Re: Another silly question
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 7 Jun 2006 17:29:34 GMT
In article <1149671980.084231.49940@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Ross Clement (Email address invalid - do not use) <clemenr@xxxxxxxxxx> wrote:
I tried to prove the general case last night and didn't get there.
Looking at Robert's proof, the step that really stopped me was x^n =
k^(n+2) => x = g^(n+2). I got as far as the left hand side but not to
the right.
As I said, this follows when n is odd.
First we need to say that if n is odd, then n and (n+2) have no common
factors.
Hint: any common factor of A and B is also a factor of B-A.
for x^n = k^(n+2) => x = g^(n+2), we have:
More generally, if gcd(n,m) = 1 and x^n is an m'th power, then x
is an m'th power.
If x = product_p p^(m_p), the product being over distinct primes,
x^n = product_p p^(n m_p). This is an m'th power if and only if
all n m_p are divisible by m. But since gcd(n, m) = 1,
that implies m_p are divisible by m, and x = g^m where
g = product_p p^(m_p/m).
We should also consider the possibility that x = - product_p p^(m_p).
If m is even, n must be odd, but then x^n is not an m'th power.
If m is odd, x is an m'th power iff -x is an m'th power.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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