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
 A: You are trying to find $x,y$ such that $x^2 - 13y^2=59$ is a perfect square. This is not even possible.
The reason for this is as follows: $x^2-59 = 13y^2$. Now, take the equation modulo $13$, and we get $x^2 - 7 =0$. The question boils down to whether there are any squares congruent to $7$ modulo $13$. It is easy to see this is not the case (you can check $x=1,2,3,4,5,6$, this is sufficient to show the above).
The squares $1,4,9,16,25,36$ are congruent to $1,4,9,3,12,10$ mod $13$, hence all squares will be congruent to these numbers modulo $13$. This information tells you, for example, that even $57,58,60$ will not split in this field. However, $56,62$ etc. may split (you still can't be certain).
A: 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}
A: 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 $$\left(\frac{9 - \sqrt{13}}{2}\right) \left(\frac{9 + \sqrt{13}}{2}\right) = 17.$$
