Particular solution to a diophantine in terms of $n$

Question

Given the linear diophantine $3^n x - 2^n y = 1$, ($n$ being a natural parameter) it's well known that it has infinitely many solutions.

My question is whether we can find a generic particular solution of it in terms of $n$.

I inspected some values using the online linear diophantine solver > https://www.math.uwaterloo.ca/~snburris/htdocs/linear.html yet found no pattern.

I also tried to conjecture the inverse of $3^n$ modulo $2^n$ in terms of $n$ assuming a formula could be found but in vain.

Any ideas are welcome. Thanks.

Answer 1

Lets see the result needs to be odd ( 1 mod 2) . That requires x be 1 mod 2 . The result is also, 1 mod 3 leading to if n is odd y is 2 mod 3, if n is even y is 1 mod 3 . These two combine because the result is Also 1 mod 6 (1 mod 2 and 1 mod 3 together) etc. It's Also 1 mod n but unless n is prime we don't get much. if we say n is prime, we get x and y both being 1 mod n works ( may not be the only way though).

Answer 2

Hint

Since it is $$ 3^{\,n} x - 2^{\,n} y = 3^{\,n} x + 2^{\,n} \left( { - y} \right) = 1 = \gcd \left( {2^{\,n} ,3^{\,n} } \right) $$ then the Bezout Identity assures that we can find two integers such that the identity above is satisfied.

The two integers $x,y,$ will come by applying the Extended Euclidean Algorithm to the couple $3^n , \, 2^n$, or by expanding $(3/2)^n$ into a Continued Fraction.

But in fact, the process does not look to show any discernible path at the varying of $n$.