Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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


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 $\mathbb F=\mathbb Q$".

[1] On the complexity of cubic forms

share|cite|improve this question
up vote 2 down vote accepted

The process is homogenization: we switch to integers by introducing a new variable.

Suppose that we are interested in the solvability in rationals $(x,y)$ of $$\alpha x^2+\beta y^2=\gamma,\tag{$1$}$$ where without loss of generality $\alpha$, $\beta$, and $\gamma$ are integers. Assume $\gamma\ne 0$. There is a rational solution of $(1)$ iff there are integers $u$, $v$, $w$, with $w\ne 0$ such that $\alpha \left(\frac{u}{w}\right)^2+\beta\left(\frac{v}{w}\right)^2=\gamma$, or equivalently $$\alpha u^2+\beta v^2=\gamma w^2.\tag{$2$}$$

Equation $(2)$ has an integer solution with $w\ne 0$ iff the equation $(\gamma\alpha)u^2+(\gamma\beta)v^2=\gamma^2w^2$ has such a solution.

Or equivalently, Equation $(2)$ has an integer solution with $w\ne 0$ iff the equation $(\gamma\alpha)u^2+(\gamma\beta)v^2=z^2$ has an integer solution with $z\ne 0$. (The $\gamma$ terms on the left force divisibility of $z$ by $\gamma$ since $\gamma$ has no square-part without loss of generality).

share|cite|improve this answer
@born: Thanks for fixing the typo. I should give up on $u$ and $v$, they look too much alike. – André Nicolas Jul 11 '12 at 18:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.