
Re: Fermat's last theorem
Posted:
Oct 20, 2017 3:23 PM


On Friday, October 20, 2017 at 7:50:06 AM UTC5, 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^(n1)E^(n2)*F +... E*F^(n2)+F^(n1)) > > 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^(knk)÷2 < S < P^(knk) > > These inequalities mean that (E+F) can be a multiple of P^(k1) 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^(nkk1) at most. > > This means however that the product of those two parts can only be a pruduct of P^(nk2) 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

