Gaussian integers and divisibility

Question

Let $(x,y,z)$ be a primitive Pythagorean triple with $x$ not congruent to $y$ modulo $2$.

Why do we say that $1+i$ does not divide $x+yi$?

Is there a lemma that connect the congruences with the divisiblity here?

Answer 1

Suppose that $1+i|x+yi$, so exist $a,b \in \mathbb{Z}$ that $(a+bi)(1+i)=x+yi$ i.e. $(a-b) + (a+b)i=x+yi,$ so we have $x=a-b$ and $y=a+b$, hence $x-y=-2b \equiv 0 \ (\text{mod } 2)$. Contradiction.

Answer 2

Suppose $x+yi=(1+i)g$, with $g$ a Gaussian integer. Then, multiplying by the conjugate, $$ x^2+y^2=2g\bar{g} $$ so $x^2+y^2\equiv0\pmod 2$, which implies $x\equiv y\pmod{2}$.

Answer 3

\begin{eqnarray*} \frac{x+\jmath y}{1+\jmath} &=& \frac{x+\jmath y}{2}{(1-\jmath)} \\ &=& \frac{x+y}{2}-\jmath \frac{x-y}{2} \end{eqnarray*}

For this to be a Gaussian integer, $x+ y$ and $x- y $ should be even, which is not true, by definition ($x \ne y (\mod 2)$.