I'm trying to prove this property from this Wikipedia page.
Let $\mathbb Z[i]$ denote the ring of Gaussian integers and $N(z) = z \overline{z}$ for $z \in \mathbb Z[i]$. For $p,q \in \mathbb Z[i]$ such that $N(q)>0$, there exist $s,r \in \mathbb Z[i]$ such that $p=sq+r$ and $N(r) \le N(q)/2$.
My below attempt contains heavy calculation which is very sensitive to errors, even small ones.
Could you please have a check on it?
Is there any cleaner way to obtain the result?
Let $a,b,c,d \in \mathbb Z$ such that $cd\neq0$. It suffices to show that there are $x,y \in \mathbb Z$ such that $N((a+bi)-(c+di)(x+yi)) \le N(c+di)/2$, or equivalently $N(a-cx+dy+(b-cy-dx)i) \le N(c+di)/2$. This inequality simplifies to $(a-cx+dy)^2 + (b-cy-dx)^2 \le (c^2+d^2)/2$.
First, let $e = a+dy$ and $f = b-cy$. Then the inequality becomes $(cx-e)^2 + (dx-f)^2 \le (c^2+d^2)/2$, or equivalently $$(c^2+d^2)x^2-2(ce+df)x+e^2+f^2-(c^2+d^2)/2 \le 0$$
Consider the map $F: x \mapsto (c^2+d^2)x^2-2(ce+df)x+e^2+f^2-(c^2+d^2)/2$. It suffices to show that the equation $F(x)=0$ has $2$ different solutions $x_1,x_2$ such that $|x_1-x_2| \ge 1$. This means $$\begin{cases} \Delta & > 0 \\ (x_1+x_2)^2-4x_1 x_2 &\ge 1\end{cases}$$ or equivalently $$\begin{cases} (ce+df)^2-(c^2+d^2)[e^2+f^2-(c^2+d^2)/2] \ge 0 & > 0 \\ \left( \frac{2(ce+df)}{c^2+d^2} \right )^2 -4 \frac{e^2+f^2-(c^2+d^2)/2}{c^2+d^2} &\ge 1\end{cases}$$ or equivalently $$\begin{cases} (ce+df)^2-(c^2+d^2)[e^2+f^2-(c^2+d^2)/2] & > 0 \\ (ce+df)^2-(c^2+d^2)[e^2+f^2-(c^2+d^2)/2] &\ge (c^2+d^2)/4\end{cases}$$
As such, it suffices to show that there is $y \in \mathbb Z$ such that $(ce+df)^2-(c^2+d^2)[e^2+f^2-(c^2+d^2)/2] \ge (c^2+d^2)/4$, or equivalently $(c^2+d^2)^2 \ge 4(cf-de)^2$. Substituting back $e = a+dy$ and $f = b-cy$, this inequality becomes $(c^2+d^2)^2 \ge 4[(c^2+d^2)y+ad-bc]^2$, which is equivalent to $$4(c^2+d^2)^2y^2+8(c^2+d^2)(ad-bc)y+4(ad-bc)^2-(c^2+d^2)^2\le 0$$
Let $k = c^2+d^2$ and $t=ad-bc$. The inequality becomes $4k^2y^2 +8kty+4t^2-k^2 \le 0$. Consider the map $H: y \mapsto 4k^2y^2 +8kty+4t^2-k^2$. It suffices to show that the equation $H(y)=0$ has $2$ different solutions $y_1,y_2$ such that $|y_1-y_2| \ge 1$.
This means $$\begin{cases} \Delta & > 0 \\ (y_1+y_2)^2-4y_1y_2 &\ge 1\end{cases}$$ or equivalently $$\begin{cases} (4kt)^2-4k^2(4t^2-k^2) & > 0 \\ \left (\frac{-8kt}{4k^2} \right)^2-4 \frac{4t^2-k^2}{4k^2} &\ge 1\end{cases}$$ or equivalently $$\begin{cases} 4k^4 & > 0 \\ \left (\frac{-8kt}{4k^2} \right)^2-4 \frac{4t^2-k^2}{4k^2} &\ge 1\end{cases}$$ or equivalently $k = c^2+d^2 \neq 0$. This completes the proof.