<s>[INST] Find a positive integer $n$ (or prove that none exists) such that $\gcd(3^{n+1}-2,2^n-3) > 1$ 
Find a positive integer $n$ (or prove that none exists) such that $\gcd(3^{n+1}-2,2^n-3) > 1$.

The greatest common divisor here seems to be equal to one for most values (I checked up to $7000$), but I didn't see  an easy way of finding an $n$ for which this it is greater than $1$. How could we go about solving this?
 [/INST] This is not correct!
If $n$ is such that $\gcd(3^{n+1}-2,2^n-3)>1$, then there exists a prime $p$ such that $3^{n+1}=2\bmod p$ and $2^n=3\bmod p$. But this implies
$$3^{n+1}=(2^n)^{n+1}=2^{n^2+n}=2\bmod p,$$
hence $2^{n^2+n-1}=1\bmod p$, and hence $\varphi(p)=p-1$ divides $n^2+n-1$, no, all we can conclude is $\gcd(p-1,n^2+n-1)>1$. Considering the possibilities modulo 4 shows that this is not possible.
Elaboration on the mod 4 argument: If $p-1$ divides $n^2+n-1$ then
$$n^2+n-1=k(p-1)$$
for some integer $k$. Note that $p$ is odd since both $3^{n+1}-2$ and $2^n-3$ are odd. Thus $p=1,3\bmod 4$, so
$$k(p-1)=0,2\bmod 4.$$
On the other side, if $n$ is odd or $n=2\bmod 4$ then
$$n^2+n-1=1\bmod 4,$$
and if $n=0\bmod 4$ then
$$n^2+n-1=3\bmod 4.$$
Therefore $n^2+n-1\ne k(p-1)$, so $p-1$ does not divide $n^2+n-1$.
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