-3
$\begingroup$

What are the rational points on the elliptic curve $y^2 = x^3 - t^{2}z^3$ ? I seem not to find any besides the trivial ones whereby $txyz=0$ or $x= \pm z$.

ADDENDUM 1. I have just noticed that if $z^3 = x$ then there do exist some non-trivial rational point(s) provided that $t$ is a congruent number. Therefore, to avoid this and related scenarios, we impose the condition that the numerators of the reduced forms of $x \neq \pm 1$ and $z \neq \pm 1$ are relatively prime.

ADDENDUM 2. It is usually interesting to generalise questions. So i would also ask for rational points on the general surface $y^2 = x^{3} -t^{2n}z^3$, $n$ being some positive integer.

Improve this question
$\endgroup$
7
  • 2
    $\begingroup$ Over $\mathbb Q(t)$? $\endgroup$
    – Will Sawin
    Commented Aug 12, 2020 at 2:31
  • $\begingroup$ But then '$t$ is a congruent number' does not make sense to me ― it is never evaluated as a number, but always functions as a formal variable... Or are you asking for $\mathbf Q$-rational points on this elliptic surface? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Aug 12, 2020 at 2:50
  • $\begingroup$ First of all, you probably mean $y^2z$ rather than $y^2$. Secondly, from the question it appears that $t$ is allowed to be algebraic over the rationals. In that case, there are many choices of $t$ which will yield rational points. $\endgroup$
    – Kapil
    Commented Aug 12, 2020 at 2:53
  • 2
    $\begingroup$ (1) Please edit your question to indicate that you are looking for 4-tuples of rational numbers satisfying your equation. (2) Which "elliptic curves" are you talking about. If you plug in a value for $t$, you get a surface, not an elliptic curve. It's a rational surface, as one of the answers indicates. But probably you meant to write $y^2z=x^3-t^2z^3$, and then for $t\ne0$ you get an elliptic curve sitting in $\mathbb P^2$. These curves are related to the congruent number problem, so you'll find lots of information if you search on that term. $\endgroup$
    – Joe Silverman
    Commented Aug 16, 2020 at 15:35
  • 1
    $\begingroup$ 12 versions within five hours of first posting. $\endgroup$
    – Gerry Myerson
    Commented Aug 16, 2020 at 22:23

2 Answers 2

Reset to default
1
$\begingroup$

If we rewrite the equation as $y^2+t^{2}z^3 = x^3$, and let $v=tz$, then $y^2+zv^2=x^3$.

If you factorize over $Q[\sqrt{-z}]$, then $(y+v\sqrt{-z})(y-v\sqrt{-z})=x^3 $.

Let $x=(a+b\sqrt{-z})(a-b\sqrt{-z})$, then $y+v\sqrt{-z}=(a+b\sqrt{-z})^3=(a^3-3ab^2z)+(3a^2b-b^3z)\sqrt{-z}$

Thus $y=a^3-3ab^2z, v=3a^2b-b^3z, x=a^2+b^2z$, and $t=\frac{v}{z}=\frac{3a^2b-b^3z}{z}$

We then have the general parametric solution: $(a^3-3ab^2z)^2=(a^2+b^2z)^3-(\frac{3a^2b-b^3z}{z})^2z^3$

Edit: Of course, if you take $t=\frac{1}{x^3-y^2}, z=x^3-y^2$, you get trivial solutions, but I'm assuming you don't want those.

Really, your equation has too many variables, so writing boring solutions is easy. Are you wanting solutions that are functions of t? If so, you can rearrange my original solution above to make z a function of t, and go from there. From there, your other equations (in Addendum 2) follow by changing $t$ to $t^n$.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ It's not (just) about the number of variables, but rather that the equation defines a rational variety (which is in part caused by the fact that it is a degree $3$ equation in $4$ variables). After the change of variables in the first line, can't you finish by writing $z = \frac{x^3-y^2}{v^2}$ as a function of the other variables? (so $t = \frac{v^3}{ x^3-y^2}$ completes the parametric solution.) $\endgroup$
    – Will Sawin
    Commented Aug 16, 2020 at 15:23
  • $\begingroup$ True. I left that up to the OP which way (s)he wanted to organize the solutions $\endgroup$
    – Thomas
    Commented Aug 16, 2020 at 22:08
0
$\begingroup$

Above equation shown below:

$y^2=x^3-t^2z^3$ ----(1)

Equation $(1)$ has parametric solution given below:

$x=m^4-3m^2+3$

$y=m(m^4-3m^2+3)^2$

$z=(m^2-1)(3m^2-m^4-3)$

$t=1$

For, $m=2$, we get:

$(x,y,z,t)=(98,7,-21,1)$

Improve this answer
$\endgroup$
2
  • $\begingroup$ I think you have mixed up $x$ and $y$ in the last line; I also suspect that the method can be generalized rather easily to a 3-parameter family: choose $t, x_0, z_0$; then let $(x, y, z, t) = (x_0 (x_0^3 - t z_0^3), (x_0^3 - t z_0^3)^2, z_0 (x_0^3 - t z_0^3), t)$. $\endgroup$
    – user44191
    Commented Aug 16, 2020 at 19:25
  • $\begingroup$ @user44191. Thanks for catching my typo. You have a typo too. In your value's for (x,y,z) the variable 't ' need's to have the power two. There is a difference in your method & my method. I took the sum of two cubes [ (m^2-1)^3+1^3] . Hence we get it= m^2(m^4-3m^2+3). Thus we just need to multiply the LHS by (m^4-3m^2+3) to make the LHS a square. $\endgroup$
    – Sam
    Commented Aug 17, 2020 at 14:48

You must log in to answer this question.

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