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Let $a$ and $b$ be two non-zero rational numbers such that the equation $ax^2+by^2=0$ has a non-zero solution in rational numbers . Prove that for any rational number $t$ , there is a solution of the equation $ax^2+by^2=t$.

I tried by defining $b$ as $-b$ and have a solution as $(x_0,y_0)$ and taking $x=x_0+k$ and $y=y_0+\sqrt{\frac{a}{b}}\cdot k$ but it comes back in a circular. By defining $b$ as $-b$ I could show that $\sqrt{\frac{a}{b}}$ is a rational number because $x$ and $y$ are one but I don't know how this might help.

Please give me an idea of how to approach this problem.

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    $\begingroup$ it does not change the problem to demand $a,b$ integers. After that, there is no damage by demanding $\gcd(a,b) = 1.$ Next, $\frac{-b}{a}$ is a rational square, so..... $\endgroup$
    – Will Jagy
    Commented Apr 25 at 0:28
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    $\begingroup$ I don't want to derail from you working with Will here, so this is something to look at later if you're interested. You're essentially given an isotropic vector in a nondegenerate quadratic space, so it is universal. This actually corresponds to an explicit construction here, but you'll have to do a bit of legwork. Here's where I first learned it in case it's helpful to you: Quadratic forms and the local global-principle - swc-math.github.io/aws/2021/index.html#2021Chan $\endgroup$
    – Merosity
    Commented Apr 25 at 1:30
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    $\begingroup$ FYI, using an Approach0 search, there's the AoPS threads 2-B-7 polynomials and B.Math - rational solution. Also, Two problems on solving equations in rational and integers has your problem as a second part, but neither of the $2$ answers there deal with it. $\endgroup$
    – John Omielan
    Commented Apr 25 at 2:12
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    $\begingroup$ Thanks @JohnOmielan didn't know such a search engine existed. I see that the problem reduces to $x^{2}-y^{2}=t$ as pointed out by Will Jagy. $\endgroup$
    – Ayan Bhowmik
    Commented Apr 25 at 2:29
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    $\begingroup$ You're welcome. And, yes, the problem can basically reduce to $x^2 - y^2 = t$, as Will Jagy pointed out earlier. You should consider answering this question yourself with the details. FYI, the solution that I worked out myself is basically the same as that in Post #$7$ of the first AoPS thread I linked to above. $\endgroup$
    – John Omielan
    Commented Apr 25 at 2:32

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This is not a personal solution but from AoPS.

We take $(u,v)$ such that $au^{2}+bv^{2}=0$.

Proceed forward with $b=-a\frac{u^{2}}{v^{2}}$. Then the problem simply reduces to:

$x^{2}-(\frac{u}{v}y)^{2}=\frac{t}{a}$

So it has infinitely many rational solutions.

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    $\begingroup$ alright, you have reduced it to $x^2 - y^2 = t.$ Can you give a convincing proof that this can be done in rationals? A beginning: $x^2 - y^2 = (x+y)(x-y) $ Tell you what: you might as well demand $x-y = 1$ and $x+y = t$ $\endgroup$
    – Will Jagy
    Commented Apr 25 at 3:25

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