Diophantine equation $x^2 + xy − 3y^2 = 17$ Determine all integer solutions to the equation $x^2 + xy − 3y^2 = 17$.
The previous part of the question was finding the fundamental unit in $\mathbb{Q}(\sqrt{13})$, which is $\varepsilon = \frac{3+\sqrt{13}}{2}$, so my guess is that I should factorise the equation in $\mathbb{Q}(\sqrt{13})$ and then use the uniqueness of factorisation of ideals into prime ideals. (But it is also possible that the parts of the question are unrelated, because they certainly seem unrelated.)
 A: Conway's Topograph, with some artwork improvements of mine. I found that I really wanted to draw the "river" as a straight line. When there is not really enough room to draw in the representations of a number (17) I realized that drawing a separate (rooted) tree as though growing from the river allowed for a more legible rendition of that number representation. For that matter, while correctly drawing the river alternates among Conway's rules, once one of these trees is correctly begun, it is just one small collection of rules to expand it outward. Now that I think of it, there are intricate rules about the direction of the little blue arrows along the river, but in a tree they all go the same way, away from the tree root when the pink values in the tree are positive. The really big item was simply that the generator of the automorphism group, which is a 2 by 2 matrix, appears as  explicit columns (in green) in the diagram. Let's see, I don't know any book that consistently draws in all of Conway's numbers, all of which have meaning; I started using four colors once I saw the legibility problems.
Oh, forgot this part. From the automorphism matrix, Cayley-Hamilton says that we get two $x$ orbits and two $y$ orbits, trace is $11$ so each  is
$$  x_{n+2 } = 11 x_{n+1} - x_n  $$
$$  y_{n+2 } = 11 y_{n+1} - y_n  $$
One orbit of pairs is
$$   19,4,25,271,...$$
$$ -8,1,19,208...$$
The other orbit of pairs
$$ 5, 11, 116, 1265, ...  $$
$$   -1,8, 89, 971, ...   $$
Since there is a mixed $xy$  term in this one, the orbits should be extended backwards as well as forwards. One way to write that is
$$  x_n = 11 x_{n+1} - x_{n+2}  $$
$$  y_n = 11 y_{n+1} - y_{n+2}  $$





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A: The two parts of that problem are absolutely related and are not some random pairing of questions.
The ring of integers of $\mathbf Q(\sqrt{13})$ is $\mathbf Z[(1+\sqrt{13})/2] = \mathbf Z + \mathbf Z(1+\sqrt{13})/2$. For $x, y \in \mathbf Z$,
$$
{\rm N}_{K/\mathbf Q}\left(x + y\frac{1+\sqrt{13}}{2}\right) = x^2 + xy - 3y^2,
$$
so the question is asking you to find all elements in $\mathbf Z[(1+\sqrt{13})/2]$ with norm $17$. (Don't write the equation as a Pell equation with right side $68 = 4 \cdot 17$, since the extra factor of $4$ that is inserted is very misleading: you want to work in $\mathbf Z[(1+\sqrt{13})/2]$, not its subring $\mathbf Z[\sqrt{13}]$.) You already found solutions $(\pm 4,\pm 1)$, which are related to the factorization $17 = (4 + (1+\sqrt{13})/2)(4 + (1 - \sqrt{13})/2)$.  Show $\mathbf Z[(1+\sqrt{13})/2]$ is a UFD (equivalently in this case, a PID), so once you know one prime factorization you know all others can be obtained from it using multiplication by units.
A: Hint:
Complete the square:  $x^2+xy+\frac14y^2-(3+\frac{1}4)y^2=17\implies(x+\frac12y)^2-\frac{13}4y^2=17$
$\implies (2x+y)^2-13y^2=68$.  That's a Pell-type equation.
