54 A^{2} + 18 A - 8) (B - 9 A^{3} - 6 A)$ {\cal D}: = \left({3B + 54{A^2} + 18A + 8} \right)\left({3B - 54{A^2} + 18A - 8} \right)\left({B - 9{A^3} - 6A} \right) is squarefree, then the septinomial ℱ_n,A,B(x) is monogenic for all n ≥ 1.

Corollary 1.2

Let ℱ_n,A,B(x) be as defined in (1.2). Then, for any u ∈ ℤ,

(1)
there exist infinitely many primes q such that ℱ_n,4u+1,12q+1(x) is monogenic for all n ≥ 1,
(2)
there exist infinitely many primes q such that ℱ_n,4u+3,12q+7(x) is monogenic for all n ≥ 1.

2.

Preliminaries

Definition 2.1 ([1])

Let ℛ be an integral domain with quotient field K, and let K̅ be an algebraic closure of K. Let f (x), g(x) ∈ ℛ[x], and suppose that $f (x) = a \prod_{i = 1}^{M} (x - α_{i}) \in \bar{K} [x]$ f\left(x \right) = a\mathop \prod \nolimits_{i = 1}^M (x - {\alpha_i}) \in \bar K\left[ x \right] and $g (x) = b \prod_{i = 1}^{N} (x - β_{i}) \in \bar{K} [x]$ g\left(x \right) = b\mathop \prod \nolimits_{i = 1}^N (x - {\beta_i}) \in \bar K\left[ x \right] . Then the resultant R(f, g) of f and g is: $R (f, g) = a^{N} \prod_{i = 1}^{M} g (α_{i}) = {(- 1)}^{MN} b^{M} \prod_{i = 1}^{N} f (β_{i}) .$ R\left({f,g} \right) = {a^N}\mathop \prod \limits_{i = 1}^M g\left({{\alpha_i}} \right) = {(- 1)^{MN}}{b^M}\mathop \prod \limits_{i = 1}^N f\left({{\beta_i}} \right).

Theorem 2.2

Let f (x) and g(x) be polynomials in ℚ[x], with respective leading coefficients a and b, and respective degrees M and N. Then $Δ (f \circ g) = {(- 1)}^{M^{2} N (N - 1) / 2} \cdot a^{N - 1} b^{M (MN - N - 1)} Δ {(f)}^{N} R (f \circ g, g^{'}) .$ \Delta \left({f \circ g} \right) = {(- 1)^{{M^2}N\left({N - 1} \right)/2}} \cdot {a^{N - 1}}{b^{M\left({MN - N - 1} \right)}}\Delta {(f)^N}R(f \circ g,\,g').

Remark 2.3

As far as we can determine, Theorem 2.2 is originally due to John Cullinan [2]. A proof of Theorem 2.2 can be found in [5].

The following theorem, known as Dedekind's Index Criterion, or simply Dedekind's Criterion if the context is clear, is a standard tool used in determining the monogenicity of a polynomial.

Theorem 2.4 (Dedekind [1])

Let K = ℚ(θ) be a number field, T (x) ∈ ℤ[x] the monic minimal polynomial of θ, and ℤ_K the ring of integers of K. Let q be a prime number and let denote reduction of ∗ modulo q (in ℤ, ℤ[x] or ℤ[θ]). Let $\bar{T} (x) = \prod_{i = 1}^{k} \bar{τ_{i}} {(x)}^{e_{i}}$ \bar T\left(x \right) = \mathop \prod \limits_{i = 1}^k \overline {{\tau_i}} {(x)^{{e_i}}} be the factorization of T (x) modulo q in 𝔽_q[x], and set $g (x) = \prod_{i = 1}^{k} τ_{i} (x),$ g\left(x \right) = \mathop \prod \limits_{i = 1}^k {\tau_i}\left(x \right), where the τ_i(x) ∈ ℤ[x] are arbitrary monic lifts of the $\bar{τ_{i}} (x)$ \overline {{\tau_i}} \left(x \right) . Let h(x) ∈ ℤ[x] be a monic lift of T̅ (x)/g̅(x) and set $F (x) = \frac{g (x) h (x) - T (x)}{q} \in ℤ [x] .$ F\left(x \right) = {{g\left(x \right)h\left(x \right) - T\left(x \right)} \over q} \in {\mathbb Z}\left[ x \right]. Then $\begin{array}{l} [ℤ_{K} : ℤ [θ]] ≢ 0 & (mod q) \Leftrightarrow gcd (\bar{F}, \bar{g}, \bar{h}) = 1 in 𝔽_{q} [x] . \end{array}$ \matrix{{\left[ {{{\mathbb Z}_K}:{\mathbb Z}\left[ \theta \right]} \right] \nequiv 0} \hfill & {\left({\bmod q} \right) \Leftrightarrow \gcd \left({\bar F,\overline g,\bar h} \right) = 1\,\,in\,\,{{\mathbb F}_q}[x].} \hfill \cr}

The next theorem follows from Corollary 2.10 in [10].

Theorem 2.5

Let K and L be number fields with K ⊂ L. Then $Δ {(K)}^{[L : K]} | Δ (L)$ \Delta {(K)^{\left[ {L:K} \right]}}|\Delta \left(L \right)

Theorem 2.6

Let G(t) ∈ ℤ[t], and suppose that G(t) factors into a product of distinct irreducibles, such that the degree of each irreducible is at most 3. Define $N_{G} (X) = | {p \leq X : p is prime and G (p) is squarefrree} | .$ {N_G}\left(X \right) = \left| {\{p\, \le \,X:\,p\,is\,prime\,and\,G(p)\,is\,squarefrree\}} \right|. Then, $N_{G} (X) \sim C_{G} \frac{X}{log (X)},$ {N_G}(X) \sim {C_G}{X \over {\log (X)}}, where $C_{G} = \prod_{ℓ_{prime}} (1 - \frac{ρ_{G} (ℓ^{2})}{ℓ (ℓ - 1)})$ {C_G} = \prod\limits_{{\ell_{prime}}} {\left({1 - {{{\rho_G}\left({{\ell^2}} \right)} \over {\ell \left({\ell - 1} \right)}}} \right)} and ρ_G(ℓ²) is the number of z ∈ (ℤ/ℓ²ℤ)^∗ such that G(z) ≡ 0 (mod ℓ²).

Remark 2.7

Theorem 2.6 follows from work of Helfgott, Hooley and Pasten [6, 7, 11]. For more details, see [8].

Definition 2.8

In the context of Theorem 2.6, for G(t) ∈ ℤ[t] and a prime ℓ, if G(z) ≡ 0 (mod ℓ²) for all z ∈ (ℤ/ℓ²ℤ)^∗, we say that G(t) has a local obstruction at ℓ.

