6
$\begingroup$

I came along the problem of finding three perfect cubes that are consecutive numbers of an arithmetic progression, i.e: $a^3-b^3=b^3-c^3$, where $a>b>c$ (to avoid trivial solutions). Clearly it is equivalent to solve $x^3+y^3=2$ over the rationals.

This is something I tried:

Consider the curve $x^3+y^3=2$ in te plane. I looked for a rational parametrisation of it. Ok, maybe looking for a rational parametrisation won't give all solutions (I'm not sure about that) but it's a start and we'll see what it leads us to.

I tried considering the intersection points with the line $y=tx+(1-t)$, since we now $(1,1)$ is such a point. (And I had seen a similar approach to find solutions to $x^2+y^2=1$.)

Solving the system for $x$ gave me (after factoring out the superfluous factor $x-1$) a quadratic in $x$ with discriminant $-3(t+1)^4+36t^2-4$. So it suffices to figure out when this is the square of a rational.

Let's write it as $-3(p+q)^4+36p^2q^2-4q^2=d^2$, where $p=qt$.

And here's where I can't continue. Any proceeding of this method or other methods are welcome.

P.S: This is not homework, it's just a question out of curiosity.

$\endgroup$
5
$\begingroup$

Euler proved that if $x^3+y^3=2z^3,x,y,z\in\mathbb Z$ then $x=\pm y.$

Hence if $x^3+y^3=2,x,y\in\mathbb Q$ then $x=y=1.$

$\endgroup$
  • $\begingroup$ Euler gave, $$(p+q)^3 + (p-q)^3 = (r+s)^3 + (r-s)^3$$ $p = 3(bc-ad)(c^2+3d^2)$, $q = (a^2+3b^2)^2 - (ac+3bd)(c^2+3d^2)$ $r = 3(bc-ad)(a^2+3b^2)$, $s = -(c^2+3d^2)^2 + (ac+3bd)(a^2+3b^2)$ $\endgroup$ – Ethan Sep 21 '13 at 7:54
  • $\begingroup$ Thanks. Is there any name for this theorem or do you know where I can find some more information about it? $\endgroup$ – punctured dusk Sep 21 '13 at 7:59
  • 2
    $\begingroup$ @barto You can see here (Chapter II, 14) matwbn-old.icm.edu.pl/kstresc.php?wyd=10&tom=42&jez=en $\endgroup$ – Next Sep 21 '13 at 8:03
4
$\begingroup$

The curve $x^3+y^3=2$ has genus one, so it doesn't admit a rational parametrization. You could waste a lot of time looking for one!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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