Re: 7-th power
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 30 Jun 2006 02:13:19 GMT
In article <1151603590.953912.111410@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Deep <deepkdeb@xxxxxxxxx> wrote:
Please don't top-post.
Thank you very much for your comments. Perhaps you have overlooked the
fact that none of a, b, m, n is a pefact square.
I have not.
Therefore it excludes
thev case a = 1.
Yes. I wanted solutions of the equation r^7 = 5/a^3 - 20/a^2 + 16/a
where a and r are rational and a is not a square. But even without
that restriction, the only rational solution I could find was a = 1.
Also how did you come to the conclusion that numerator
and denominator are < 1000?
I didn't. I did a brute-force search for solutions with
numerator and denominator < 1000. Since I didn't find any, other
than the trivial a=1, you can conclude that a nontrivial solution
has numerator or denominator >= 1000.
A bit more can be said. Suppose p is a prime (other than 2 or 5)
such that the p-adic order of a is k <> 0 (i.e. a = p^k q where
the numerator and denominator of the rational number q are
coprime to p). Then the p-adic order of R = 5/a^3 - 20/a^2 + 16/a
is -k if k < 0 or -3k if k > 0.
Since the p-adic order of r^7 must be a multiple of 7, that implies
k is divisible by 7.
For p = 2, if the 2-adic order of a is k, then the 2-adic
order of R is -k+4 if k <= -2 and -3k if k > -2. This implies that the
2-adic order of a must be a negative number == 4 mod 7 or a nonnegative
number divisible by 7.
Similarly, the 5-adic order of a must be either a nonpositive number
divisible by 7 or a positive number == 5 mod 7.
I still don't know any solution other than a=1, and I suspect none exists.
Your statement is equivalent to saying there is at most one solution
where a is not a square.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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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