Re: 3^k + 2^k revisited
From: Ray Steiner (steiner_at_math.bgsu.edu)
Date: 10/15/04
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Date: 15 Oct 2004 10:08:47 -0700
Gottfried Helms <helms@uni-kassel.de> wrote in message news:<ckls0p$v55$04$1@news.t-online.com>...
> Am 14.10.04 12:10 schrieb Gottfried Helms:
> >>
> >>0 < y log(x) - k log(3) < (2/3)^k
> >
> > (*)
> > Is there a reason, why rhs is not logarithmed?
> >
> Well stupid question. Got it myself after writing things
> down with pencil&paper...
>
> Gottfried Helms
Hello again!
First, I am solving this problem for odd k only.
Next, using
k/ (log(k))^2 < 66.09 * log (x)
from my previous post, we can resolve one simple case.
Suppose x <=k.
Then we get
k/ (log(k))^3 < 66.09,
which yields k < 246500.
Thus
5^(y-1)<246500,
y-1 < 8
y < 9.
But we know from previous posts that 3^k + 2^k is not a square, cube, 5th power
or 7th power.
So the equation has no solution for odd k and x <= k.
But I'm completely clueless on how to resolve the case x > k.
Does anyone have any ideas?
Regards,
Ray Steiner
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