GCD of $n^a\,\prod\limits_{i=1}^k\,\left(n^{b_i}-n\right)$ for $n\in\mathbb{Z}$

Let $a$ be a nonnegative integer. For a given positive integer $k$, let $b_1,b_2,\ldots,b_k$ be odd integers greater than $1$. Using this result, it can be shown that, for each integer $n$, $$f_{a;b_1,b_2,\ldots,b_k}(n):=n^a\,\prod_{i=1}^k\,\left(n^{b_i}-n\right)$$ is divisible by $$\Gamma\left(a;b_1,b_2,\ldots,b_k\right):=2^{\min\left\{k+a\,,\,2k+\sum\limits_{i=1}^k\,v_2\left(b_i-1\right)\right\}}\,\prod_{\substack{{p\in\mathbb{Z}_{>2}}\\p\text{ prime}}}\,p^{t_p}\,,$$ where $$t_p:={\min\left\{k+a\,,\,\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\,\big(1+v_p\left(b_i-1\right)\big)\right\}} $$ for each prime $p>2$. Here, $v_q(m)$ denotes the largest exponent $\nu\in\mathbb{Z}_{\geq0}$ such that the $\nu$-th power of the given prime $q$ divides the integer $m$.

Question. Fix $a$ and $\left(b_1,b_2,\ldots,b_k\right)$. What is the greatest common divisor $G\left(a;b_1,b_2,\ldots,b_k\right)$ of all integers of the form $f_{a;b_1,b_2,\ldots,b_k}(n)$ (i.e., $n$ runs over all the integers)? Is it equal to $\Gamma\left(a;b_1,b_2,\ldots,b_k\right)$? If not, what is a counterexample?

At least, we know that $G\left(a;b_1,b_2,\ldots,b_k\right)=\Gamma\left(a;b_1,b_2,\ldots,b_k\right)$ in the following examples.

1. Case $k=1$: See this link.

2. Case $a=0$, $k=2$, and $\left(b_1,b_2\right)=\left(13,17\right)$: See this link.

Here is another related problem.

Solution 1:

It is true that$$G\left(a;b_1,b_2,\ldots,b_k\right)=\Gamma\left(a;b_1,b_2,\ldots,b_k\right)$$

Proof :

We can write $$f_{a;b_1,b_2,\ldots,b_k}(n)=n^{k+a}\prod_{i=1}^{k}\bigg(n^{b_i-1}-1\bigg)$$

Firstly, let us prove that $$v_2\bigg(G\left(a;b_1,b_2,\ldots,b_k\right)\bigg)=\min\left\{k+a\,,\,2k+\sum\limits_{i=1}^k\,v_2\left(b_i-1\right)\right\}\tag1$$

• If $n$ is even, then $v_2(f_{a;b_1,b_2,\ldots,b_k}(n))=v_2(n^{k+a})\ge k+a$ where the equality is attained when $n=2$.

• If $n$ is odd, then using the following lemma (Lifting The Exponent for $p=2$) : $$v_2(n^{b_i-1}-1)=\begin{cases}v_2(n - 1) + v_2(b_i-1)&\text{if $n\equiv 1\pmod 4$} \\\\v_2(n + 1) + v_2(b_i-1)&\text{if $n\equiv 3\pmod 4$}\end{cases}$$ we get $$\small v_2(f_{a;b_1,b_2,\ldots,b_k}(n))=v_2\bigg(\prod_{i=1}^{k}\bigg(n^{b_i-1}-1\bigg)\bigg)=\sum_{i=1}^{k}v_2\bigg(n^{b_i-1}-1\bigg)\ge \sum_{i=1}^{k}\bigg(2 + v_2(b_i-1)\bigg)$$ where the equality is attained when $n=5$.

Therefore, $(1)$ follows.

Secondly, let us prove that $$v_p\bigg(G\left(a;b_1,b_2,\ldots,b_k\right)\bigg)={\min\left\{k+a\,,\,\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\,\big(1+v_p\left(b_i-1\right)\big)\right\}}\tag2$$ where $p$ is an odd prime.

• For $n$ such that $n\equiv 0\pmod p$, $v_p(f_{a;b_1,b_2,\ldots,b_k}(n))=v_p(n^{k+a})\ge k+a$ where the equality is attained when $n=p$.

• For $n$ such that $n\not\equiv 0\pmod p$, using the followig lemma (Lifting The Exponent) :
  
  If $p\mid x-1$, then $v_p(x^{m} - 1) = v_p(x- 1) + v_p(m)$
  
  and letting $b_i-1=(p-1)p^{m_i}u_i$ where $m_i,u_i$ are non-negative integers with $\gcd(p,u_i)=1$, we have$$v_p(n^{b_i-1}-1)=v_p((n^{p-1})^{p^{m_i}u_i}-1)=v_p(n^{p-1}-1)+v_p(p^{m_i}u_i)\ge 1+v_p(b_i-1)$$So, it follows that$$\begin{align}v_p(f_{a;b_1,b_2,\ldots,b_k}(n))&=\underbrace{\sum\limits_{\substack{{1\leq i\leq k}\\{p-1\ \not\mid\ b_i-1}}}\,v_p(n^{b_i-1}-1)}_{\ge 0}+\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\,v_p(n^{b_i-1}-1) \\\\&\ge\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\,\big(1+v_p\left(b_i-1\right)\big)\end{align}$$In the following, let us prove that the equality of this inequality is attained either when $n=g$ or when $n=(1+p)g$ where $g$ is a primitive root of $p$. (I learned this idea from Chris Wuthrich.)
  
  If $v_p(g^1-1)=v_p(g^2-1)=\cdots=v_p(g^{p-2}-1)=0$ and $v_p(g^{p-1}-1)\color{red}{=1}$, then$$\begin{align}v_p(f_{a;b_1,b_2,\ldots,b_k}(g))&=\underbrace{\sum\limits_{\substack{{1\leq i\leq k}\\{p-1\ \not\mid\ b_i-1}}}v_p\left(g^{b_i-1}-1\right)}_{=0}+\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}v_p\left(g^{b_i-1}-1\right) \\\\&=\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\bigg(v_p(g^{p-1}-1)+v_p(b_i-1)\bigg) \\\\&=\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\bigg(1+v_p(b_i-1)\bigg)\end{align}$$
  If $v_p(g^1-1)=v_p(g^2-1)=\cdots=v_p(g^{p-2}-1)=0$ and $v_p(g^{p-1}-1)\color{red}{\ge 2}$, then there are integers $m_1,m_2$ such that$$\begin{align}&v_p(f_{a;b_1,b_2,\ldots,b_k}((1+p)g)) \\\\&=\underbrace{\sum\limits_{\substack{{1\leq i\leq k}\\{p-1\ \not\mid\ b_i-1}}}v_p\left(((1+p)g)^{b_i-1}-1\right)}_{=0}+\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}v_p\left(((1+p)g)^{b_i-1}-1\right) \\\\&=\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\bigg(v_p((g+pg)^{p-1}-1)+v_p(b_i-1)\bigg) \\\\&=\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\bigg(v_p\bigg(g^{p-1}+(p-1)g^{p-2}pg+p^2m_1-1\bigg)+v_p(b_i-1)\bigg) \\\\&=\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\bigg(v_p(-p+p^2m_2)+v_p(b_i-1)\bigg) \\\\&=\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\bigg(1+v_p(b_i-1)\bigg)\end{align}$$

Therefore, $(2)$ follows.

It follows from $(1)(2)$ that$$G\left(a;b_1,b_2,\ldots,b_k\right)=\Gamma\left(a;b_1,b_2,\ldots,b_k\right)$$