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.)

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.
