Re: x^n + y^n = k (mod m)
- From: Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Sun, 04 Nov 2007 21:52:41 GMT
In article <j1eri39vlnds5itga57pvsclrf0eld0j3c@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
Does there exist an integer n > 2, and an integer k, such that
(1) x^n + y^n = k has no integer solutions.
(2) For all integers m > 1, the congruence x^n + y^n = k (mod m) has
integer solutions.
Here are two possible routes to finding such a thing.
1. There is no solution in integers to x^3 + y^3 = 6,
but there is a solution in rationals,
so x^3 + y^3 = 6 (mod m) has a solution for every m
avoiding prime divisors of the denominators. Maybe it
has solutions for those primes, too.
2. There are only finitely many primes p for which
there exists k such that x^3 + y^3 = k (mod p)
has no solution. Use Chinese Remainder Theorem
to find k such that x^3 + y^3 = k (mod p) has solution
for all those k; then x^3 + y^3 = k (mod m) ought to have
a solution for all m. Just make sure you choose k so there
are no solutions in Z.
It occurs to me we're asking whether a local-global principle
applies, and that ought to be known. Just one step more
complicated, it's known there are equations of the form
a x^3 + b y^3 + c z^3 = 0
that have solutions (mod m) for all m, but no solutions in
the rationals. Selmer was the first to give examples, I think
his example was 3 x^3 + 4 y^3 + 5 z^3 = 0.
--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.
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