I want to understand an algorithm from [1] to solve
$\alpha x^2+\beta y^2=\gamma \text{ over } \mathbb{Q}$
with $\alpha, \beta, \gamma\in\mathbb{Q}$. As far is I understood the process the following happens: Multiply the equation by the $\gcd$ of the denominators of $\alpha, \beta, \gamma$ to obtain an equation
$\alpha x^2+\beta y^2=\gamma$
where $\alpha, \beta, \gamma\in\mathbb Z$. We may furthermore assume that $\alpha$ and $\beta$ are square free, since else we can solve the equation
$\bar{\alpha}x'^2+\bar{\beta}y'^2=\gamma$
where $\bar{\alpha}, \bar{\beta}$ are the nonsquare-parts of $\alpha$ and $\beta$ and with new variables $x'=(\tilde{\alpha}x)^2$ and $y'=(\tilde{\alpha}y)^2$ where $\tilde{\alpha}$ and$\tilde{\beta}$ are the square parts of $\alpha$ and $\beta$. So without loss of generalty, we are left with an equation
$\alpha x^2+\beta y^2=\gamma \text{ over } \mathbb{Q}$
where $\alpha, \beta, \gamma\in\mathbb Z$ and $\alpha, \beta$ are square free.
But now the magic happens: This equation should be solvable if and only if an equation
$ax^2+by^2=z^2$
is solvable over $\mathbb Z$ with coprime $x,y,z$. Unfortunately I don't see how the coefficients $\alpha$, $\beta$ and $\gamma$ should be related to the coefficient $a$ and $b$ and therefore am not able to understand the equivalence of solving these two equations.
I think that there might be a very easy number-theoretic argument that I don't know and of course it would be awesome if there is indeed an elementary argument for this.
As a remark, that might or might not help: A friend gave me the idea that $z$ could have something to do with the square part $\tilde{\gamma}$ of $\gamma$.
The argument I'm referring to is on page 22 in[1], the middle after "Suppose $\mathbb F=\mathbb Q$".