The following immediate corollary of Theorem 2.6 is used to establish Corollary 1.2.

Corollary 2.9

Let G(t) ∈ ℤ[t], and suppose that G(t) factors into a product of distinct irreducibles, such that the degree of each irreducible is at most 3. To avoid the situation when C_G = 0, we suppose further that G(t) has no local obstructions. Then there exist infinitely many primes q such that G(q) is squarefree.

We make the following observation concerning G(t) from Corollary 2.9 in the special case when each of the distinct irreducible factors of G(t) is of the form a_it+b_i with gcd(a_i, b_i) = 1. In this situation, it follows that the minimum number of distinct factors required in G(t) so that G(t) has a local obstruction at the prime ℓ is 2(ℓ − 1). More precisely, in this minimum scenario, we have $\begin{array}{l} G (t) = \prod_{i = 1}^{2 (ℓ - 1)} (a_{i} t + b_{i}) \equiv C {(t - 1)}^{2} {(t - 2)}^{2} \dots {(t - (ℓ - 1))}^{2} & (mod ℓ) \end{array},$ \matrix{{G\left(t \right) = \mathop \prod \limits_{i = 1}^{2\left({\ell - 1} \right)} ({a_i}t + {b_i}) \equiv C{{(t - 1)}^2}{{(t - 2)}^2} \cdots {{(t - (\ell - 1))}^2}} \hfill & {(\bmod \,\ell)} \hfill \cr}, where C ≢ 0 (mod ℓ). Then each zero r of G(t) modulo ℓ lifts to the ℓ distinct zeros $r, r + ℓ, r + 2 ℓ, \dots \dots, r + (ℓ - 1) ℓ \in {(ℤ / ℓ^{2} ℤ)}^{*}$ r,\,\,\,\,\,r + \ell,\,\,\,\,\,r + 2\ell,\,\,\,\,\, \ldots \ldots,\,\,\,\,\,r + \left({\ell - 1} \right)\ell \in {({\mathbb Z}/{\ell^2}{\mathbb Z})^*} of G(t) modulo ℓ² [3, Theorem 4.11]. That is, G(t) has exactly ℓ(ℓ − 1) = ϕ(ℓ²) distinct zeros z ∈ (ℤ/ℓ²ℤ)^∗. Therefore, if the number of factors k of G(t) satisfies k < 2(ℓ − 1), then there must exist z ∈ (ℤ/ℓ²ℤ)^∗ for which G(z) ≢ 0 (mod ℓ²), and we do not need to check such primes ℓ for a local obstruction. Consequently, only finitely many primes need to be checked for local obstructions. They are precisely the primes ℓ such that ℓ ≤ (k + 2)/2.

The following proposition, which follows from a generalization of a theorem of Capelli, is a special case of the results in [4], and gives simple necessary and sufficient conditions for the irreducibility of polynomials of the form w(x^{2^k}) ∈ ℤ[x], when w(x) is monic and irreducible.

Proposition 2.10 ([4])

Let w(x) ∈ ℤ[x] be monic and irreducible, with deg(w) = m. Then w (x^{2^k}) is reducible if and only if there exist S₀ (x), S₁ (x) ∈ ℤ[x] such that either ${(- 1)}^{m} w (x) = {(S_{0} (x))}^{2} - x {(S_{1} (x))}^{2},$ {(- 1)^m}w\left(x \right) = {({S_0}(x))^2} - x{({S_1}(x))^2}, or $k \geq 2 and w (x^{2}) = {(S_{0} (x))}^{2} - x {(S_{1} (x))}^{2} .$ k \ge 2\,\,\,\,\,and\,\,\,\,\,w({x^2}) = {({S_0}(x))^2} - x{({S_1}(x))^2}.

3.

The proof of Theorem 1.1

For the proof of Theorem 1.1, we require the following lemma, which is of some independent interest.

Lemma 3.1

Let n, A, B ∈ ℤ, with n ≥ 1, and let ℱ_n,A,B(x) be as defined in (1.2). Then ℱ_n,A,B(x) is irreducible for all n ≥ 1 if and only if $(\hat{A}, \hat{B}) \in Γ = {(0, 2), (1, 1), (2, 2), (3, 3)},$ (\hat A,\hat B) \in \Gamma = \{(0,2),(1,1),(2,2),(3,3)\}, where $\hat{*} \in {0, 1, 2, 3}$ \hat * \in \left\{{0,1,2,3} \right\} is the reduction modulo 4 of ∗.

Proof

Suppose first that (Â, B̂) ∈ Γ. This direction of the proof is composed of several steps, each of which involves a proof by contradiction. For most of the steps, the procedure is the same. We assume that two particular polynomials are equal, equate coefficients on the two polynomials and show that there is no solution to the resulting system of equations. To see that there is no solution, we view the system of equations arising from equating the coefficients of the two polynomials as a system of congruences modulo 4. We then use a computer to verify, for every possible viable set of values of the variables modulo 4, that there is always at least one congruence that is impossible. Because there are so many possibilities, we do not provide all details of the computer calculations.

