Does the Bunyakovsky conjecture apply for every cyclotomic polynomial?

Question

The Bunyakovsky conjecture states that for every irreducible polynomial $\ f\in \mathbb Z[x]\ $ with positive leading coefficient, there are infinite many primes of the form $\ f(m)\ $ , where $\ m\ $ is a positive integer , if $\ \gcd(f(1),f(2),f(3),\cdots)=1\ $

The cyclotomic polynomials are well known to be irreducible and the leading coefficient is $\ 1\ $.

But what about the condition $\ \gcd(f(1),f(2),f(3),\cdots)=1\ $ ? Does it hold for all cyclotomic polynomials ? If yes, how can we prove it ?

Answer 1

Let $n\geq2$ be a positive integer, and let $\Phi_n$ denote the $n$-th cyclotomic polynomial. Then $$\Phi_n(1)=\begin{cases}p&\text{ if }n=p^k\\1&\text{ otherwise}\end{cases},$$ where in the first case $p$ is a prime number and $k$ is a positive integer, i.e. $n$ is a prime power. For a proof, see this question. It immediately follows that if $n$ is not a prime power, then $$\gcd(f(1),f(2),f(3),\ldots)=1.$$ Finally, if $n=p^k$ then $\Phi_n(p)\equiv1\pmod{p}$ and so $\gcd(\Phi_n(1),\Phi_n(p))=1$.

Answer 2

The $\gcd(f(1),f(2),f(3),\dots)$ is equal to $\gcd$ of values of $f$ at any consecutive $\deg f + 1$ integers. This is a result by K. Hensel (see for example A Survey on Fixed Divisors Theorem 2.1). This tells us we can look at $f$'s values also at $0$ (or negative for that matter), which in case of Cyclotomic polynomials are very simple. In fact, Möbius inversion formula $\Phi_n(x) = \prod_{d|n} (x^d-1)^{\mu(n/d)}$ implies $|\Phi_n(0)|=1$ for $n \geq 1$, and the result follows.