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Topic: number theory quickie
Replies: 16   Last Post: May 14, 2014 6:02 PM

 Messages: [ Previous | Next ]
 Paul Posts: 780 Registered: 7/12/10
Re: number theory quickie
Posted: May 13, 2014 2:10 PM

On Tuesday, May 13, 2014 11:21:03 AM UTC+1, quasi wrote:
> Here's an improved version ...
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> My last version was OK, but this one's simpler.
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> Proposition:
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> There are no rational numbers a,b,c such that a/b + b/c = c/a.
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> Proof:
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> Suppose a,b,c are rational numbers such that a/b + b/c = c/a.
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> Let r = a/b, s = b/c.
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> a/b + b/c = c/a
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> => r + s = 1/(rs)
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> => (rs)(r + s) = 1
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> Let d be the least common denominator for the rationals r,s.
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> Thus, write r = x/d, s = y/d where d is a positive integer,
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> x,y are nonzero integers, and gcd(x,y,d) = 1.
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> Then
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> (rs)*(r + s) = 1
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> => (xy)*(x + y) = d^3
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> Claim gcd(x,y) = 1.
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> If not, let p be a common prime factor of x,y.
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> Then (xy)*(x + y) = d^3 => p|d, contrary to gcd(x,y,d) = 1.
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> Hence gcd(x,y) = 1, as claimed.
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> Since gcd(x,y) = 1, it follows that gcd(x,x+y) = 1 and
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> gcd(y,x+y) = 1.
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> Since x,y,x+y are pairwise coprime and their product is the
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> cube of a nonzero integer, it follows that each of x,y,x+y
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> is the cube of a nonzero integer.
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> Writing x = u^3, y = v^3, x+y = w^3 yields
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> u^3 + v^3 = w^3
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> where u,v,w are nonzero integers, contrary to FLT for
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> exponent 3.
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quasi,

This looks incomplete (or wrong) to me. Positivity is not mentioned in the hypotheses, and I don't think FLT prevents an equality of the form x ^ 3 - y ^ 3 = z ^ 3.

You definitely seem to have proved it if all quantities are positive, though.

Paul

Date Subject Author
5/10/14 davidlpetry@gmail.com
5/10/14 Bart Goddard
5/11/14 quasi
5/12/14 quasi
5/13/14 Jeremy Ess
5/13/14 quasi
5/13/14 quasi
5/13/14 quasi
5/13/14 quasi
5/13/14 Paul
5/13/14 Bart Goddard
5/13/14 Paul
5/13/14 Virgil
5/12/14 Chris M. Thomasson
5/12/14 Bart Goddard
5/13/14 Chris M. Thomasson
5/14/14 Chris M. Thomasson