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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?

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sjm23
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  • Note the set equality means that as you change $c$ and $d$ leaving $a$ and $b$ the same, you get all the $k+il$ numbers with $k$ and $l$ both even or both odd. Do you know about vector subspaces and basis spans? – aschepler Nov 19 '21 at 13:56
  • @aschepler I have taken a course in linear algebra 3 years ago, so I am familiar with it. So I can see that we are asking whether the $span{a,b}+span{ia,ib}={k+i l \mid k, l \text { even or odd }}$ – sjm23 Nov 19 '21 at 14:24
  • However, I don't see how using the span helps me find such a,b that satisfy this? – sjm23 Nov 19 '21 at 14:26
  • So you want the second set to have all $k+il$ where $k\equiv l \pmod 2$? For example $1+i$ must be in that set? If so, this is already too restrictive. – Sil Nov 19 '21 at 15:20
  • I don't suppose it's supposed to be $ac-bd+i(ad+bc)$? That would make a big difference. – aschepler Nov 19 '21 at 22:01
  • @aschepler Yes, in fact it is $ac-bd+i(ad+bc)$ but that shouldn't make a difference since d can be chosen arbitrarily among the integers? – sjm23 Nov 20 '21 at 11:59
  • It does, though. With the $+bd$ it's not possible, and with the $-bd$ it is. Probably better not to make the existing answer incorrect, so maybe you could open another question about that. – aschepler Nov 20 '21 at 12:35

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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.

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