How do I find a,b∈Z s.t. {ac−bd+i(ad+bc)∣c,d∈Z} have real and imaginary parts both even or both odd?

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How do I find a,b∈Z s.t.  {ac−bd+i(ad+bc)∣c,d∈Z}  have real and imaginary parts both even or both odd?

Beverly Smith · Accepted Answer

Step 1 Conceptually it boils down to: αβ is even ⇔α or β is even, for α,β∈Z[i], once we generalize the notion of &#039;&#039;even&#039;&#039; appropriately for Gaussian integers. Hint for p=(2,i−1) we have Z[i]p=∼Z2 so p is &#039; &amp;amp; induces a parity structure on Z[i] via α=a+bi is even ⇔p∣α⇔2∣a+b⇔a,b are equal parity. Hence αβ is even, for β=c+di ⇔p∣αβ ⇔p∣α or p∣β, by p &#039; ⇔α is even or β is even ⇔a≡b or c≡d (mod 2)

Terry Ray · Accepted Answer

Step 1  Show that  I={k+li:2∣k−l}  is an ideal of Z[i].  To show I absorbs, the following characterization of I is convenient:  I={k+li:2∣k2+l2}={z|z∣2 is even}  Step 2  As Z[i] is a PID, I does have a generator a+bi.  1+i∈I,  and  (1+i)Z[i]  is a maximal ideal of Z[i] (by e.g. Z[i](1+i)=∼Z2Z is a field), hence (1+i)Z[i]=I.  Since {±1,±i} are all the units of Z[i], a+bi can possibly be ±(1±i).

How do I find a,b\in\mathbb{Z} s.t. \{ac-bd+i(ad+bc)|c,d\in\mathbb{Z}\} have real and imaginary parts

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