We begin by showing that (3.1) $ℱ_{1, A, B} (x) = x^{6} + 9 A x^{5} + (27 A^{2} + 3) x^{4} + 3 B x^{3} + (27 A^{2} + 3) x^{2} + 9 Ax + 1$ {{\cal F}_{1,A,B}}(x) = {x^6} + 9A{x^5} + (27{A^2} + 3){x^4} + 3B{x^3} + (27{A^2} + 3){x^2} + 9Ax + 1 is irreducible. Assume, by way of contradiction, that ℱ_1,A,B(x) is reducible. Observe that if $ℱ_{1, A, B} (1) = 54 A^{2} + 18 A + 3 B + 8 = 0,$ {{\cal F}_{1,A,B}}\left(1 \right) = 54{A^2} + 18A + 3B + 8 = 0, then B = −18A² − 6A− 8/3 ∉ ℤ. Similarly, ℱ_1,A,B(−1) ≠ 0. Hence, ℱ_1,A,B(x) has no linear factors by the Rational Zero Theorem. Suppose then that (3.2) $ℱ_{1, A, B} (x) = (x^{2} + a_{1} x + a_{0}) (x^{4} + b_{3} x^{3} + b_{2} x^{2} + b_{1} x + b_{0}),$ {{\cal F}_{1,A,B}}(x) = ({x^2} + {a_1}x + {a_0})({x^4} + {b_3}{x^3} + {b_2}{x^2} + {b_1}x + {b_0}), for some a_i, b_i ∈ ℤ. Expanding the right-hand side of (3.2) and equating coefficients with ℱ_1,A,B(x) in (3.1), we arrive at the system of equations: (3.3) $\begin{array}{l} constant term : & a_{0} b_{0} = 1 \\ x : & a_{0} b_{1} + a_{1} b_{0} = 9 A \\ x^{2} : & b_{0} + a_{1} b_{1} + a_{0} b_{2} = 27 A^{2} + 3 \\ x^{3} : & b_{1} + a_{1} b_{2} + a_{0} b_{3} = 3 B \\ x^{4} : & a_{0} + a_{1} b_{3} + b_{2} = 27 A^{2} + 3 \\ x^{5} : & a_{1} + b_{3} = 9 A . \end{array}$ \matrix{{{\rm{constant}}\,{\rm{term}}:} & {{a_0}{b_0} = 1} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {x:} & {{a_0}{b_1} + {a_1}{b_0} = 9A} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^2}:} \hfill & {{b_0} + {a_1}{b_1} + {a_0}{b_2} = 27{A^2} + 3} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^3}:} \hfill & {{b_1} + {a_1}{b_2} + {a_0}{b_3} = 3B} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^4}:} \hfill & {{a_0} + {a_1}{b_3} + {b_2} = 27{A^2} + 3} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^5}:} \hfill & {{a_1} + {b_3} = 9A.} \hfill \cr} As previously mentioned, to see that the system (3.3) has no solutions, we view (3.3) as a system of congruences modulo 4. Then we use a computer to check every (Â, B̂) ∈ Γ and every possible value for $\hat{a_{i}}$ \widehat {{a_i}} and $\hat{b_{i}}$ \widehat {{b_i}} , noting that either a₀ = b₀ = 1 or a₀ = b₀ = −1. In every situation, there is at least one impossible congruence. For example, if $\begin{matrix} (\hat{A}, \hat{B}) = (0, 2) & and & (\hat{a_{0}}, \hat{a_{1}}, \hat{b_{0}}, \hat{b_{1}}, \hat{b_{2}}, \hat{b_{3}}) = (3, 0, 3, 0, 2, 1) \end{matrix},$ \matrix{{\left({\hat A,\hat B} \right) = \left({0,2} \right)} & {{\rm{and}}} & {(\widehat {{a_0}},\widehat {{a_1}},\widehat {{b_0}},\widehat {{b_1}},\widehat {{b_2}},\widehat {{b_3}}) = (3,0,3,0,2,1)} \cr}, then the left-hand side of the congruence corresponding to x² reduces to 1 (mod 4), while the right-hand side reduces to 3 (mod 4). Also, the left-hand side of the congruence corresponding to x³ reduces to 3 (mod 4), while the right-hand side reduces to 2 (mod 4).

Hence, it must be that (3.4) $ℱ_{1, A, B} (x) = (x^{3} + a_{2} x^{2} + a_{1} x + a_{0}) (x^{3} + b_{2} x^{2} + b_{1} x + b_{0}),$ {{\cal F}_{1,A,B}}(x) = ({x^3} + {a_2}{x^2} + {a_1}x + {a_0})({x^3} + {b_2}{x^2} + {b_1}x + {b_0}), for some a_i, b_i ∈ ℤ. However, when we apply the same procedure to (3.4), we also arrive at a contradiction in every possible scenario. For example, if $\begin{matrix} (\hat{A}, \hat{B}) = (2, 2) & and & (\hat{a_{0}}, \hat{a_{1}}, \hat{a_{2}}, \hat{b_{0}}, \hat{b_{1}}, \hat{b_{2}}) = (1, 1, 1, 1, 1, 1) \end{matrix},$ \matrix{{\left({\hat A,\hat B} \right) = \left({2,2} \right)} & {{\rm{and}}} & {(\widehat {{a_0}},\widehat {{a_1}},\widehat {{a_2}},\widehat {{b_0}},\widehat {{b_1}},\widehat {{b_2}}) = (1,1,1,1,1,1)} \cr}, then the left-hand side of the congruence corresponding to x³, which is $a_{0} + a_{1} b_{2} + a_{2} b_{1} + b_{0} \equiv 3 B (mod 4),$ {a_0} + {a_1}{b_2} + {a_2}{b_1} + {b_0} \equiv 3B\,\,\,\,\,\left({\bmod 4} \right), reduces to 0 (mod 4), while the right-hand side reduces to 2 (mod 4). We remark that this is the only contradictory congruence for this example. Thus, we deduce that ℱ_1,A,B(x) is irreducible.

