I'm trying to research elliptic curves, and I always get the generic equation $$y^2 = a_0 x^3 + a_1 x^2 + a_2 x + a_3.$$ However, I'm looking for information on an equation like $$y^3 = a_0 x^3 + a_1 x^2 + a_2 x + a_3$$ or an equation with cubes on both sides. I can't seem to find anything... are they called something else? Are there any papers I could read on them? Thanks!
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Why are you interested in curves of that form? – Tobias Kildetoft May 12 '15 at 18:13
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@TobiasKildetoft Personal research. I'm looking into when the sum of consecutive cubes equals another cube (i.e. $11^3+12^3+13^3+14^3 = 20^3$) and this can be written as $ ax^3 + bx^2 + cx + d = y^3$ for certain coefficients. – Derek Orr May 12 '15 at 18:16
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Isn't the sum of $n$ consecutive cubes a polynomial of degree $4$ in $n$? – Tobias Kildetoft May 12 '15 at 18:20
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Yes. But when you look at $\sum_{i=k}^{k+n-1} i^3 = \sum_{i=1}^{k+n-1} i^3 - \sum_{i=1}^{k} i^3 $, the fourth power cancels. For the above example, $k=11, n=4$. – Derek Orr May 12 '15 at 18:22
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Ahh, of course. Neat. – Tobias Kildetoft May 12 '15 at 18:25
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@DerekOrr: The special case $x^3+(x+y)^3+(x+2y)^3 = z^3$ is discussed in this post. – Tito Piezas III May 13 '15 at 06:23
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Have a look at Silverman and Tate's "Rational Points on Elliptic Curves". There, in page 22, they tell you how to transform any non-singular cubic into a Weierstrass form. The reason why you don't see much work on curves of the form $y^3=x^3+\cdots$ is that we first bring it to a Weierstrass form and then work there.
Álvaro Lozano-Robledo
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Yes, I believe I know of it. However, if I remember correctly, it uses a substitution that introduces rationals. I wasn't sure if there was a way to stick with integers or not. – Derek Orr May 13 '15 at 17:52
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Ah, I see. So if you want to study integral points, then you need to find all the S-integral points on the Weierstrass equation (i.e., rationals with denominators, but only allowing certain primes in the denominators), and then bring those back to the original equation. The integral points will be among those. – Álvaro Lozano-Robledo May 13 '15 at 17:55
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So, I should stick with Silverman's transformation and look for rational points? Then try to find integral points from these rationals? For example, if one transformation had "Y = 1/(x+1)" and a rational point on the curve was at Y = 4/5 (for example), I wouldn't use that (because x = 1/4 and I'm looking for integers only) but if another rational point on the curve was at Y = 1/7, I can say x = 6. Is that right? – Derek Orr May 13 '15 at 18:02
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Such equations are called Cubic plane curves; references are given here. The projective version is given by $F(x,y,z)=0$ where $F$ is a non-zero linear combination of the third-degree monomials $$ x^3, y^3, z^3, x^2y, x^2z, y^2x, y^2z, z^2x, z^2y, xyz. $$ For $z=1$ we obtain the affine version. Any non-singular cubic curve can be transformed into the Weierstrass equation of an elliptic curve.
Dietrich Burde
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Okay, I see, thanks. They don't seem to have much information on here though... It looks like they just have many abstract examples to me. Do you know of other places that might be more basic or involved? – Derek Orr May 12 '15 at 18:28
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There are effective, albeit large, upper bounds for the integer solutions to any non-singular cubic $$ ax^3+by^3+c+dx^2y+exy^2+fx^2+gx+hy^2+iy+jxy=0. $$ Such bounds are proven using linear forms in logarithms. I don't know the current best results, but the original result is in
[1] A. Baker and J. Coates, Integer points on curves of genus 1, Proc. Cambridge Philos. Soc. 67 (1970), 595-602.
Let $H$ be an upper bound for the coefficients of the cubic. They prove that any integer solution $(x,y)$ satisfies $$ \max\{|x|,|y|\} \le \exp\exp\exp\left((2H)^{10^{3^{10}}}\right). $$
On the other hand, there are much better bounds known for equations of the form $$ y^m = f(x), $$ which are sometimes call superelliptic equations (a term due to Lang, I believe). Anyway, here's one reference to get you started on practical methods for solving superelliptic euqations:
[2] Yuri F. Bilu, Guillaume Hanrot, Solving superelliptic Diophantine equations by Baker's method, Compositio Mathematica 112 (1998), 273-312.
Joe Silverman
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