Search All of the Math Forum:

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

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

Topic: rewriting my earlier proof of FLT: please critique
Replies: 10   Last Post: Jan 19, 2014 4:19 PM

 Messages: [ Previous | Next ]
 Brian Q. Hutchings Posts: 6,427 Registered: 12/6/04
Re: rewriting my earlier proof of FLT: please critique
Posted: Jan 18, 2014 3:37 PM

how, so, p | c+d ??

> B. c^n + d^n > p^n .... (13)
>
> so p divides c+d, let us write c + d = Mp.
>
> adding (5) and (6) gives us, a + b = p^n (j+k) + Mp ...... (14)
>
> Implies p divides a+b, which violates our assumption (5)
>
> This logic will work even if we assume a = jp^n +/- c and b = kp^n +- d
>
>
>
> C. c^n + d^n = 0 possible when c and d are both zero (violates (9))
>
> or c = -d and adding (7) and (8) will give us a+b = p^n (j+k)
>
> and so p divides a+b, which violates our assumption (5)
>
>
>
> D. Let us assume p divides(both a+b) and Q.
>
> We know that the only number that divides a+b and Q is n.
>
>
>
> C. Q = n (so n^(n-1) divides a+b)
>
> Q = a^(n-1) b + a^(n-2)b^2 + .... + b^(n-1)
>
> Q = n possible if both a and b are equal to 1.
>
>
>
> D. Q = n^(n-1) [and n divides a+b]
>
> then n divides a or b and hence n divides d in (2), violates (1)
>
>
>
> C. Q has a factor q (not equal to n) that is not a factor of (a+b).
>
> This is not possible as shown in (12), (13) and (14)
>
>
>
> I think this rounds out the proof.

Date Subject Author
1/18/14 sharma.kunapalli@gmail.com
1/18/14 Brian Q. Hutchings
1/18/14 quasi
1/18/14 sharma.kunapalli@gmail.com
1/19/14 quasi
1/19/14 sharma.kunapalli@gmail.com
1/19/14 quasi
1/19/14 Port563
1/19/14 Port563
1/19/14 Port563
1/18/14 Port563