I am stuck with part of my homework.
Let R be the following set of complex numbers: $R=\{a+bi\sqrt5 : a,b\;\text{are ordinary integers}\}$
Can you find distinct irreducibles $\pi_1,\,\pi_2,\,\pi_3,\,\pi_4,\,\pi_5,\,\pi_6$ in R with the property that $\pi_1\pi_2=\pi_3\pi_4=\pi_5\pi_6$?
cf) 'irreducible' here means irreducible into a multiple form of elements of R.
Could anyone give me a start on how to deal with this?
Thank you very much for your help.
Let $\pi_1=2+i\sqrt 5,\pi_2=-2+i\sqrt 5,\pi_3=3,\pi_4=-3,\pi_5=-2-i\sqrt 5,\pi_6= 2-i\sqrt 5$. Note $$\pi_1\pi_2=\pi_3\pi_4=\pi_5\pi_6=-9$$
The map $\eta :R\to\Bbb N\cup\{0\}$ defined by $(x+iy)\mapsto x^2+y^2$ is multiplicative ; i.e. $\eta(zz')=\eta(z)\eta(z')$ with $\eta(a+(b\sqrt5)i)=1\iff a^2+5b^2=1\iff a=\pm1\land b=0$. Moreover, note $\eta(z)=3$ has no solutions in $R$ as we'd certainly have $\text{Im}(z)=0$ but $3$ is not a perfect square. Therefore, if $\eta(z)=9$ then $z$ must be irreducible because $z=z'z''$ would then imply $\eta(z)=\eta(z')\eta(z'')=9$ implying one of $z',z''$ is a unit.