Find $\gcd (3 + \sqrt{13},\ 2 + 5\sqrt{13})$ in $\mathbb Z[(1+\sqrt{13})/2]$ My task is find $\gcd (3 + \sqrt{13},\ 2 + 5\sqrt{13})$ in $\mathbb Z[(1+\sqrt{13})/2]$.
Can you give me some advice?
Any help is highly appreciated.
 A: \begin{align*}
\text{Let}\;\;R &= \mathbb{Z}[(1+\sqrt{13})/2]\\[4pt]
a &=3 + \sqrt{13}\\[4pt]
b &=2 + 5\sqrt{13}\\[8pt]
\end{align*}
and suppose that $d \in R$ is a common divisor of $a,b$.

By definition,
\begin{align*}
& N(a) = a\bar{a} = \left(3 + \sqrt{13}\right)\left(3 - \sqrt{13}\right) = -4\\[4pt]
& N(b) = b\bar{b} = \left(2 + 5\sqrt{13}\right)\left(2 - 5\sqrt{13}\right)= -321\\[4pt]
\end{align*}
so in the ring $R$, we have
\begin{align*}
&d\,{\mid}\,a\;\;\text{and}\;\;a\,{\mid}\,(-4)\implies d\,{\mid}\,(-4)\\[4pt]
&d\,{\mid}\,b\;\;\text{and}\;\;b\,{\mid}\,(-321)\implies d\,{\mid}\,(-321)\\[4pt]
\end{align*}
Since $-4$ and $321$ are relatively prime in $\mathbb{Z}$, there is an ordinary integer linear combination of $-4$ and $-321$ which equals $1$. For example
$$(80)(-4) + (-1)(-321) = 1$$
Since in the ring $R$, $d\,{\mid}\,(-4)$ and $d\,{\mid}\,(-321)$, it follows that $d$ divides any integer linear combination of $-4$ and $-321$.

In particular, $d\,{\mid}\,1$, so $d$ is a unit in $R$.

Therefore $\gcd(a,b)=1$.
