Find integers $x,y \in \mathbb{Z}[i]$ such that $x^2 + y^2 = 8 + 5i$.

Question

I want to know if the Gaussian prime $\mathfrak{p} = 5 + 8i$ is the sum of two squares in the ring of Gaussian integers $\mathbb{Z}[i]$. So I can write down an equation:

\begin{eqnarray*} x^2 + y^2 &=& \mathfrak{p} \\ (a+bi)^2 + (c+di)^2 &=& 5 + 8i\end{eqnarray*}

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

\begin{eqnarray*} (a^2 - b^2) + (c^2 - d^2) &=& 5 \\ 2(ab+cd) &=& 8\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

You have one solution with $a=3$, $b=4$, $c=4$ and $d=-2$.

$(3^2-4^2)+(4^2-(-2)^2) = 3^2-(-2)^2 = 9-4 = 5$

$2(3*4+4*(-2)) = 2(12-8) = 2*4 = 8$

Answer 2

Your equation is $$(x+iy)(x-iy)=5+8i$$ and you claim to know the right side is a prime. Therefore one of these systems is true: $$\begin{align} x+iy&=i^n(5+8i)&x-iy&=i^n(5+8i)\\ x-iy&=i^{-n}&x+iy&=i^{-n} \end{align}$$

where $n$ can range $0$ through $3$. That makes $8$ systems of linear equations, and the solutions end up that one of $x$, $y$ is $\pm(3+4i)$, and the other is $\pm(4-2i)$.