$a^n-1$ and $b^n-1$ have the same set of prime factors for each $n\in\Bbb{Z}^+$, show that $a=b$. Let $a,b$ be two positive integers, if $a^n-1$ and $b^n-1$ have the same set of prime factors for each $n\in\Bbb{Z}^+$, then $a=b$.
It seems to be not hard, but I have no idea. Any hint is welcome.
Edit: As the comment points out, it is probably not easy. I thought it is easy because it is kind of intuitively correct. I tried to study the Zsigomondy primes as $n$ increases, but it is not successful.
 A: The condition that $a^n-1$ and $b^n-1$ have the same prime factors for every $n$ is equivalent to $a$ and $b$ having the same order mod $p$ for every $p$. Equivalently, for every positive integer $k$ and every prime $p$, $a$ is a $k$-th power mod $p$ if and only if $b$ is a $k$-th power mod $p$. I will argue that for fixed $a\neq b$ and every sufficiently large prime value of $k$, there is a positive proportion of primes $p$ for which $a$ is a $k$-th power mod $p$ but $b$ is not.
Suppose $2\leq a< b$. Let $q>b$ be a prime, and consider the field $K=\mathbb{Q}(\sqrt[q]{a},\sqrt[q]{b},\zeta_q)$, where $\zeta_q$ is a primitive $q$-th root of unity. Then
$$
G:=\operatorname{Gal}(L/\mathbb{Q})\cong \mathbb{F}_q^\times \ltimes\left(\mathbb{F}_q\times\mathbb{F}_q\right).
$$
Then $\mathbb{Q}(\sqrt[q]{a})$ is the fixed field of $H_1:=\mathbb{F}_q^\times \ltimes\left(\{0\}\times\mathbb{F}_q\right)$, and $\mathbb{Q}(\sqrt[q]{b})$ is the fixed field of $H_2:=\mathbb{F}_q^\times \ltimes\left(\mathbb{F}_q\times \{0\}\right)$. If $p$ is a rational prime coprime to $a$, $b$, and $q$, then $a$ is a $q$-th power mod $p$ if and only if the Frobenius conjugacy class of $p$ in $G$ intersects $H_1$, and $b$ is a $q$-th power mod $p$ if and only if the Frobenius conjugacy class of $p$ in $G$ intersects $H_2$. Now, by the Chebotarev density theory, there exist a positive proportion of primes $p$ whose Frobenius conjugacy class contains $(1,0,1)\in G$. Then the conjugacy class intersects $H_1$ but not $H_2$, so for such $p$, $a$ is a $q$-th power mod $p$ but $b$ is not.
