Fermat's Last Theorem approach?
in [Math]
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From: Gerry Myerson on 7 May 2010 01:39 In article <1029487796.80205.1273124763670.JavaMail.root(a)gallium.mathforum.org>, ThinkTank <ebiglari(a)gmail.com> wrote: > > In article > > <1595104331.80090.1273121129807.JavaMail.root(a)gallium. > > mathforum.org>, > > ThinkTank <ebiglari(a)gmail.com> wrote: > > > > > Suppose Fermat's Last Theorem could be proven for > > all n and a,b,c of the > > > form: > > > > > > sum(i=0,k) d_i*2^(m*i), for all k>=0, and one > > m=n^3, > > > > > > where d_i is an element of {0,1}, and q_n is an > > integer constant for each n. > > > > What? > > > > "q_n is an integer constant for each n" but there's > > no other mention > > of q_n. > > > > "one m = n^3" but there is only one m. Maybe m*i was > > supposed > > to be m_i? > > > > Sorry, editing error. I changed q_n to m to make it more > readable. Also I removed the restriction of one m. There > is one m per n, not just one m. That is, m=n^3 is the > correct statement. You can ignore any statement > regarding q_n. So you're talking about a, b, and c being of the form d_0 + d_1 2^(n^3) + d_2 2^(2 n^3) + ... + d_k 2^(k n^3) with each d_i being 0 or 1. That's a mighty sparse set of integers. Even for the case n = 3, we're talking about d_0 + d_1 2^(27) + d_2 2^(54) + ... + d_k 2^(27 k). The number of such integers less than 100,000,000 is...wait for it...2, and those 2 numbers are 0 and 1. You really reckon such a sparse set could tell you something about the full set of natural numbers? And have you looked, as I suggested, at the case n = 2? -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: ThinkTank on 9 May 2010 05:54
> > So you're talking about a, b, and c being of the form > > > d_0 + d_1 2^(n^3) + d_2 2^(2 n^3) + ... + d_k 2^(k > k n^3) > > with each d_i being 0 or 1. > > That's a mighty sparse set of integers. > Yes, true. > > You really reckon such a sparse set could tell you > something > about the full set of natural numbers? It just might be dense enough. Consider, if one can show that: (a_i f_i)^p + (b_i f_i)^p != (c_i f_i)^p for p>2, a_i b_i c_i f_i != 0, then FLT is proven. And those integers (a_i f_i, etc...) can be infinitely more sparse than the set I described. That is, for any set more spare than the set I described, one may find an even more sparse set which can still prove FLT by factorization of (f_i)^p. So, sparsity alone cannot be an argument against reduction. You would need to prove basically lack of prime density in the set, which I'm fairly certain you can't do. In fact, I'm fairly certain you could do the exact opposite, which would then be the equivalent of an extension to all integers. > > And have you looked, as I suggested, at the case n = > 2? > Yes, I have looked at n=2. It's usually the first thing I look at. I have been trying to solve FLT for nearly 15 years, and I've learned quite a bit in the process. I'm not one of those cranks that thinks proving FLT is an easy thing to do, but I can understand the frustration you must feel when there are so many cranks out there. I want you to know I highly respect professional mathematicians like yourself. I know that you probably know quite a good deal more than me. That said, I do hold a BSCS degree (which, as you probably know, is mathematics intensive) and, I have taught myself quite a bit of additional number theory and group theory (although, I have taken courses that have covered both of these). So, I am by no means a crank. The general approach of my proof should work for n=2 (that is, it gives the expected result consistent with FLT), and the set of numbers above. Namely, it is fairly easy to show that there exists a particular bijection between the integers and the set of integers I described that creates an approximate isomorphism (I don't know the technical term for an approximate isomorphism), under which the inequality relation (but not the equality relation) holds. In any case, I think at least for the time being I am going to take a break and think about it a little more. I will not be posting back until I am certain I have a complete proof. |