<s>[INST] Is there any integer $n>1$ such that $3^n - 1$ is divisible by $2^n - 1$? Is there any integer $n>1$ such that $3^n - 1$  is divisible by $2^n - 1$?
I guess not. For every even integer $n$, we can show that $3^n - 1$  is not divisible by $2^n - 1$ because $2^n -1$ is a multiple of $3$, but $3^n-1$ is not. Would anyone give some tips for the other case (case for $n$ odd integer)?
 [/INST] Here is a proof by contradiction. Assume there exists such $n$. 
If $n$ is even, notice $2^n-1 \equiv 0 \pmod 3$. Thus, as $3^n-1$ is not dividible by $3$, $n$ is odd.  $$2^n-1 \equiv  7 \pmod {12}$$So there exists such a $p$ that $$p \equiv \pm 7 \pmod {12}, 2^n-1 \equiv 0 \pmod p \Rightarrow 3^n-1 \equiv 0 \pmod p$$ So notice if $x=3^{\frac{n-1}{2}}$, $$3^{n} \equiv 3x^2 \equiv 1 \pmod {p} \Rightarrow (3x)^2 \equiv 3 \pmod {p}$$ This implies $3$ is a quadratic residue mod $p$. A contradiction, as $p \equiv \pm 7 \pmod {12}$. 
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