Find all solutions for $x \cdot y = 5$ in $\mathbb{Z}[i] $

Question

Let $\mathbb{Z}[i] = \{a+ib \in \mathbb{C}, a, b \in \mathbb{Z} \}. $ Find all $x, y \in \mathbb{Z} [i] \backslash \{i, - i, 1,-1\} $ such that $$x \cdot y = 5.$$ I've denoted $x=a+ib, y=c+id $ and substituted that into the equation. By the equality of complex numbers it follows that $$ac-bd=5$$ $$ad+bc=0.$$ I'm not sure how to use the assumption that all numbers are integers and that the solution is not from $i, - i, 1,-1.$

Any hints would be appreciated!

Answer 1

If $xy=5$, then $|xy|^2=25$. In other words, $|x|^2|y|^2=25$. But the numbers $|x|^2$ and $|y|^2$ are natural numbers. And you can express $25$ as the product of two natural numbers in two ways and only two ways: as $1\times25$ or as $5\times5$. Can you take it from here?

Answer 2

Hint: The equation $ad-bc=0$ corresponds to the zero-determinant of the real-valued matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$. In this case, the columns of the matrix must be collinar.

Answer 3

Do you know all prime elements in the UFD $\mathbb{Z}[i]$? Moreover $5=1^2 + 2^2$. Now it's your turn.