Paul
Posts:
763
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 ... > > > > My last version was OK, but this one's simpler. > > > > Proposition: > > > > There are no rational numbers a,b,c such that a/b + b/c = c/a. > > > > Proof: > > > > Suppose a,b,c are rational numbers such that a/b + b/c = c/a. > > > > Let r = a/b, s = b/c. > > > > a/b + b/c = c/a > > > > => r + s = 1/(rs) > > > > => (rs)(r + s) = 1 > > > > Let d be the least common denominator for the rationals r,s. > > > > Thus, write r = x/d, s = y/d where d is a positive integer, > > x,y are nonzero integers, and gcd(x,y,d) = 1. > > > > Then > > > > (rs)*(r + s) = 1 > > > > => (xy)*(x + y) = d^3 > > > > Claim gcd(x,y) = 1. > > > > If not, let p be a common prime factor of x,y. > > > > Then (xy)*(x + y) = d^3 => pd, contrary to gcd(x,y,d) = 1. > > > > Hence gcd(x,y) = 1, as claimed. > > > > Since gcd(x,y) = 1, it follows that gcd(x,x+y) = 1 and > > gcd(y,x+y) = 1. > > > > Since x,y,x+y are pairwise coprime and their product is the > > cube of a nonzero integer, it follows that each of x,y,x+y > > is the cube of a nonzero integer. > > > > Writing x = u^3, y = v^3, x+y = w^3 yields > > > > u^3 + v^3 = w^3 > > > > where u,v,w are nonzero integers, contrary to FLT for > > exponent 3. > >
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

