Sextic reciprocal monogenic dihedral polynomials (Q2054721)

A polynomial \(f(x)\in\mathbb{Z}[x]\) with integer coefficients is said to be \textit{reciprocal} (elsewhere \textit{self-reciprocal}, or \textit{palindromic} or \textit{symmetrical})), if \(f(1/x)=f(x)/x^{deg(f)}\). A polynomial \(f\) is called \textit{monogenic}, if \(f\) is irreducible over the rationals and \(\mathbb{Z}_K=\mathbb{Z}[\theta]\), where \(\theta\) is a root of \(f\) and \(K=\mathbb{Q}(\theta)\); equivalently, the respective discriminants coincide: \(\Delta(f)=\Delta(K)\). The central result is as follows: Let Gal(\(f\)) denote the Galois group over the rationals \(\mathbb{Q}\) of the polynomial \(f(x)\in\mathbb{Z}[x]\). Denote the dihedral group of order \(2n\) by \(D_n\). Then the following hold: \begin{itemize} \item[1.] Let \(q\neq 7\) be a prime such that \(q\not\equiv\pm 1\) (mod 7). Then there exist infinitely many primes \(p\) such that the polynomial \(f(x)=x^6+x^5+(pq+1)x^4+(2pq+1)x^3+(pq+1)x^2+x+1 \) is monogenic. Furthermore, if \(q\equiv 3\) (mod 7) or \(q\equiv 5\) (mod 7), then \(\mathrm{Gal}(f)\cong D_6\). \item[2.] There are infinitely many primes \(p\) such that \(f(x)=x^6+px^3+1\) is monogenic and \(\mathrm{Gal}(f)\cong D_6\). \item[3.] There exist infinitely many primes \(p\) such that \(f(x)=x^6+3x^5+(p+6)x^4+(2p+7)x^3+(p+6)x^2+3x+1\) is monogenic and \(\mathrm{Gal}(f)\cong D_3\). \end{itemize} The emphasis in all three parts of the result is on the existence of infinitely many primes. The appropriately long proof is aided by computations in computer packages MAGMA, Maple, and Sage.