Prove that $\gcd(3^n-2,2^n-3)=\gcd(5,2^n-3)$ 
Prove that $\gcd(3^n-2,2^n-3)=1$ if and only if $\gcd(5,2^n-3)=1$ where $n$ is a natural number.

I didn't see an easy way to prove this using the Euclidean algorithm, but it seems true that both gcd's are not $1$ only if $n = 3+4k$. Is there an easy way to prove the statement?
 A: This does not completely answer the question.  This provides some more information than those given by previous answers and comments. Also, the method I provide here slightly differs from the following reference, but main ideas are from the paper. 
According to the reference http://www.fq.math.ca/Papers1/43-2/paper43-2-6.pdf
We obtain common prime divisors of $2^n-3$ and $3^n-2$ in the following way: 


*

*Find a prime number $p$ satisfying 
$$
\gcd(\mathrm{ord}_p (3\cdot 2^{-1}), \mathrm{ord}_p (6)) = 1 \ \mathrm{or} \ 2.
$$

*For such prime $p$, let $\ell = \mathrm{ord}_p(3\cdot 2^{-1})$ and $k=\mathrm{ord}_p(6)$. Find integer solutions to:
$$
\ell x - k y = 2.$$

*We want $x\geq 1$ from Step 2. Starting from $1$, find:
$$
2^{\ell x -1} \ \mathrm{mod} \ p, \ \mathrm{and} \ 3^{\ell x -1} \ \mathrm{mod} \ p
$$

*If they are $3$ and $2$ respectively, then the prime $p$ is the common divisor of $2^n-3$ and $3^n-2$ where  $n=\ell x -1$. 
A SAGE program that I wrote, doesn't exactly follow the above algorithm, but it was successful in finding at least the three primes $5$, $5333$, $18414001$. 
for p in primes(4,40000000): 
a=2.inverse_mod(p); 
b=Mod(Mod(3,p)*a,p).multiplicative_order(); 
c=Mod(6,p).multiplicative_order(); 
if gcd(b, c)==1: 
    g,s,t=xgcd(b, -c); 
    h=(2*s)%c; 
    print p, Mod(b*h-1,4), power_mod(3,b*h-1,p), power_mod(2,b*h-1,p);

with the following results:
5 3 2 3
5333 1 5331 5330
18414001 3 2 3

The second line result may be due to the program lacking to proceed Step 2 and Step 3. With more care, it might be possible to implement those and get a correct result 5333 3 2 3. 
It was not possible for me to completely settle the (equivalent) problem:
$\bullet$ If $p|\gcd(2^n-3,3^n-2)$, then $n\equiv 3 \ \mathrm{mod} \ 4$. 
But, I now have one more conjecture:
$\bullet$ If $\gcd(\mathrm{ord}_p (3\cdot 2^{-1}), \mathrm{ord}_p (6)) = 1$, then there exists $n$ such that $p|\gcd(2^n-3,3^n-2)$. 
Both of these questions still remain open. 
