Prove that if $m,n\in \mathbb{N}$ and $\gcd(2^m+1,2^n-1)=1$, then $\gcd (2^m+1,2^{3n}-1)=1$ .

Question

Let $m,n\in \mathbb{N}$. Prove that if $\gcd(2^m+1,2^n-1)=1$， then $\gcd(2^m+1,2^{3n}-1)=1$. It is equivalent to proving that if $\gcd(2^m+1,2^n-1)=1$, then $\gcd(2^m+1,2^{2n}+2^n+1)=1$.

I have verified the above question for $1\leq m,n \leq 8000$ by Matlab. Let $i, j$ be positive integers. I know that if $i$ is odd, then $\gcd(2^j+1,2^i-1)=1$. Therefore, considering the above question, we need to shou that if $n$ is even and $\gcd(2^m+1,2^n-1)=1$， then $\gcd(2^m+1,2^{3n}-1)=1$. However, I have not find an effective method to solve the above question. I know that $\gcd(a,b)=1$ if and only if there exists integers $x$ and $y$ such that $ax+by=1$.

Answer 1

We will use that $$ \gcd(2^a - 1, 2^b - 1) = 2^{\gcd(a, b)} - 1 $$ for all non-negative integers $a$ and $b$. See, for example, this question: Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$

Since $\gcd(2^m + 1, 2^n - 1) = 1$, we have that $\gcd(2^{2m} - 1, 2^n - 1) = \gcd(2^m - 1, 2^n - 1)$, and so it follows that $$ 2^{\gcd(2m, n)} - 1 = 2^{\gcd(m, n)} - 1 $$ which implies that $\gcd(2m, n) = \gcd(m, n)$. This implies that the largest power of $2$ that divides $m$ is at least as large as the largest power of $2$ that divides $n$. Since $3$ is odd, the same is true for the largest powers of $2$ that divide $m$ and $3n$ respectively, and so it follows that $\gcd(2m, 3n) = \gcd(m, 3n)$, which gives us that $$ \gcd(2^{2m} - 1, 2^{3n} - 1) = 2^{\gcd(2m, 3n)} - 1 = 2^{\gcd(m, 3n)} - 1 = \gcd(2^{m} - 1, 2^{3n} - 1). $$ Since $\gcd(2^m - 1, 2^m + 1) = 1$, we know that $$ \gcd(2^{2m} - 1, 2^{3n} - 1) = \gcd(2^m - 1, 2^{3n} - 1) \cdot \gcd(2^m + 1, 2^{3n} - 1) $$ and so combining the above observations we obtain that $$ \gcd(2^m + 1, 2^{3n} - 1) = 1. $$