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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.

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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 $x=y=1.$

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  • $\begingroup$ 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? $\endgroup$ – Bart Michels Sep 21 '13 at 7:59
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    $\begingroup$ @barto You can see here (Chapter II, 14) matwbn-old.icm.edu.pl/kstresc.php?wyd=10&tom=42&jez=en $\endgroup$ – lsr314 Sep 21 '13 at 8:03
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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!

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