Re: 7-th power
- From: "Deep" <deepkdeb@xxxxxxxxx>
- Date: 29 Jun 2006 10:53:11 -0700
Thank you very much for your comments. Perhaps you have overlooked the
fact that none of a, b, m, n is a pefact square. Therefore it excludes
thev case a = 1. Also how did you come to the conclusion that numerator
and denominator are < 1000?
A liile clarifcication from you will be very helpful.
Greetings.
Robert Israel wrote:
In article <1151541469.122083.204670@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Deep <deepkdeb@xxxxxxxxx> wrote:
Consider the following two equations under the given conditions.
A = 5x -20x^3 + 16x^5 (1)
A = x (5 - 20 x^2 + 16 x^4)
B = 5y - 20y^3 + 16y^5 (2)
where x = (a)^1/2 and y = (b)^1/2, a, b are real rational and none of a
or b is a perfect square.
A = a^(1/2) (5 - 20 a + 16 a^2)
Statement:
If A = (M)^7 and B = (N)^7 where M = (m)^1/2 and N = (n)^1/2; m, n are
real, rational and are not perfect squares then m = n which in turn
implies x = y.
Any comment upon the correctness of the above statement will be
appreciated.
So you want a^(1/2) (5 - 20 a + 16 a^2) = m^(7/2).
(m/a)^(1/2) = (5 - 20 a + 16 a^2)/m^3 is rational. Call it r.
So m = r^2 a and
r^7 = 5/a^3 - 20/a^2 + 16/a.
One solution is a=1, r=1, m=1. I don't know if that is the only
rational solution. If it is, your statement is vacuously true.
Unless I've made a mistake, there are no other rational solutions
where the absolute value of the numerator and denominator of a are both
less than 1000.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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