I'm trying to find some numbers, $a,b\in\mathbb{Z}$ s.t. the following equation is satisfied. \begin{equation} \{a c+b d+i(a d+b c) \mid c, d \in \mathbb{Z}\}=\{k+i l \mid k, l \text { even or odd }\} \end{equation}
So I need to find a number $a$ and $b$ s.t. the first set consist of complex numbers, where both the real part and imaginary part is even or the real and imaginary part is odd. Is it possible to find such numbers and how do I approach this?
Suppose $a,b\in\Bbb{Z}$ are such that \begin{equation} \{a c+b d+i(a d+b c) \mid c, d \in \mathbb{Z}\}=\{k+i l \mid k\equiv l\pmod{2}\}. \end{equation} First note that the set on the left hand side can be expressed as $$\{(c(a+bi)+d(b+ai)\mid c,d,\in\Bbb{Z}\}.$$ Next note that $$(2,0),(0,2),(1,1)\in\{k+i l \mid k\equiv l\pmod{2}\}.$$ In particular there exist $c,d\in\Bbb{Z}$ such that $$ac+bd=1,$$ and so $a$ and $b$ are coprime. There also exist $c'$ and $d'$ such that $$ac'+bd'=0\qquad\text{ and }\qquad bc'+ad'=2.$$ The former shows that $(c',d')=(kb,-ka)$ for some integer $k$, because $a$ and $b$ are coprime. Then the latter shows that $$2=b(kb)+a(-ka)=k(b^2-a^2).$$ Then $(a,b)=(0,\pm1)$ or $(a,b)=(\pm1,0)$. But then the set generated by $a$ and $b$ is all of $\Bbb{Z}[i]$, a contradiction.