TITLE: How to split 59 in $\mathbb{Q}(\sqrt{13})$
QUESTION [5 upvotes]: 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

REPLY [2 votes]: You can save yourself a lot of time by availing yourself to the Legendre symbol. Does the prime $p$ split in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ (where $d$ is positive and squarefree)? If $$\left(\frac{d}{p}\right) = d^{\frac{p - 1}{2}} \equiv -1 \bmod p$$ then the answer is absolutely not. But if it's $1$ and $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ is a unique factorization domain, then the answer is yes, and if it doesn't have unique factorization, then the answer is maybe. The Legendre symbol is JacobiSymbol[d, p] in Wolfram Mathematica and Wolfram Alpha.
So with $59$ we see that $13^{29} \equiv 28 \equiv -1 \bmod 59$, which means that $59$ definitely does not split in $\mathcal{O}_{\mathbb{Q}(\sqrt{13})}$.
This works even when the domain contains "half" integers. For example, $$\left(\frac{13}{17}\right) = 1$$ and $$\left(\frac{9 - \sqrt{13}}{2}\right) \left(\frac{9 + \sqrt{13}}{2}\right) = 17.$$