Find integers $x,y \in \mathbb{Z}[i]$ such that $x^2 + i\,y^2 = 9+ 4i$.

Question

I want to know if the Gaussian prime $\mathfrak{p} = 9 + 4i$ is the sum of two squares [square] $+$ $i \,\times $ [square] in the ring of Gaussian integers $\mathbb{Z}[i]$. So I can write down an equation:

\begin{eqnarray*} x^2 + i\, y^2 &=& \mathfrak{p} \\ (a+bi)^2 + i\,(c+di)^2 &=& 9 + 4i\end{eqnarray*}

These can reduce to a pair of equations over $\mathbb{Z}$ which could also be solved:

\begin{eqnarray*} (a^2 - b^2) + 2cd &=& 9 \\ 2ab + (c^2 - d^2) &=& 4\end{eqnarray*}

Now we have $2$ equations and $4$ unknowns, so ostensibly we have $4 - 2 = 2$ degrees of freedom. I am looking for at least one solution, or to show there are finitely many.

Answer 1

The obvious solutions are the real special cases $x=\pm 3,\,y=\pm 2$.

To prove only finitely many solutions exist note $|x|^4+|y|^4=97$, writing $97$ as the sum of two squares. This gives $a^2+b^2=|x|^2,\,c^2+d^2=|y|^2$ only finitely many possible values, each of which only has finitely many solutions in $a,\,b,\,c,\,d$.

Answer 2

Just try $x=3+0i$ and $y=2+0i$.