Originally Posted by

**georgel** Here's what I did so far:

I try to find GCD$\displaystyle (1+3i, 3+11i, -1-11i)$ .

Firstly, I find GCD$\displaystyle (1+3i, 3+11i)$

$\displaystyle N(1+3i)=10$

$\displaystyle N(3+11i)=130$

$\displaystyle 3+11i=(1+3i)*4 -1-i$

$\displaystyle 1+3i=(-1-i)*(-2-i)+0$

GCD$\displaystyle (1+3i, 3+11i)=-1-i$

Next, GCD$\displaystyle (-1-i, -1-11i)$

$\displaystyle -1-11i=(-1-i)*(6-5i)+0$ (*)

So GCD$\displaystyle (-1-i, -1-11i)=-1-i$

Now I check is (-1-i) irreducible.

$\displaystyle N(-1-i)=1+1=2$

$\displaystyle 2=(a^2+b^2)(c^2+d^2)$

$\displaystyle (a^2+b^2)\in\{+-1, +-2\}, (c^2+d^2)\in\{+-1, +-2\}$

So in any case, one of them is +-1, which means that (-1-i) is irreducible, so (1+3i, 3+11i, -1-11i) is generated by an irreducible element so it is prime.

I'm pretty sure I made mistakes, could someone please correct me?