From: Gerry Myerson Subject: Re: abc and fermat Date: Fri, 11 Aug 2000 13:19:35 +1000 Newsgroups: sci.math Summary: How to prove Fermat's Last Theorem from the ABC conjecture? In article <8mqum5$9n$1@news.ccit.arizona.edu>, "Santiago Canez" wrote: > Could someone explain, in simplest terms, how Fermat's Last Theorem is > implied by the abc Conjecture? A form of the abc-conjecture is that if a, b, and c are positive integers and a + b = c then c < (N(abc))^2, where N(x) is the product of all the prime divisors of x (e.g., N(360) = 30). Let a = x^n, b = y^n, c = z^n, with n at least 6. Assume a + b = c, that is, assume you have a counterexample to Wiles' Theorem. Note that N(abc) = N(x^n y^n z^n) = N(xyz) doesn't exceed xyz, which doesn't exceed z^3. So c = z^n is at least (z^3)^2 which is at least (N(abc))^2, contradicting abc. Gerry Myerson (gerry@mpce.mq.edu.au) ============================================================================== From: kramsay@aol.commangled (Keith Ramsay) Subject: Re: abc and fermat Date: 13 Aug 2000 05:51:52 GMT Newsgroups: sci.math In article , Gerry Myerson writes: |A form of the abc-conjecture is that if a, b, and c are positive |integers and a + b = c then c < (N(abc))^2, where N(x) is the product |of all the prime divisors of x (e.g., N(360) = 30). Not if a=b=16 and c=32. (2+2=4 is also a counterexample for that matter). ;-) The way I've always seen the conjecture stated, is that for each epsilon>0 there exists a constant C such that if a+b=c *and a, b, and c are coprime*, then |a|, |b|, |c|< C(N(abc))^{1+epsilon}. Perhaps it's believed that when epsilon=1, we can take C=1? We don't need to worry about the coprimeness condition for the application to Fermat, because if there were a Fermat counterexample there would clearly be one with relatively prime summands. However, from the abc conjecture as I've stated it, one only gets "asymptotic Fermat", and this is the implication which I've usually seen claimed. Keith Ramsay ============================================================================== From: Gerry Myerson Subject: Re: abc and fermat Date: Mon, 14 Aug 2000 09:10:57 +1000 Newsgroups: sci.math In article <20000813015152.18821.00004080@nso-df.aol.com>, kramsay@aol.commangled (Keith Ramsay) wrote: => In article , => Gerry Myerson writes: => |A form of the abc-conjecture is that if a, b, and c are positive => |integers and a + b = c then c < (N(abc))^2, where N(x) is the => |product of all the prime divisors of x (e.g., N(360) = 30). => => Not if a=b=16 and c=32. (2+2=4 is also a counterexample for that => matter). ;-) => => The way I've always seen the conjecture stated, is that for each => epsilon>0 there exists a constant C such that if a+b=c *and a, b, => and c are coprime*, then |a|, |b|, |c|< C(N(abc))^{1+epsilon}. You are quite right. It appears my source & I were careless in omitting the coprimeness condition. => Perhaps it's believed that when epsilon=1, we can take C=1? Van der Poorten writes, Certainly, the exponent does seem always to be less than 2. The current record, and the corresponding exponent, is 3^10 x 109 + 2 = 23^5, requiring exponent 1.62991..., due to Reyssat. Gerry Myerson (gerry@mpce.mq.edu.au) ============================================================================== From: Gerry Myerson Subject: Re: abc and fermat Date: Fri, 11 Aug 2000 13:21:15 +1000 Newsgroups: sci.math In article <8mqum5$9n$1@news.ccit.arizona.edu>, "Santiago Canez" wrote: > Could someone explain, in simplest terms, how Fermat's Last Theorem is > implied by the abc Conjecture? I should have mentioned that the discussion in my previous answer was based on that in Alf van der Poorten, Notes on Fermat's Last Theorem, p. 144. Gerry Myerson (gerry@mpce.mq.edu.au)