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Charles Matthews's user avatar
Charles Matthews
  • Member for 14 years, 7 months
  • Last seen more than 9 years ago
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Solution of a special class of Diophantine Equations
So we have x, f(x), g(x), h(x) ... where f, g, h are certain functions. What now? Which equations are concerned? I have a feeling you want something about implicit algebraic functions.
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Solution of a special class of Diophantine Equations
The question really isn't well posed. The theory of diophantine equations for all curves of genus 0 (i.e. all "space curves" that are rational curves from a geometric point of view) was written down by Hilbert and Hurwitz. It is in Mordell's books. If that is not essentially the question you are asking, I'm not really understanding the question.
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Proof theory and primitive roots
Contradiction, yes. If I'm not simplifying too much, the situation comes down to three subgroups of residues mod p, of index q, r and 1 (under multiplication). The three numbers are constrained each to generate such subgroup. If you rule out index 1, then by a pigeonhole argument two generate the same subgroup, which will be contradictory.
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Research trends in geometry of numbers?
WADR to Minkowski, the field should have been renamed "geometric number theory" in the 1950s. The part of geometric number theory that should be called "lattice theory" is not called that, because of overloading. Diophantine approximation is another rather questionable name of subdiscipline, but at least is named after a cluster of problems, rather than techniques or objects of study. The answers below suggest that some shifts of perspective are overdue.
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The Riemann Hypothesis and the Langlands program
@Chandan Singh Dalawat: I am sometimes a trivial character.
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The Riemann Hypothesis and the Langlands program
The Langlands program has always and explicitly had a lot to do with Artin L-functions, and in particular with the Artin conjecture on their meromorphy. So in some sense it is a question of how adjacent the Artin conjecture is to GRH. The obstruction to using Brauer's theorem to prove the Artin conjecture is partly to do with not being able to take rational powers of meromorphic functions at will, absent control of zeroes (and poles). The quote from the book review may have overstated the point, to some extent.
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