Re: x^3-x=6*y^3
- From: john_ramsden@xxxxxxxxxxxxxx
- Date: 30 Sep 2005 11:25:52 -0700
Ricardo Alfaro wrote:
>
> Has anyone reference to this Diophantine equation?
> or a proof that it has only the trivial solutions
> (0,0),(1,0),(-1,0)?
The guy who said this has genus 2 is wrong, because
for any given integer N in place of 6, a rational
solution can be derived from a rational solution
to any one of a finite number of equations of the
form c.x^3 + d.y^3 = 2^f.b where N = b.c.d.
In particular taking c = d = 1, the latter equation
often has an infinite number of non-trivial rational
solutions, which can be obtained by the Bachet form
of the Mordell-Weill chord-tangent process, applied
to x^3 + y^3 = N', i.e. using pairs of solutions
(or one solution doubled) to leap-frog recursively
to another.
For the stated equation, with N = 6, any rational
solutions can be derived from integer solutions
to one of the following equal the cube of an
integer:
X^3 + 6.Y^3
X^3 + 12.Y^3
2.X^3 + 3.Y^3
4.X^3 + 3.Y^3
The first two have no non-trivial solutions, by the
congruence conditions of Example 3.1 (on page 3) in
http://emis.library.cornell.edu/journals/JIS/VOL6/Broughan/broughan25.pdf
and I strongly suspect that neither of the other
two does either.
It's easy to prove this, by starting with the
homogenized version X.(X^2 - Y^2) = 6.Z^3 for
integers X, Y, Z (with reassigned X, Y we can
assume coprime), which implies the following
for integers a, b p, q, r, s with a.b = 6
and p, q squarefree:
X, X^2 - Y^2 = b.p.q^2.r^3, a.p^2.q.s^3
Plugging the first in the second we conclude
successively:
p | Y so that p = 1 (as also p | X)
q | s and hence q | Y so that q = 1
Thus b^2.r^6 - Y^2 = a.s^3, and as the factors
b.r^3 +/- Y have GCD 2^e where e = 0 or 1, this
implies there are integers c, d, u, v with
c.d = a (and hence b.c.d = 6) so that:
b.r^3 + Y, b.r^3 - Y = 2^e.c.u^3, 2^e.d.v^3
Finally, adding these gives the following, from
where the cases listed above were derived:
2^(1-e).b.r^3 = c.u^3 + d.v^3
(Note that if e = 0 then c.d must be odd.)
Cheers
John R Ramsden (jhnrmsdn@xxxxxxxxxxxx)
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