Divisibility of integers Let $n > 1$ be an integer. Then $2^n - 1\nmid 3^n - 1$. I don't know how to prove it. Can anybody help me, please?
In general, for a fixed positive integer $a > 1$, has $a^n - 1|(a +1)^n - 1$ any integer solutions?

As @AQP said, if $n$ is even then $3\mid 2^n-1$ so $2^n-1\nmid 3^n-1$.
If $n=2k-1$
then
$2^n-1 \equiv 1 \pmod{3}$ so $2^n-1$ is a quadratic residue mod 3.
$3(3^n-1)=3^{2k}-3$ so $2^n-1 \mid 3^n-1$ would require that $3^{2k}\equiv 3 \pmod{2^n-1}
$,
i.e. that 3 is a quadratic residue mod $2^n-1$.
But $2^n-1\equiv 3 \pmod{4}$ so is divisible by an odd number of primes $p\equiv 3\pmod{4}$. By quadratic reciprocity 3 cannot be a quadratic residue mod $2^n-1$, hence $2^n-1\nmid 3^n-1$.
