In the integral domain $D = \{r+s \sqrt{17}: r,s \in\Bbb Z\}$, which element is irreducible?

Question

In the integral domain $D = \{r + s \sqrt{17} | r,s \in\Bbb Z\}$, which is irreducible?

$3 - \sqrt{17}$

$9 - 2\sqrt{17}$

$7 + \sqrt{17}$

$13 + \sqrt{17}$

I got all of them are irreducible, if you try to make it, for example, $3 - \sqrt{17} = (a + b \sqrt {17}) (c + d \sqrt{17})$, where $a, b,c,d\in\Bbb Z$ are integers, you will see $a = b = 0 $ or $c = d = 0$. and this apply to all the above four.

Answer 1

No, they are not all irreducible. For example, we have $$ 13+\sqrt{17}=(6-\sqrt{17})(5+\sqrt{17}). $$ The norm of $13+\sqrt{17}$ is $152=8\cdot 19$, and $N(6-\sqrt{17})=19$, $N(5+\sqrt{17})=8$.