# Particular solution to a diophantine in terms of $n$

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.

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 $$n$$.