<s>[INST] How to split 59 in $\mathbb{Q}(\sqrt{13})$ How to split 59 in $\mathbb{Q}(\sqrt{13})$
as i know that  59 is prime and we can write $(x+y\sqrt{13})(x-y\sqrt{13})=59$
$x^2-13y^2=59$ but i cant find the x,y
 [/INST] Agree with Parcly Taxel.
In fact for squarefree $-50 \leq d \leq 50$, $59$ splits in $\Bbb{Q}[\sqrt{d}]$ only for \begin{align}
d &=-43  &  59 &= 4^2 + 43 \cdot 1^2  \\
d &=-34  &  59 &= 5^2 + 34 \cdot 1^2  \\
d &=-23  &  59 &= 6^2 + 23 \cdot 1^2  \\
d &=-11  &  59 &= \left(\frac{15}{2}\right)^2 + 11 \cdot \left(\frac{1}{2}\right)^2  &  [11 \cong 3 \pmod{4}]  \\
d &=-10  &  59 &= 7^2 + 10 \cdot 1^2  \\
d &= -2  &  59 &= 3^2 + 2 \cdot 5^2  \\
d &=  1  &  59 &= 30^2 - 1 \cdot 29^2  &  [\text{"splits" as $59 \cdot 1$}]  \\
d &=  5  &  59 &= 8^2 - 5 \cdot 1^2  \\
d &= 17  &  59 &= 22^2 - 17 \cdot 5^2  \\
d &= 22  &  59 &= 9^2 - 22 \cdot 1^2  \\
d &= 29  &  59 &= \left(\frac{31}{2}\right)^2 - 29 \cdot \left(\frac{5}{2}\right)^2  \\
d &= 41  &  59 &= 10^2 - 41 \cdot 1^2  \\
d &= 46  &  59 &= 75^2 - 46 \cdot 11^2
\end{align}
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