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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.

2132123
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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.

markvs
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  • could you elaborate a little bit? From what I am reading then, $u^2+3v^2=b$ has a solution iff $b$ is a norm in $Q(\sqrt{-3})$. I am not sure what it even means to be a norm. Is there a simple way to check for $Q$? – 2132123 Feb 27 '22 at 02:01
  • Being a norm in that extension means that $b=(u+\sqrt{-3}b)(u-\sqrt{-3}v)$, nothing more. The method explained in the answer is the simplest I know. – markvs Feb 27 '22 at 02:09
  • Eric Wofsey's answer in the question I linked to is a geometric version, perhaps that could be easier. – markvs Feb 27 '22 at 02:14
  • The existence question is addressed in the Complement 2 of the answer by @nguyenquangdo. It is not easy, involves computing the Hilbert symbol, etc. – markvs Feb 27 '22 at 02:25
  • I don't see how you found the coefficient $3a^2$. From $x^2 - axy + y^2 = b$ I reached $ (2x-ay)^2 + (4-a^2) , y^2 = 4b $ – Will Jagy Feb 27 '22 at 19:38
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    @WillJagy: Yes, you are right. So the equation is $u^2+cv^2=4b$ where $c=4-a^2$. It is basically the same only we cannot ignore $a$, and the field is $\Bbb Q[\sqrt{-c}]$. – markvs Feb 27 '22 at 20:24