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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?

Sorry about the sign variation on my prior post. But I didn't realise that the sign in front of bd was important.

Bill Dubuque
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sjm23
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  • There is another way to write $ac-bd + i(ad+bc)$ that might make things a little clearer. – Arthur Nov 21 '21 at 11:26
  • If $c,d$ both even, then the real and imaginary portions must both be even (i.e. neither one can be odd). Therefore, without loss of generality, it is not the case that $c,d$ both even. – user2661923 Nov 21 '21 at 11:36
  • For all candidate values $(a,b)$, if both $a$ and $b$ are even, then the real and imaginary parts must both be even. Therefore, without loss of generality, it is not the case that $a,b$ are both even. – user2661923 Nov 21 '21 at 11:38
  • Thus, you have $3$ possible odd/even parities for $c,d$ and $3$ possible odd/even parities for $a,b$. Therefore, you can form a chart of the $9$ possible odd/even parity combinations. – user2661923 Nov 21 '21 at 11:39
  • By the way: you can (for example) let $(a = 0, b = 1)$ represent the case of $a$ even, $b$ odd. That is, all odd numbers may be represented by $(1)$ and all even numbers may be represented by $(0)$. This simplifies your computations. – user2661923 Nov 21 '21 at 11:44
  • Don't know if $(ac-bd)+(ad+bc)i = (a+bi)(c+di)$ helps. – Gargantuar Nov 21 '21 at 11:45
  • Try $a=b=1$.... – Gerry Myerson Nov 21 '21 at 11:54
  • Any thoughts on the answers and comments, sjm? – Gerry Myerson Nov 23 '21 at 12:24

2 Answers2

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Conceptually it boils down to: $\,\ \alpha\beta\,$ is even $\iff \alpha\,$ or $\,\beta\,$ is even, $ $ for $\,\alpha,\beta\in\Bbb Z[i],\,$ once we generalize the notion of "even" appropriately for Gaussian integers.

Hint for $\,p := (2,i\!-\!1)$ we have $\Bbb Z[i]/p \cong \Bbb Z/2\,$ so $\,p\,$ is prime & induces a parity structure on $\,\Bbb Z[i]\,$ via $\,\alpha := a+bi\,$ is $\rm\color{#c00}{even}$ $\iff p\mid\alpha \iff 2\mid a+b\iff a,b\,$ are $\rm\color{#c00}{equal\ parity}$. Hence

$$\begin{align} &\alpha\beta\,\ \text{is even},\ {\rm for}\,\ \beta = c+di\\[.2em] \iff\ &p\mid \alpha\beta\\[.2em] \iff\ & p\mid \alpha\,\ {\rm or}\ \,p\mid \beta,\,\ \text{by $\,p\,$ prime}\\[.2em] \iff\ &\alpha\,\ \text{is even or}\,\ \beta\,\ \text{is even}\\[.2em] \iff \ & a\equiv b\,\ {\rm or}\,\ c\equiv d\!\!\pmod{\!2}\end{align}\qquad\qquad$$

Bill Dubuque
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(1) Show that $I=\{k+li: 2|k-l\}$ is an ideal of $\mathbb Z[i]$. To show $I$ absorbs, the following characterization of $I$ is convenient: $$I=\{k+li: 2|k^2+l^2\}=\{z: |z|^2 \text{ is even } \}$$

(2) As $\mathbb Z[i]$ is a PID, $I$ does have a generator $a+bi$.

(3) $1+i\in I$, and $(1+i)\mathbb Z[i]$ is a maximal ideal of $\mathbb Z[i]$ (by e.g. $\mathbb Z[i]/(1+i)\cong \mathbb Z/2\mathbb Z$ is a field), hence $(1+i)\mathbb Z[i]=I$.

(4) Since $\{\pm 1, \pm i\}$ are all the units of $\mathbb Z[i]$, $a+bi$ can possibly be $\pm (1\pm i)$.

Just a user
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