When does $x^2+y^2-axy-b=0$ have a rational solution?

Question

As in the title, I am looking for conditions on $ a$ and $b$ that guarantee existence of a rational solution (which then means there aqre infinitely many rational solutions and that those solutions are dense). $a,b$ are of course rational. This is number theory which I have no background in, so I am not sure how to approach this. Only thing that comes to mind is completing the square but that leads nowhere.

Answer 1

Your equation is equivalent to $(x-ay/2)^2+ 3(ya)^2=b$. So $a$ is irrelevant as long as it is rational, and $b$ is represented as $u^2+3v^2$ with rational $u,v$. You can see an answer here. It treats the case when the field of rationals is replaced by any field $K$.

Edit As @WillJagy noticed, the first line is not correct: the correct reduction is $(x-ay/2)^2+(4-a^2)y^2=4b$. Then you are interested in representing $4b$ by the quadratic form $u^2+cv^2$, where $c=4-a^2$. This cab be further simplified a little.

If $c$ is odd then $x,y$ are even and you get $(u/2)^2+c(v/2)^2=b$. If $c$ is even then $a$ is even and $c$ is divisible by $4$, hence $u$ is even, so $(u/2)^2+c/4v^2 = b$, so the question is about representing $b$ by a quadratic form $w^2+dt^2$. The Complement 2 of the answer by @nguyenquangdo in the question I linked to explains how to do it.