Observing that ℱ_n,A,B(x) = ℱ_1,A,B(x^2ⁿ⁻¹) for n ≥ 1, we apply Proposition 2.10 with w(x) = ℱ_1,A,B(x) and m = 6. We first address the case n = 2, which corresponds to k = 1 in Proposition 2.10. By way of contradiction, we assume that ℱ_2,A,B(x) = ℱ_1,A,B(x²) is reducible. Then, by Proposition 2.10, we have that there exist S₀(x), S₁(x) ∈ ℤ[x] such that $ℱ_{1, A, B} (x) = {(S_{0} (x))}^{2} - x {(S_{1} (x))}^{2} .$ {{\cal F}_{1,A,B}}(x) = {({S_0}(x))^2} - x{({S_1}(x))^2}. Since deg(ℱ_1,A,B) = 6, it follows that $\begin{matrix} S_{0} (x) = x^{3} + a_{2} x^{2} + a_{1} x + a_{0} & and & S_{1} (x) = b_{2} x^{2} + b_{1} x + b_{0} \end{matrix}$ \matrix{{{S_0}(x) = {x^3} + {a_2}{x^2} + {a_1}x + {a_0}} & {{\rm{and}}} & {{S_1}(x) = {b_2}{x^2} + {b_1}x + {b_0}} \cr} for some a_i, b_i ∈ ℤ. Then (3.5) ${(S_{0} (x))}^{2} - x {(S_{1} (x))}^{2} = x^{6} + (2 a_{2} - b_{2}^{2}) x^{5} + (2 a_{1} - 2 b_{1} b_{2} + a_{2}^{2}) x^{4} + (2 a_{0} - b_{1}^{2} - 2 b_{0} b_{2} + 2 a_{1} a_{2}) x^{3} + (a_{1}^{2} + 2 a_{0} a_{1} - 2 b_{0} b_{1}) x^{2} + (2 a_{0} a_{1} - b_{0}^{2}) x + a_{0}^{2} .$ {({S_0}\left(x \right))^2} - x{({S_1}\left(x \right))^2} = {x^6} + \left({2{a_2} - b_2^2} \right){x^5} + \left({2{a_1} - 2{b_1}{b_2} + a_2^2} \right){x^4} + \left({2{a_0} - b_1^2 - 2{b_0}{b_2} + 2{a_1}{a_2}} \right){x^3} + \left({a_1^2 + 2{a_0}{a_1} - 2{b_0}{b_1}} \right){x^2} + \left({2{a_0}{a_1} - b_0^2} \right)x + a_0^2. We equate coefficients on (3.5) and (3.1), which yields the system of equations: (3.6) $\begin{array}{l} constant term : & a_{0}^{2} = 1 \\ x : & 2 a_{0} a_{1} - b_{0}^{2} = 9 A \\ x^{2} : & a_{1}^{2} + 2 a_{0} a_{2} - 2 b_{0} b_{1} = 27 A^{2} + 3 \\ x^{3} : & 2 a_{0} - b_{1}^{2} + 2 a_{1} a_{2} - 2 b_{0} b_{2} = 3 B \\ x^{4} : & 2 a_{1} + a_{2}^{2} - 2 b_{1} b_{2} = 27 A^{2} + 3 \\ x^{5} : & 2 a_{2} - b_{2}^{2} = 9 A . \end{array}$ \matrix{{{\rm{constant}}\,{\rm{term}}:} & {a_0^2 = 1} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {x:} & {2{a_0}{a_1} - b_0^2 = 9A} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^2}:} & {a_1^2 + 2{a_0}{a_2} - 2{b_0}{b_1} = 27{A^2} + 3} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^3}:} & {2{a_0} - b_1^2 + 2{a_1}{a_2} - 2{b_0}{b_2} = 3B} \hfill \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^4}:} & {2{a_1} + a_2^2 - 2{b_1}{b_2} = 27{A^2} + 3} \cr \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {{x^5}:} & {2{a_2} - b_2^2 = 9A.}\hfill } Noting that a₀ = ±1, and applying the same procedure as before to the system (3.6), we see that every possibility provides a contradiction. For example, if $\begin{matrix} (\hat{A}, \hat{B}) = (3, 3) & and & (\hat{a_{0}}, \hat{a_{1}}, \hat{a_{2}}, \hat{b_{0}}, \hat{b_{1}}, \hat{b_{2}}) = (1, 0, 0, 0, 1, 1) \end{matrix},$ \matrix{{(\hat A,\hat B) = (3,3)} & {{\rm{and}}} & {(\widehat {{a_0}},\widehat {{a_1}},\widehat {{a_2}},\widehat {{b_0}},\widehat {{b_1}},\widehat {{b_2}}) = (1,0,0,0,1,1)} \cr}, then the left-hand side of the congruence corresponding to x reduces to 0 (mod 4), while the right-hand side reduces to 3 (mod 4).

Now suppose that n ≥ 3, which corresponds to k ≥ 2 in Proposition 2.10. Assume, by way of contradiction, that ℱ_n,A,B(x) is reducible. Then, by Proposition 2.10, there exist S₀(x), S₁(x) ∈ ℤ[x] such that (3.7) $w (x^{2}) = ℱ_{1, A, B} (x^{2}) = ℱ_{2, A, B} (x) = {(S_{0} (x))}^{2} - x {(S_{1} (x))}^{2},$ w\left({{x^2}} \right) = {{\cal F}_{1,A,B}}\left({{x^2}} \right) = {{\cal F}_{2,A,B}}\left(x \right) = {({S_0}\left(x \right))^2} - x{({S_1}\left(x \right))^2}, where $\begin{matrix} S_{0} (x) = x^{6} + \sum_{j = 0}^{5} c_{j} x^{j} & and & S_{1} (x) = \sum_{j = 0}^{5} d_{j} x^{j} \end{matrix},$ \matrix{{{S_0}\left(x \right) = {x^6} + \mathop \sum \nolimits_{j = 0}^5 {c_j}{x^j}} & {{\rm{and}}} & {{S_1}\left(x \right) = \mathop \sum \nolimits_{j = 0}^5 {d_j}{x^j}} \cr}, for some c_j, d_j ∈ ℤ. Noting that c₀ = ±1, we equate the other coefficients in (3.7) and use the same procedure as before to verify that there is no solution to the resulting system of equations. For example, if (Â, B̂) = (1, 1) and $(\hat{c_{0}}, \hat{c_{1}}, \hat{c_{2}}, \hat{c_{3}}, \hat{c_{4}}, \hat{c_{5}}, \hat{d_{0}}, \hat{d_{1}}, \hat{d_{2}}, \hat{d_{3}}, \hat{d_{3}}, \hat{d_{3}}) = (1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 1, 1),$ (\widehat {{c_0}},\widehat {{c_1}},\widehat {{c_2}},\widehat {{c_3}},\widehat {{c_4}},\widehat {{c_5}},\widehat {{d_0}},\widehat {{d_1}},\widehat {{d_2}},\widehat {{d_3}},\widehat {{d_3}},\widehat {{d_3}}) = (1,0,0,0,0,0,2,0,0,1,1,1), then the left-hand side of the congruence corresponding to x³, which is $c_{1}^{2} + 2 c_{0} c_{2} - 2 d_{0} d_{1} \equiv 9 A (mod 4),$ c_1^2 + 2{c_0}{c_2} - 2{d_0}{d_1} \equiv 9A\,\,\,\,\,\left({\bmod 4} \right), reduces to 0 (mod 4), while the right-hand side reduces to 1 (mod 4). Hence, we conclude, by Proposition 2.10, that ℱ_n,A,B(x) is irreducible for all n ≥ 1, and the proof of the lemma is complete in this direction.

For the other direction of the proof, suppose that (Â, B̂) ∉ Γ. That is, assume (Â, B̂) is an element of the set (3.8) ${(0, 0), (0, 1), (0, 3), (1, 0), (1, 2), (1, 3), (2, 0), (2, 1), (2, 3), (3, 0), (3, 1), (3, 2)} .$ \{(0,0),\,(0,1),\,(0,3),\,(1,0),\,(1,2),\,(1,3),\,(2,0),\,(2,1),\,(2,3),\,(3,0),\,(3,1),\,(3,2)\,\}. For each (Â, B̂) in (3.8), we provide in Table 1 an explicit example of (A, B) such that ℱ_n,A,B(x) is reducible, not only for some n, but for all n ≥ 1. In Table 1, we let Φ_N (x) denote the cyclotomic polynomial of index N.

Table 1.

