0
$\begingroup$

Would it be possible to use the circle method to prove that there are no solutions to certain diophantine equations. For example, could one use the circle method to prove the fact that there are no solutions in the integers to $x^2 = y^3 + 7$? (This fact on its own is quite easy to prove)

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I'm guessing you've missed the point of the Hardy-Littelwood method. Say you want to count the number of rational points on some alg. variety (up to say - given height). You would take a proper delta-function combination on say the intersection of a lattice with your variety, and then try to expand the expression spectrally. Given some "equidistribution" motivation, and some exponential sums techniques, you will (usually) be able to bound this sum from above by the "correct" asymptotics. To bound it from bellow and show that the asymptotics agree, is much more subtle problem. $\endgroup$
    – Asaf
    Apr 28, 2014 at 10:16
  • 1
    $\begingroup$ In any case, in order for those "techniques" to work, some algebro-geometrical constraints have to be satisfied (even for the upper bound) whence they are not met in your situation. Even more, your question basically amount to - show that the counting function is zero (exactly). This is hardly doable analytically and won't gain anything from spectral analysis. P.S. for the question in the post, this is a very basic exercises in algebraic number theory (i.e. no analysis behind some easy solution of unit equations or so). $\endgroup$
    – Asaf
    Apr 28, 2014 at 10:19

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.