Are these polynomials associated to prime numbers irreducible? (The case for $p-1=2^k,k\ge2$, $p$ prime)

Question

Define

$$ g_n(x) = \begin{cases} 1, & \text{if } n = 1, \\ x + g_{n-1}(x), & \text{if } n > 1 \text{ is prime}, \\ \prod_{p \mid n} \left( g_p(x) \right)^{v_p(n)}, & \text{if } n > 1 \text{ is composite}. \end{cases} $$

Here are some of these polynomials:

1 1
2 x + 1
3 2*x + 1
4 x^2 + 2*x + 1
5 x^2 + 3*x + 1
6 2*x^2 + 3*x + 1
7 2*x^2 + 4*x + 1
8 x^3 + 3*x^2 + 3*x + 1
9 4*x^2 + 4*x + 1
10 x^3 + 4*x^2 + 4*x + 1
11 x^3 + 4*x^2 + 5*x + 1
12 2*x^3 + 5*x^2 + 4*x + 1
13 2*x^3 + 5*x^2 + 5*x + 1
14 2*x^3 + 6*x^2 + 5*x + 1
15 2*x^3 + 7*x^2 + 5*x + 1
16 x^4 + 4*x^3 + 6*x^2 + 4*x + 1
17 x^4 + 4*x^3 + 6*x^2 + 5*x + 1
18 4*x^3 + 8*x^2 + 5*x + 1
19 4*x^3 + 8*x^2 + 6*x + 1
20 x^4 + 5*x^3 + 8*x^2 + 5*x + 1
21 4*x^3 + 10*x^2 + 6*x + 1
22 x^4 + 5*x^3 + 9*x^2 + 6*x + 1
23 x^4 + 5*x^3 + 9*x^2 + 7*x + 1

It seems that empirically we have $\forall n \ge 2$ the polynomial $g_n(x)$ is a Hurwitz polynomial, wich by definition means it has real coefficients and all of its roots have real value $<0$.

This observation with the Lemma which comes afterwards would explain why for $p$ prime we have $g_p(x)$ is irreducible.

For prime $p$ it seems that $g_p(x)$ is always irreducible:

This could be proved if $\operatorname{Re}(z)<0<\frac{1}{2}$ for all roots $z$ of the polynomial $g_p(x)$. Here is a picture of collected roots of some $g_p(x)$ with $p$ prime $\le 100000$. If all roots have real value $<1-1/2$ then the following Lemma could be applied to the polynomial $P(x):=g_p(x)$ with $n=1$, since $p=g_p(1) = g_p(n)$ is a prime number and $g_p(n-1) = g_p(1-1) = g_p(0) = 1 \neq 0 $:

Lemma (B. Sury, Polynomials with Integer Values, Lemma 1.9):

Let $ P(X) \in \mathbb{Z}[X] $ be a polynomial, and suppose there exists an integer $ n $ such that:

1. The zeros of $P$ lie in the half-plane $ \text{Re}(z) < n - \frac{1}{2} $.
2. $P(n - 1) \neq 0$.
3. $P(n)$ is a prime number.

Then $P(X)$ is irreducible.

Edit: With the comment (for $p\ge 3,$ prime and $\operatorname{Re}(z)\ge 1/2$ we have $|g_p(z)|>2|z|$, which would prove by Lemma of Sury, the irreducibility.) from Jonathan Love, I can verify:

For $p = 2$: If $\mathrm{Re}(z) \geq 1/2$ and $z = a + b i$, then

$$ |g_p(z)| = |z+1| = \sqrt{(a+1)^2 + b^2} \geq \max(3/2, |z|). $$

For $p = 3$: If $\mathrm{Re}(z) \geq 1/2$ and $z = a + b i$, then

$$ |g_p(z)| = |2z+1| = \sqrt{(2a+1)^2 + (2b)^2} > \sqrt{(2a)^2 + (2b)^2} = 2|z|. $$

Let now $p > 3$ and suppose that a prime $q > 3$ divides $p-1$. Then we get:

$$ |g_p(z)| = |z + g_{p-1}(z)| \geq \text{(reverse triangle inequality)} \; |g_{p-1}(z)| - |z|. $$

But

$$ |g_{p-1}(z)| = |z+1|^{v_2(p-1)} \cdot |g_q(z)|^{v_q(p-1)} \cdot \prod_{r | (p-1), r \neq q} |g_r(z)|^{v_r(p-1)}, $$

and from the case $p = 2$ and induction on $p$, we get:

$$ |z+1| \geq 3/2, \quad |g_q(z)| > 2|z|. $$

After multiplying the terms, we have:

$$ |g_{p-1}(z)| > \frac{3}{2} \cdot 2|z| = 3|z|, $$

and therefore

$$ |g_p(z)| \geq |g_{p-1}(z)| - |z| > 3|z| - |z| = 2|z|. $$

This proves the inequality in this case.

It remains to discuss the case where $p-1 = 2^k$ with $k \geq 2$. which I do not see now, and any help is appreciated.

Related question: https://mathoverflow.net/questions/485027/are-these-polynomials-g-nx-for-n-ge-2-hurwitz-polynomials

Answer 1

This is answer to what seems to be the question per the initial part of the post - showing that $g_p(x)$ is irreducible when $p=2^k+1$ is a prime. I've also included an argument that shows $\operatorname{Re}(z)<\frac{1}{2}$ for its roots, which seems to be something you are after.

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Let $p=2^k+1$ be a prime, $k\geq 2$. In particular, $k$ must be a power of $2$. Now using the definition, $g_p(x)=x+(x+1)^k$. However, $g_p(x)$ is irreducible iff $g_p(x-1)=x^k+x-1$ is. Hence, it is sufficient to prove that $$x^{2^n}+x-1$$ is irreducible for $n\geq 1$. Using methods from this answer, it can be shown that $x^m+x-1$ is irreducible iff $m\not\equiv 5\pmod 6$, which is clearly satisfied for $m=2^n$. The result follows.

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For your approach, we can show that roots $z$ of $g_p(x)$ with $p=2^k+1$ prime satisfy $\operatorname{Re}(z)<\frac{1}{2}$. Again as above, note that $h(x)=g_p(x-1)=x^k+x-1$ and $k=2^n$. For $n=1$ we can just check the roots directly. For $n \geq 2$ assume $h(\alpha)=0$ such that $|\alpha| \geq \frac{3}{2}$, then $$1=|\alpha^k+\alpha|=|\alpha||\alpha^{k-1}+1|\geq \frac{3}{2}|\alpha^{k-1}+1|.$$
So $|\alpha^{k-1}+1|\leq \frac{2}{3}$ and since it is a distance between $-1$ and $\alpha^{k-1}$, it follows $$|\alpha|^{k-1}\leq 1+\frac{2}{3}=\frac{5}{3}.$$ But $k=2^n\geq 4$ and so $|\alpha|\leq\sqrt[3]{\frac{5}{3}}<\frac{3}{2}$, a contradiction. Hence there is no root with $|\alpha|\geq \frac{3}{2}$. In turn, all roots of $h(x+1)=g_p(x)$ satisfy $\operatorname{Re}(z)<\frac{1}{2}$.