Examples for (3.8) and their factorizations

(Â, B̂)	(A, B)	Factorization of ℱ_n,A,B(x)
(0, 0)	(4, 24)	(x^2ⁿ⁺¹ + 36x^2ⁿ⁻¹ ³ + 434x^2ⁿ + 36x²ⁿ⁻¹ + 1)Φ_2ⁿ⁺¹ (x)
(0, 1)	(4, 357)	(x^2ⁿ + 3x^2ⁿ⁻¹ + 1)(x^2ⁿ⁺¹ + 33x^{2ⁿ⁻¹ 3} + 335x^2ⁿ + 33x^2ⁿ⁻¹ + 1)
(0, 3)	(4, 591)	(x^2ⁿ + 9x^2ⁿ⁻¹ + 1)(x^2ⁿ⁺¹ + 27x^2ⁿ⁻¹3 + 191x^2ⁿ + 27x^2ⁿ⁻¹ + 1)
(1, 0)	(1, 24)	(x^2ⁿ + 6x^2ⁿ⁻¹ + 1)(x^2ⁿ⁺¹ + 3x^2ⁿ⁻¹3 + 11x^2ⁿ + 3x^2ⁿ⁻¹ + 1)
(1, 2)	(1, 6)	(x^2ⁿ⁺¹ + 9x^2ⁿ⁻¹3 + 29x^2ⁿ + 9^2ⁿ⁻¹ + 1)Φ_2ⁿ⁺¹ (x)
(1, 3)	(1, 15)	(x^2ⁿ + 3x^2ⁿ⁻¹ + 1)³
(2, 0)	(2, 12)	(x^2ⁿ⁺¹ + 18x^2ⁿ⁻¹3 + 110x^2ⁿ + 18x^2ⁿ⁻¹ + 1)Φ_2ⁿ⁺¹ (x)
(2, 1)	(2, 93)	(x^2ⁿ + 9x^2ⁿ⁻¹ + 1)(x^2ⁿ⁺¹ + 9x^2ⁿ⁻¹3 + 29x^2ⁿ + 9x^2ⁿ⁻¹ + 1)
(2, 3)	(2, 75)	(x^2ⁿ + 3x^2ⁿ⁻¹ + 1)(x^2ⁿ⁺¹ + 15x^2ⁿ⁻¹3 + 65x^2ⁿ + 15x^2ⁿ⁻¹ + 1)
(3, 0)	(3, 252)	(x^2ⁿ + 6x^2ⁿ⁻¹ + 1)(x^2ⁿ⁺¹ + 21x^{2ⁿ⁻¹ 3} + 119x^2ⁿ + 21x^2ⁿ⁻¹ + 1)
(3, 1)	(3, 189)	(x^2ⁿ + 3x^2ⁿ⁻¹ + 1)(x^2ⁿ⁺¹ + 24x^{2ⁿ⁻¹ 3} + 173x^2ⁿ + 24x^2ⁿ⁻¹ + 1)
(3, 2)	(3, 18)	(x^2ⁿ⁺¹ + 27x^{2ⁿ⁻¹ 3} + 245x^2ⁿ + 27x^2ⁿ⁻¹ + 1)Φ_2ⁿ⁺¹ (x)

Proof of Theorem 1.1

Observe first that since Ψ ⊂ Γ, ℱ_n,A,B(x) is irreducible for all n ≥ 1 by Lemma 3.1. To complete the proof of monogenicity, we examine the prime divisors of Δ(ℱ_n,A,B).

We begin with the case n = 1. Suppose that ℱ_1,A,B(θ) = 0. A computation in Maple produces (3.9) $Δ (ℱ_{1, A, B}) = 3^{10} (3 B + 54 A^{2} + 18 A + 8) (3 B - 54 A^{2} + 18 A - 8) {(B - 9 A^{3} - 6 A)}^{4} .$ \Delta \left({{{\cal F}_{1,A,B}}} \right) = {3^{10}}\left({3B + 54{A^2} + 18A + 8} \right)\left({3B - 54{A^2} + 18A - 8} \right){(B - 9{A^3} - 6A)^4}. We use Theorem 2.4 with T (x) := ℱ_1,A,B(x) to show that [ℤ_K : ℤ[θ]] ≢ 0 (mod q), for every prime q dividing Δ(ℱ_1,A,B), where ℤ_K is the ring of integers of K = ℚ(θ). Because 𝒟 is squarefree, it follows from (3.9) and (1.1) that no prime dividing $(3 B + 54 A^{2} + 18 A + 8) (3 B - 54 A^{2} + 18 A - 8)$ (3B + 54{A^2} + 18A + 8)(3B - 54{A^2} + 18A - 8) can divide [ℤ_K : ℤ[θ]]. Hence, we only need to focus on the prime 3 and primes dividing B − 9A³ − 6A.

Suppose first that q = 3. Then, in Theorem 2.4, we have T̅ (x) = (x² + 1)³, so that we can let $g (x) = x^{2} + 1 and h (x) = {(x^{2} + 1)}^{2} .$ g\left(x \right) = {x^2} + 1\,\,{\rm{and}}\,\,h\left(x \right) = {({x^2} + 1)^2}. Then $\bar{F} (x) = \bar{(\frac{{(x^{2} + 1)}^{3} - T (x)}{3})} = 2 B x^{3} ≢ 0 (mod 3)$ \bar F\left(x \right) = \overline {\left({{{{{({x^2} + 1)}^3} - T\left(x \right)} \over 3}} \right)} = 2B{x^3} \nequiv 0\,\,\,\,\left({\bmod 3} \right) since B ≢ 0 (mod 3). Thus, it is easy to see that gcd(F̅, g̅) = 1, and consequently, [ℤ_K : ℤ[θ]] ≢ 0 (mod 3) by Theorem 2.4.

Next, suppose that q ≠ 3 is a prime divisor of B − 9A³ − 6A. Then $\bar{T} (x) = {(x^{2} + 3 Ax + 1)}^{3} = \bar{τ} {(x)}^{3} .$ \bar T\left(x \right) = {({x^2} + 3Ax + 1)^3} = \bar \tau {(x)^3}. By the quadratic formula, there are three cases to consider:

(1)
$\bar{τ} (x) \equiv {(x + 3 A / 2)}^{2} (mod q)$ \bar \tau \left(x \right) \equiv {(x + 3A/2)^2}\,\,\left({\bmod q} \right) ,
(2)
$\bar{τ} (x)$ \bar \tau \left(x \right) is irreducible over 𝔽_q,
(3)
$\bar{τ} (x) \equiv (x - (- 3 A + w) / 2) (x - (- 3 A - w) / 2) (mod q)$ \bar \tau \left(x \right) \equiv \left({x - \left({- 3A + w} \right)/2} \right)\left({x - \left({- 3A - w} \right)/2} \right)\,\,\,\left({\bmod q} \right) , where w² ≡ 9A² − 4 (mod q).

