May 16th, 2017, 12:34 AM  #21 
Banned Camp Joined: Dec 2012 Posts: 1,028 Thanks: 24 
Is more far as you think since the most simple proff comes just looking to the numbers of what I've posted above so to the inequality: So we prove we can give a solution to the inequality: \[ Y_P' > Y'_{MP} > Y'_{BP} \] so to: \[ Y_P' > Y'_{MP} \] so to: \[ [2n(X_f)^{(1/n)}]^{(n1)} > [(2nA^{(n1)}+((nB^{(n1)}nA^{(n1)}))] \] or: \[ 2n[(A^n+B^n)^{(1/n)}]^{(n1)} > [(2nA^{(n1)}+((nB^{(n1)}nA^{(n1)}))] \] That luckily for me is well sorted, so onve we prove (by comutation) the case $n=3$ with $A=5$, $B=6$ any other Bigger combination will return RISING DIFFERENCE, so we can be sure there are no zeros somewhere in the middle, just using the comparison criterion, and knowing that, viceversa going infimus in the other direction we always get a solution when $A=B$. Infact if we try to say somethink on $X$, we can't say somethink since the value of $X_f$ can be Bigger or Lower $X_MP$ so we have in case to prove: \[ X_f \neq X_{BP} \] so to: \[((B^n+A^n)/2]^{(1/n)} \neq [(A^{(n1)}+((B^{(n1)}A^{(n1)})/2)]^{(1/{(n1)}} \] That again fells in knowing FLT solution by Wiles proof... I'm still working in a good definition of all the points... will be a long work... but now the main "subject" of the proof comes out... I know that from a long time by numbers, but to prove in "math" was a veery long way... and I admit just question of luckyness to find a well sorted Ordinate... The proof involves Ordinal Numbers can holds too, but will be very hard for all is not aware of that concepts. Last edited by complicatemodulus; May 16th, 2017 at 12:50 AM. 
May 17th, 2017, 09:40 AM  #22 
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20  Yes. When n=2 assumption that (c+a) and (ca) have common divisor k means it divides each term c, a, and b and must be reduced. When n>2 the only possible common divisor of (ca) and [c^(n1)+c^(n2)b+... +b^(n1)] is n. Division of latter by former gives remainder na^(n1). So they cannot have common divisor k. 
May 18th, 2017, 03:55 AM  #23  
Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory  Quote:
If a=3, ca=2 c+a=8. In this case, k=2, x1^2=1, x2^2=4 If a=4, ca=1 c+a=9. so k=1, x1^2=1, x2^2=9  
May 18th, 2017, 04:54 AM  #24 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra 
It is incorrect to suggest that $b^n$ must be a product of integers $(kx_1^n)(kx_2^n)(kx_3^n)\cdots (kx_n^n)$. Or rather, there are other ways to form $b^n$. For example $$b^3=(x_1^3)(kx_2^3)(k^2x_3^3)$$

May 18th, 2017, 04:57 AM  #25 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 Math Focus: Mainly analysis and algebra 
The suggestion that Fermat's idea for the general case was the same as his proof for $n=4$ doesn't follow. If his general case had worked, he'd never have needed a proof for $n=4$.

May 18th, 2017, 08:59 PM  #26  
Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory  Quote:
same as $k^3x_1^3x_2^3x_3^3$ for the two cases. Also still in both cases, $b^3$ is a product of integers that can be expressed in three groups. Hence we have to show $c^3a^3$ is not a product of any 3 combinations/operations of $c$ and $a$ such that the combinations are integers. Though the proof has to be corrected and that point noted down. Last edited by Mariga; May 18th, 2017 at 09:19 PM.  
May 18th, 2017, 09:14 PM  #27 
Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory  He might have followed a procedure related to the case of $n=4$. I guess it would have been lengthy. We just wonder why he never wrote it down yet the only limiting factor would have been time.. not the margin. Anyway he will forever remain to be the only person who knew whether he really had the proof. He could have his own reasons for not sharing.

May 18th, 2017, 09:22 PM  #28  
Senior Member Joined: Aug 2012 Posts: 1,998 Thanks: 569  Quote:
There's a false proof that would work if only some ring or other were a unique factorization domain but it isn't. Some people think Fermat THOUGHT he had a proof but actually didn't. On the other hand he never mentioned the matter again over the next 30 years. It really is unlikely he had a proof and he probably did know he didn't have it.  
May 19th, 2017, 04:03 AM  #29  
Senior Member Joined: May 2016 From: USA Posts: 1,116 Thanks: 462  Quote:
 
May 25th, 2017, 06:36 AM  #30 
Banned Camp Joined: Mar 2017 From: . Posts: 338 Thanks: 8 Math Focus: Number theory 
Take a look at this


Tags 
fermat, proof, simplest, theorem 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Fermat was correct... he had a truly marvelous simple proof of his last theorem.  RanberSingh  Number Theory  2  April 27th, 2017 08:06 AM 
Simple Proof of Fermat's Last Theorem and Beal's Conjecture  MrAwojobi  Number Theory  1  September 29th, 2015 07:22 AM 
Simple Fermat Proof a^n+b^n=c^n  M_B_S  Number Theory  15  July 29th, 2015 05:14 AM 
Analytical simple proof of Fermat's last theorem ????  RanberSingh  New Users  3  August 10th, 2014 01:44 PM 
Simple Proof of Fermat's Last Theorem and Beal's Conjecture  MrAwojobi  Number Theory  21  January 8th, 2011 09:11 AM 