0
$\begingroup$

What is the parametric form of the rational solutions of the equation $y^2 = x^3 - z^3 ?$

Improve this question
$\endgroup$
2
  • $\begingroup$ math.stackexchange.com/questions/2414088/… $\endgroup$
    – individ
    Commented Apr 8, 2021 at 7:18
  • $\begingroup$ Hello @PRIMES is in P., and welcome to MO. Please have a look at mathoverflow.net/help/on-topic . Mathoverflow is a website for questions about research level mathematics. As pointed out in individ 's comment, this question as already been asked (and answered) on MSE. $\endgroup$
    – DamienC
    Commented Apr 8, 2021 at 20:23

1 Answer 1

Reset to default
11
$\begingroup$

This may be (probably is) too elementary for this site, it would be better on MathStackExchange. But the basic idea is to make a change of variables to simplify. (In fancy algebraic geometry terms, it's blowing up a singularity.) In this case, let $$ x = y u \quad\text{and}\quad z = y v. $$ Then your equation becomes $$ y^{-1} = u^3-v^3, $$ so you get the parametrization $$ x = \frac{u}{u^3-v^3},\quad y=\frac{1}{u^3-v^3},\quad z=\frac{v}{u^3-v^3}. $$ This gives a rational solution for all $u,v\in\mathbb Q$ with $u\ne v$. And conversely, (almost) every rational solution gives a unique $u,v$ value.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .