The Math Forum

Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]

Posts: 11,749
Registered: 12/4/04
Re: Fermat's last theorem
Posted: Oct 20, 2017 3:23 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.