Case (1) occurs if Δ(x² + 3Ax + 1) = 9A² − 4 ≡ 0 (mod q). Then, A ≡ ±(2/3) (mod q) and, respectively, B ≡ ±(20/3) (mod q) since B ≡ 9A³ + 6A (mod q). But then, respectively, $\begin{matrix} 3 B \mp 54 A^{2} + 18 A \mp 8 \equiv 0 & (mod q), \end{matrix}$ \matrix{{3B \mp 54{A^2} + 18A \mp 8 \equiv 0} & {\left({\bmod q} \right),} \cr} contradicting, in either situation, the fact that 𝒟 is squarefree. Hence, case (1) cannot happen.

Suppose next that we are in case (2), so that we can let $\begin{matrix} g (x) = x^{2} + 3 Ax + 1 & and & h (x) = {(x^{2} + 3 Ax + 1)}^{2} \end{matrix} .$ \matrix{{g\left(x \right) = {x^2} + 3Ax + 1} & {{\rm{and}}} & {h\left(x \right) = {{({x^2} + 3Ax + 1)}^2}} \cr}. Then $\bar{F} (x) = \bar{(\frac{g (x) h (x) - T (x)}{q})} = - 3 \bar{(\frac{B - 9 A^{3} - 6 A}{q})} x^{2} ≢ 0 (mod q),$ \bar F\left(x \right) = \overline {\left({{{g\left(x \right)h\left(x \right) - T\left(x \right)} \over q}} \right)} = - 3\overline {\left({{{B - 9{A^3} - 6A} \over q}} \right)} {x^2} \ne 0\,\,\,(\bmod \,q), since B − 9A³ − 6A is squarefree and q ≠ 3. Then it is easy to see that gcd(F̅, g̅) = 1. Hence, [ℤ_K : ℤ[θ]] ≢ 0 (mod q) by Theorem 2.4, and ℱ_1,A,B(x) is monogenic in this case.

Finally, suppose that we are in case (3). Without loss of generality, assume that w ≡ 1 (mod 2). Since w² ≡ 9A² − 4 (mod q), we can write (3.10) $w^{2} = 9 A^{2} - 4 + qk,$ {w^2} = 9{A^2} - 4 + qk, for some k ∈ ℤ. Note that k ≡ 0 (mod 4). Then −3A ± w ≡ 0 (mod 2) since A ≡ 1 (mod 2). Thus, we can let $g (x) = h (x) = (x - (- 3 A + w) / 2) (x - (- 3 A - w) / 2),$ g\left(x \right) = h\left(x \right) = \left({x - \left({- 3A + w} \right)/2} \right)\left({x - \left({- 3A - w} \right)/2} \right), so that $g (x) h (x) = x^{2} + 3 Ax + (9 / 4) A^{2} - w^{2} / 4 = x^{2} + 3 Ax + 1 - qk / 4 \in ℤ [x]$ g\left(x \right)h\left(x \right) = {x^2} + 3Ax + \left({9/4} \right){A^2} - {w^2}/4 = {x^2} + 3Ax + 1 - qk/4 \in {\mathbb Z}\left[ x \right] by (3.10). Therefore, to prove that gcd(F̅, g̅) = 1, we only have to show that F̅ ((−3A ± w)/2) ≠ 0. Because the methods are the same, we give details only for x = (−3A + w)/2. Noting that $\begin{matrix} \bar{F} ((- 3 A + w) / 2) \neq 0 & if and if only & qF ((- 3 A + w) / 2) ≢ 0 (mod q^{2}) \end{matrix},$ \matrix{{\bar F((- 3A + w)/2) \ne 0} & {{\rm{if}}\,{\rm{and}}\,{\rm{if}}\,{\rm{only}}} & {qF((- 3A + w)/2) \nequiv 0\,\,\,(\bmod {q^2})} \cr}, we examine qF ((−3A + w)/2). Then, using (3.10) and the fact that q divides B − 9A³ − 6A, a straightforward calculation in Maple yields (3.11) $\begin{array}{l} 64 qF ((- 3 A + w) / 2) = - k^{3} q^{3} - 24 k (9 A - w) (9 A^{3} + 6 A - B) q \\ - 96 (27 A^{3} - 9 A - w (9 A^{2} - 1)) (9 A^{3} + 6 A - B) \\ \equiv - 96 (27 A^{3} - 9 A - w (9 A^{2} - 1)) (9 A^{3} + 6 A - B) (mod q^{2}) . \end{array}$ \matrix{{64qF\left({\left({- 3A + w} \right)/2} \right) = - {k^3}{q^3} - 24k\left({9A - w} \right)\left({9{A^3} + 6A - B} \right)q} \cr {- 96\left({27{A^3} - 9A - w\left({9{A^2} - 1} \right)} \right)\left({9{A^3} + 6A - B} \right)} \cr {\equiv - 96\left({27{A^3} - 9A - w\left({9{A^2} - 1} \right)} \right)\left({9{A^3} + 6A - B} \right)\left({\bmod {q^2}} \right).} \cr} We claim that (3.12) $27 A^{3} - 9 A - w (9 A^{2} - 1) ≢ 0 (mod q) .$ 27{A^3} - 9A - w\left({9{A^2} - 1} \right) \nequiv 0\,\,\,\,\,\left({\bmod q} \right). Assume, by way of contradiction, that (3.13) $27 A^{3} - 9 A - w (9 A^{2} - 1) \equiv 0 (mod q) .$ 27{A^3} - 9A - w\left({9{A^2} - 1} \right) \equiv 0\,\,\,\,\left({\bmod q} \right). Then, if (3.14) $9 A^{2} - 1 \equiv 0 (mod q),$ 9{A^2} - 1 \equiv 0\,\,\,\,\,\left({\bmod q} \right), it follows that $27 A^{3} - 9 A \equiv - 6 A \equiv 0 (mod q),$ 27{A^3} - 9A \equiv - 6A \equiv 0\,\,\,\,\,\left({\bmod \,q} \right), which implies that A ≡ 0 (mod q) since q ∉ {2, 3}. Consequently, $9 A^{2} - 1 \equiv - 1 (mod q),$ 9{A^2} - 1 \equiv - 1\,\,\,\,\,\left({\bmod \,q} \right), contradicting (3.14). Hence, 9A² − 1 ≢ 0 (mod q), and we have from (3.13) that $w^{2} \equiv \frac{{(27 A^{3} - 9 A)}^{2}}{{(9 A^{2} - 1)}^{2}} (mod q)$ {w^2} \equiv {{{{(27{A^3} - 9A)}^2}} \over {{{(9{A^2} - 1)}^2}}}\,\,\,\,\,\left({\bmod \,\,q} \right) Then, since w² ≡ 9A² − 4 (mod q), we arrive at the congruence $\begin{array}{l} {(27 A^{3} - 9 A)}^{2} \equiv {(9 A^{2} - 1)}^{2} (9 A^{2} - 4) & (mod q) \end{array} .$ \matrix{{{{(27{A^3} - 9A)}^2} \equiv {{(9{A^2} - 1)}^2}\left({9{A^2} - 4} \right)} \hfill & {\left({\bmod q} \right)} \hfill \cr}. which yields, after expansion, the impossible congruence 4 ≡ 0 (mod q). Hence, (3.12) is established. Since 9A³ + 6A − B is squarefree, we deduce from (3.11) that qF ((−3A+w)/2) ≢ 0 (mod q²), and the proof that ℱ_1,A,B(x) is monogenic is complete.

