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Let $a,b\in\mathbb Z$ be squarefree with $a>0$. Suppose that I know that there exist $(0,0,0)\neq (x,y,z) \in \mathbb Z^3$ s.t. $x^2-by^2-az^2=0$. Is there any known algorithm to find any such a triple?

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Maybe en.wikipedia.org/wiki/Integer_relation_algorithm but you might not get squares. –  xavierm02 Sep 2 '13 at 11:55

1 Answer 1

Hint: Since your equation is homogeneous, an integer solution can be interpreted as a rational point on the corresponding projective curve. Note that this curve has genus $0$.

Useful link: http://www.risc.jku.at/publications/download/risc_1355/Rational%20Points%20on%20Conics.pdf


In particular, you can use Magma to solve your problem, here is an example.

T<X,Y,Z>:=ProjectiveSpace(Rationals(),2);
C:=Curve(T,X^2-25*Y^2-19*Z^2);
bool,c1:=IsConic(C);
HasRationalPoint(c1);

Try this code in the magma calculator (http://magma.maths.usyd.edu.au/calc/) and find that $(50,9,5)$ is an integer solution to $x^2-25y^5-19z^2=0.$

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