Re: x^5+y^5=31

From: JoeS (jhs_at_math.brown.edu)
Date: 03/18/05 

Date: 17 Mar 2005 17:55:44 -0800

Sorry, David, that's not quite right about integral points on elliptic
curves. Siegel's original proofs (he gave two rather different ones)
for finiteness of integral points on elliptic curves were indeed not
effective. However, Baker's result on linear forms in logarithms,
although far weaker than the Thue-Siegel-Roth theorem on Diophantine
approximation, is enough to make the finiteness theorem effective. So
there are explicit formulas, in terms of the coefficients of the
equation, for the largest integer point on an elliptic curve. And I
would guess it was that application that played a significant role in
Baker receiving a Fields' medal. Of course, the bounds are huge,
conjecturally more than exponentially worse than they should be.

As various posters have indicated, Faltings' theorem (Mordell
conjecture) is not effective, neither Faltings original proof nor
Vojta's later proof. However, Vojta's proof does allow one to give an
effective upper bound for the number of solutions.

Finally, Elkies pointed out that if one had an effective version of the
ABC conjecture, then using Belyi's theorem, one could turn it into an
effective version of Faltings' theorem. Of course, it's also possible
someone will come up with an ineffective proof of the ABC conjecture,
which would give a new ineffective proof of Faltings' theorem.

So that's the current status of the general problem. To get back the
the OP's problem, someone with more computational ability than me
(meaning one of the true experts, e.g. Noam Elkies, Bjorn Poonen,
Victor Flynn, Ed Schaefer, etc) might be able to handle it. Ignoring
modular methods apres Wiles, the standard procedure is first to
determine the rank of the Jacobian, then use Chaubety-Coleman methods
to try to pin down which points in J(Q) actually lie in C(Q). In this
case, the rank of J(Q) is at least one (presumably), since the divisor
(1,-1,0)-(2,-1,1) probably has infinite order in J(Q).

One other thought. Sometimes one can map a curve to one of lower genus
and work there. That can be done with the standard Fermat curves
X^p+Y^p=Z^p.

Joe Silverman

Relevant Pages

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    ... For example, Faltings' result ... >such a bound existing for cubic curves, ... one can determine an upper bound on the _number_ of integral points, ... be and so a simple exhaustive search can never conclusively show that ...
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