[seqfan] Re: Natural numbers of the form (2^m + 3)/(2^k - 9)
maxale at gmail.com
Tue Jun 1 06:46:35 CEST 2010
On Fri, May 28, 2010 at 12:29 PM, Hagen von Eitzen <math at von-eitzen.de> wrote:
> I'm contemplating natural numbers of the form (2^m + 3)/(2^k - 9) and
> would like to hear some opinions:
> I checked up to k=31 that there is no solution if k is odd. Can this be
> proved generally?
Assume that for some odd k, (2^k - 9) divides (2^m + 3).
It is easy to see that 2^k - 9 == 23 (mod 24) and every prime divisor
of 2^k - 9 is congruent to 1, 7, 17, or 23 modulo 24.
Therefore, there exists a prime divisor p == 17 or 23 (mod 24).
For such prime, we trivially have Legendre symbols:
(2/p) = 1 and (-3/p) = -1
making impossible the congruence:
2^m == -3 (mod p).
This contradiction proves that (2^k - 9) cannot divide (2^m + 3).
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