I am considering the Diophantine Equation $X^2 + 4Y^2 = 3^n$, for some fixed $n$. I would wish to find Gaussian solutions to this Diophantine equation, by using a prime factorisation argument in the Gaussian integers. How might I proceed?
Edit: Here, $X, Y \in \mathbb{Z}$ and $n$ is a non-negative integer. (Having just realised this–– I thought I was solving for $X, Y \in \mathbb{Z}[i]$, I believe I can solve the question easily!)
Let $n$ be an integer, and let $X$ and $Y$ be integers such that $$X^2+4Y^2=3^n.$$ The left hand side is an integer so $n$ must be nonnegative. Reducing mod $3$ shows that $X\equiv Y\equiv 0\pmod{3}$, so $X=3x$ and $Y=3y$ for some integers $X_1$ and $Y_1$. Then $$X_1^2+4Y_1^2=3^{n-2},$$ where again the left hand side is an integer, so $n-2$ must be nonnegative. Repeating this $k:=\lfloor\tfrac n2\rfloor$ times yields two integers $X_k$ and $Y_k$ such that $$X_k^2+4Y_k^2=3^r,$$ where $r=0$ or $r=1$, and these two equations are easily solved by inspection.
For $r=1$ there is no integral solution, so $r=0$ meaning that $n$ is even. Clearly the only integral solutions for $r=0$ are $X_k=\pm1$ and $Y_k=0$. This corresponds to the solutions $X=\pm3^n$ and $Y=0$.