Re: 3^k + 2^k revisited

From: Gottfried Helms (helms_at_uni-kassel.de)
Date: 10/07/04 

Date: Thu, 07 Oct 2004 10:45:06 +0200

Am 07.10.04 09:09 schrieb William Elliot:

> From: Gottfried Helms <helms@uni-kassel.de>
> Newsgroups: sci.math
> Subject: Re: 3^k + 2^k revisited
>
> schrieb Ray Steiner:
> >>A while ago the problem of whether 3^k + 2^k could be a power for
> >>odd k was posed to this group. I would like to propose a special
> >>case: If k is odd, can 3^k + 2^k ever be a square?
>
> > let
> > 3^k + 2^k = s^2 (1)
>
> > then s^2 == 1 (mod 3)
> > 2^k == 1 (mod 3) if k is even.
> > 2^k == 2 (mod 3) if k is odd.
>
> > So for odd k (1) has no solution in integer k,s>0, k odd
> > (but what about even k?)
>
> 3^k + 2^k is never a square, thus never an even power.
>
> Case: k = 2j + 1
> 3^k + 2^k = 9^j 3 + 4^j 2 = 0 + 1^j 2 = 2 (mod 3)
> 0^2 = 0, 1^2 = 1, 2^2 = 1 (mod 3)
>
> Case: k = 2j
> 3^k + 2^k = 9^j + 4^j = (-1)^j + (-1)^j = +-2 (mod 5)
> 0^2 = 0, 1^2 = 1, 2^2 = -1, 3^2 = -1, 4^2 = 1 (mod 5)
>
> ----
>
>
Well, that's concise...

I stuck with the following

 since
      3^k -1 = 2^a *u // u odd
 and
   a k
 -------------------------------------
   3 2+4*i = 2,6,10,14,18,....
   4 4+8*i = 4,12,20,...
   5 8+16*i = 8,24,...
   ...
--------------------------------------
   3^k - 1 + 2^k = s^2 - 1

   2^a ( u + 2^b) = (s+1)(s-1) // b = k-a

  I first thought, "a" must be even, and so k must be divisible
  by 4 (plus some more restrictions) and the right hand must be

   2^a ( u + 2^b) = 2*p * 2^(a-1)*q

 ... but then had no way further (also "a" need not be even by
 these arguments).

Your example is *much* nicer ... :-)

Gottfried Helms