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(a) Find the fundamental unit of $\mathbb{O}_{\mathbb{Q}(\sqrt{13})}$ (the ring of integers of $\mathbb{Q}(\sqrt{13})$).
(b) Find the fundamental solution of Pell's equation $x^2-13y^2=1$.

For (a), let $\epsilon=r+s\sqrt{13}, r, s \in \mathbb{Q}, r, s >0$ be the fundamental unit of $\mathbb{O}_{\mathbb{Q}(\sqrt{13})}$.
The minimal polynomial of $\epsilon$ over $\mathbb{Q}$ is
$x^2-2rx+r^2-13s^2=(x-r-s\sqrt{13})(x-r+s\sqrt{13})$.
Hence, $[\mathbb{Q}(\epsilon) : \mathbb{Q}]=2$.
$\epsilon$ is a unit of $\mathbb{O}_{\mathbb{Q}(\sqrt{13})}$ if and only if $N(\epsilon)=(-1)^2\cdot(r^2-13s^2)=\pm1$.
We plug in $r=1, 2, 3, \cdots, s= 1, 2, 3, \cdots$ into
$r^2-13s^2=1$ and $r^2-13s^2=-1$
The smallest value of $r, s$ are $r=18, s=5$ ($18^2-13 \cdot5^2=-1$).
Hence, $\epsilon=18+5\sqrt{13}$ is the fundamental unit of $\mathbb{O}_{\mathbb{Q}(\sqrt{13})}$ .

For (b), $\epsilon^2=649+180\sqrt{13}$ and $649^2-180^2\cdot 13=1$.
Hence, $x=649, y=180$ is the fundamental solution of Pell's equation $x^2-13y^2=1$.

Is this correct? And if I got the smallest value of $r$ and $s$ from the equation $r^2-13s^2=1$, then would the fundamental unit of $\mathbb{O}_{\mathbb{Q}(\sqrt{13})}$ and the fundamental solution of Pell's equation $x^2-13y^2=1$ be same? I'm also wondering if I can always get the fundamental solution by taking the power of the fundamental unit.

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  • $\begingroup$ The fundamental unit should be $(3+\sqrt{13})/2$, see here. For more details, see this post. $\endgroup$ – Dietrich Burde Jun 6 '18 at 20:17
  • $\begingroup$ @Dietrich Burde But in my class note, definition of fundamental element is the smallest uint in $\mathbb{A} \cap \mathbb{Q}[\sqrt{d}]$, greater than 1 denoted by $\epsilon = a+b\sqrt{d}, a, b, \in \mathbb{Q}, a, b>0$ $\endgroup$ – Andrew Jun 6 '18 at 20:22
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    $\begingroup$ But unit means, in the ring of integers $\mathbb{O}_{\mathbb{Q}(\sqrt{13})}=\Bbb{Z} [(1+\sqrt{13})/2]$. $\endgroup$ – Dietrich Burde Jun 6 '18 at 20:24
  • $\begingroup$ @Dietrich Burde yeah there is a structure theorem of the ring of integers, but $(1+\sqrt{13})/2$ itself is not a fundamental element. Doesn't it has to satisfy $a, b \in \mathbb{Q}$ to be a fundamental element? $\endgroup$ – Andrew Jun 6 '18 at 20:30
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    $\begingroup$ @Andrew studying units of $\Bbb Q(\sqrt{13})$ is not very interesting (it's a field!). You're interested in studying units of $\mathcal{O}_{\Bbb Q(\sqrt{13})}$. A fundamental unit is a generator of the group of units of $\mathcal{O}_{\Bbb Q(\sqrt{13})}$. $\endgroup$ – ÍgjøgnumMeg Jun 6 '18 at 21:42

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