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Topic: Fermat's last theorem
Replies: 16   Last Post: Nov 11, 2017 7:33 AM

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magidin@math.berkeley.edu

Posts: 11,740
Registered: 12/4/04
Re: Fermat's last theorem
Posted: Oct 20, 2017 3:23 PM
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On Friday, October 20, 2017 at 7:50:06 AM UTC-5, Jarno grenson wrote:
> Hey guys, i'm new here and i would like to get some input about a possible proof of fermat's last theorem. I do think i didnt make any mistakes, but i would like to see wether i actually did. Here is it:
>
> We have
> A^n+B^n = C^n --(1)
>
> We state that
> B<A<C<A+B, --(2)
> gcd(a,b) = gcd(a,c) = gcd(b,c) = 1
> All numbers are part of N
>
> And we assume that
> n is odd and bigger than 2
> C = (P^k)*D with P prime
>
> We rewrite (1) as
>
> (E)^n + (F)^n = P^(kn)
>
>  Where E and F are A/D and B/D respectively


Note that E and F are rationals, not integers, unless D = 1, because we are assuming that gcd(a,c) = gcd(b,c) = 1 .

> We can rewrite the left side as
>
> (E+F)* (E^(n-1)-E^(n-2)*F +... -E*F^(n-2)+F^(n-1))
>
> We know from (2) that
> P^k<E+F< 2*P^k
>
> This also means however that the other part, which we replace as S, holds true to these conditions:
> P^(kn-k)÷2 < S < P^(kn-k)
>
> These inequalities mean that (E+F) can be a multiple of P^(k-1) at most,


E+F is a rational number; "multiples" in the rational numbers are... useless. Every nonzero rational number is a multiple of every nonzero rational number.



while S can be a multiple of P^(nk-k-1) at most.
>
> This means however that the product of those two parts can only be a pruduct of P^(nk-2) at most, since P is prime. We see a clear contradiction,


Unfortunately, you seem to be talking about rational numbers, not integers. Talking about divisibility among rational numbers does not give contradictions.

--
Arturo Magidin



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