Next, we address the monogenicity of ℱ_n,A,B(x) for n ≥ 2. Since ℱ_n,A,B(x) = ℱ_1,A,B(x^2ⁿ⁻¹) for n ≥ 1, we use Theorem 2.2 and Definition 2.1 to calculate (3.15) $\begin{array}{l} Δ (ℱ_{n, A, B}) & = Δ (ℱ_{1, A, B} \circ x^{2^{n - 1}}) \\ = {(- 1)}^{3^{2} \cdot 2^{n} (2^{n - 1} - 1)} Δ {(ℱ_{1, A, B})}^{2^{n - 1}} R (ℱ_{n, A, B}, 2^{n - 1} x^{2^{n - 1} - 1}) \\ = 2^{3 \cdot 2^{n} (n - 1)} Δ {(ℱ_{1, A, B})}^{2^{n - 1}} . \end{array}$ \matrix{{\Delta \left({{{\cal F}_{n,A,B}}} \right)} \hfill & {= \Delta \left({{{\cal F}_{1,A,B}} \circ {x^{{2^{n - 1}}}}} \right)} \hfill \cr {} \hfill & {= {{(- 1)}^{{3^2} \cdot {2^n}\left({{2^{n - 1}} - 1} \right)}}\Delta {{({{\cal F}_{1,A,B}})}^{{2^{n - 1}}}}R\left({{{\cal F}_{n,A,B}},{2^{n - 1}}{x^{{2^{n - 1}} - 1}}} \right)} \hfill \cr {} \hfill & {= {2^{3 \cdot {2^n}\left({n - 1} \right)}}\Delta {{({{\cal F}_{1,A,B}})}^{{2^{n - 1}}}}.} \hfill \cr} For n ≥ 1, we define $\begin{array}{l} θ_{n} : = θ^{1 / 2^{n - 1}} & and & K_{n} : = ℚ (θ_{n}) \end{array},$ \matrix{{{\theta_n}: = {\theta^{1/{2^{n - 1}}}}} \hfill & {{\rm{and}}} \hfill & {{K_n}: = {\mathbb Q}\left({{\theta_n}} \right)} \hfill \cr}, noting that θ₁ = θ and K₁ = K from the case n = 1 earlier in this proof. Furthermore, observe that ℱ_n,A,B (θ_n) = 0 and [K_n₊₁ : K_n] = 2. Thus, if ℱ_n,A,B(x) is monogenic, then Δ(ℱ_n,A,B) = Δ(K_n), and we deduce from Theorem 2.5 that $\begin{array}{l} Δ (K_{n + 1}) \equiv 0 & (mod Δ {(ℱ_{n, A, B})}^{2}) . \end{array}$ \matrix{{\Delta \left({{K_{n + 1}}} \right) \equiv 0} \hfill & {(\bmod \,\Delta {{({{\cal F}_{n,A,B}})}^2}).} \hfill \cr} By (3.15), we have that $Δ (ℱ_{n + 1, A, B}) / Δ {(ℱ_{n, A, B})}^{2} = 2^{3 \cdot 2^{n + 1}}$ \Delta \left({{{\cal F}_{n + 1,A,B}}} \right)/\Delta {({{\cal F}_{n,A,B}})^2} = {2^{3 \cdot {2^{n + 1}}}} Hence, to show that ℱ_n+1,A,B(x) is monogenic, we only have to show that (3.16) $[ℤ_{K_{n + 1}} : ℤ [θ_{n + 1}]] ≢ 0 (mod 2) .$ [{{\mathbb Z}_{{K_{n + 1}}}}: {\mathbb Z}[{\theta_{n + 1}}]] \nequiv 0\,\,\,\,\,\left({\bmod 2} \right). We apply Theorem 2.4 with T (x) := ℱ_n+1,A,B(x). Then $\bar{T} (x) = {(x^{6} + x^{5} + x^{3} + x + 1)}^{2^{n}} = {(x^{2} + x + 1)}^{3 \cdot 2^{n}} = Φ_{3} {(x)}^{3 \cdot 2^{n}},$ \bar T\left(x \right) = {({x^6} + {x^5} + {x^3} + x + 1)^{{2^n}}} = {({x^2} + x + 1)^{3 \cdot {2^n}}} = {\Phi_3}{(x)^{3 \cdot {2^n}}}, where Φ₃(x) is easily seen to be irreducible over 𝔽₂. Therefore, we can let $\begin{array}{l} g (x) = Φ_{3} (x) & and & h (x) = Φ_{3} {(x)}^{3 \cdot 2^{n} - 1} . \end{array}$ \matrix{{g\left(x \right) = {\Phi_3}\left(x \right)} \hfill & {{\rm{and}}} \hfill & {h\left(x \right) = {\Phi_3}{{(x)}^{3 \cdot {2^n} - 1}}.} \hfill \cr} A straightforward induction argument shows for n ≥ 1 that $\begin{array}{l} g (x) h (x) = Φ_{3} {(x)}^{3 \cdot 2^{n}} \equiv Q (x) & (mod 4) \end{array},$ \matrix{{g(x)h(x) = {\Phi_3}{{(x)}^{3 \cdot {2^n}}} \equiv Q(x)} \hfill & {(\bmod 4)} \hfill \cr}, where $Q (x) = x^{12 \cdot 2^{n - 1}} + 2 x^{11 \cdot 2^{n - 1}} + x^{10 \cdot 2^{n - 1}} + 2 x^{9 \cdot 2^{n - 1}} + 2 x^{8 \cdot 2^{n - 1}} + 2 x^{7 \cdot 2^{n - 1}} + x^{6 \cdot 2^{n - 1}} + 2 x^{5 \cdot 2^{n - 1}} + 2 x^{4 \cdot 2^{n - 1}} + 2 x^{3 \cdot 2^{n - 1}} + x^{2 \cdot 2^{n - 1}} + 2 x^{2^{n - 1}} + 1 .$ Q\left(x \right) = {x^{12 \cdot {2^{n - 1}}}} + 2{x^{11 \cdot {2^{n - 1}}}} + {x^{10 \cdot {2^{n - 1}}}} + 2{x^{9 \cdot {2^{n - 1}}}} + 2{x^{8 \cdot {2^{n - 1}}}} + 2{x^{7 \cdot {2^{n - 1}}}} + {x^{6 \cdot {2^{n - 1}}}} + 2{x^{5 \cdot {2^{n - 1}}}} + 2{x^{4 \cdot {2^{n - 1}}}} + 2{x^{3 \cdot {2^{n - 1}}}} + {x^{2 \cdot {2^{n - 1}}}} + 2{x^{{2^{n - 1}}}} + 1. Then, writing g(x)h(x) = Q(x) + 4E(x) for some E(x) ∈ ℤ[x], we get that $\begin{array}{l} F (x) & = \frac{g (x) h (x) - T (x)}{2} \\ = x^{11 \cdot 2^{n - 1}} + (\frac{1 - 9 A}{2}) x^{10 \cdot 2^{n - 1}} + x^{9 \cdot 2^{n - 1}} + (\frac{2 - (27 A^{2} + 3)}{2}) x^{8 \cdot 2^{n - 1}} \\ + x^{7 \cdot 2^{n - 1}} + (\frac{1 - 3 B}{2}) x^{6 \cdot 2^{n - 1}} + x^{5 \cdot 2^{n - 1}} + (\frac{2 - (27 A^{2} + 3)}{2}) x^{4 \cdot 2^{n - 1}} \\ + x^{3 \cdot 2^{n - 1}} + (\frac{1 - 9 A}{2}) x^{2 \cdot 2^{n - 1}} + x^{2^{n - 1}} + 2 E (x) . \end{array}$ \matrix{{F\left(x \right)} \hfill & {= {{g\left(x \right)h\left(x \right) - T\left(x \right)} \over 2}} \hfill \cr {} \hfill & {= {x^{11 \cdot {2^{n - 1}}}} + \left({{{1 - 9A} \over 2}} \right){x^{10 \cdot {2^{n - 1}}}} + {x^{9 \cdot {2^{n - 1}}}} + \left({{{2 - \left({27{A^2} + 3} \right)} \over 2}} \right){x^{8 \cdot {2^{n - 1}}}}} \hfill \cr {} \hfill & {+ {x^{7 \cdot {2^{n - 1}}}} + \left({{{1 - 3B} \over 2}} \right){x^{6 \cdot {2^{n - 1}}}} + {x^{5 \cdot {2^{n - 1}}}} + \left({{{2 - \left({27{A^2} + 3} \right)} \over 2}} \right){x^{4 \cdot {2^{n - 1}}}}} \hfill \cr {} \hfill & {+ {x^{3 \cdot {2^{n - 1}}}} + \left({{{1 - 9A} \over 2}} \right){x^{2 \cdot {2^{n - 1}}}} + {x^{{2^{n - 1}}}} + 2E\left(x \right).} \hfill \cr} Hence, $\bar{F} (x) = {\begin{array}{l} {(x (x^{3} + x + 1) (x^{3} + x^{2} + 1) Φ_{5} (x))}^{2^{n - 1}} & if (\hat{A}, \hat{B}) = (1, 1), \\ {(x + 1)}^{2^{n}} {(x (x^{4} + x + 1) (x^{4} + x^{3} + 1))}^{2^{n - 1}} & if (\hat{A}, \hat{B}) = (3, 3) . \end{array}$ \bar F\left(x \right) = \left\{{\matrix{{{{(x({x^3} + x + 1)({x^3} + {x^2} + 1){\Phi_5}(x))}^{{2^{n - 1}}}}} \hfill & {{\rm{if}}\,(\hat A,\hat B) = (1,1),} \hfill \cr {{{(x + 1)}^{{2^n}}}{{(x({x^4} + x + 1)({x^4} + {x^3} + 1))}^{{2^{n - 1}}}}} \hfill & {{\rm{if}}\,(\hat A,\hat B) = (3,3).} \hfill \cr}} \right. It is then apparent that gcd(F̅, g̅) = 1 in each case of (Â, B̂) ∈ Ψ, from which we conclude by Theorem 2.4 that (3.16) is valid. Therefore, ℱ_n+1,A,B(x) is monogenic, and consequently, ℱ_n,A,B(x) is monogenic for all n ≥ 1 by induction.

4.

The proof of Corollary 1.2

Proof

Since the methods used in the proof are the same for both parts, we give details only for part (2). Define the polynomial G(t) = g₁(t)g₂(2)g₃(t) ∈ ℤ[t], where $\begin{array}{l} g_{1} (t) = 36 t + 864 u^{2} + 1368 u + 569, \\ g_{2} (t) = 36 t - 864 u^{2} - 1224 u - 419 and \\ g_{3} (t) = 6 t - 288 u^{3} - 648 u^{2} - 498 u + 127 . \end{array}$ \matrix{{{g_1}\left(t \right) = 36t + 864{u^2} + 1368u + 569,} \hfill \cr {{g_2}\left(t \right) = 36t - 864{u^2} - 1224u - 419\,\,\,{\rm{and}}} \hfill \cr {{g_3}\left(t \right) = 6t - 288{u^3} - 648{u^2} - 498u + 127.} \hfill \cr} We wish to apply Corollary 2.9 to G(t). Note that $gcd (36, g_{1} (0)) = gcd (36, g_{2} (0)) = gcd (6, g_{3} (0)) = 1 .$ \gcd \left({36,{g_1}\left(0 \right)} \right) = \gcd \left({36,{g_2}\left(0 \right)} \right) = \gcd \left({6,{g_3}\left(0 \right)} \right) = 1. According to the discussion following Corollary 2.9, we only need to check for local obstructions at the primes ℓ satisfying ℓ ≤ (k + 2)/1 = 5/2. That is, we only need to check the prime ℓ = 2. Since G(1) ≡ 2u + 1 (mod 4), we see that there is no local obstruction at ℓ = 2. Hence, by Corollary 2.9, there exist infinitely many primes q such that G(q) is squarefree. Then 2G(q) is also squarefree since g_i(q) ≡ 1 (mod 2) for each i. Observe that 𝒟= 2G(q) and (Â, B̂) = (3, 3) with A = 4u + 1 and B = 12q + 7 ≢ 0 (mod 3). Thus, for any such prime q, we deduce from Theorem 1.1 that ℱ_n,4u+1,12q+7(x) is monogenic for all n ≥ 1.

Reciprocal Monogenic Septinomials of Degree 2n3

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