May 31st, 2013, 09:14 PM  #11  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A new approach to Fermat's last theorem Quote:
 
May 31st, 2013, 09:48 PM  #12  
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  Re: A new approach to Fermat's last theorem Quote:
You're right!Thousands of excuses! Adjustment:If and the natural numbers then necessarily are the sides of a triangle.In fact even for my assertion is valid and bviously . Thank you very much!  
June 1st, 2013, 02:54 AM  #13 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: A new approach to Fermat's last theorem
I'm not sure what the '*' is doing in 'N*'. Surely you don't mean that any three natural numbers form the sides of a triangle. You can't, for instance, adjust angles to make a triangle with sides 1, 2 and 10,000, just to take one screamingly obvious example. So what numbers are you talking about?

June 1st, 2013, 02:58 AM  #14 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: A new approach to Fermat's last theorem
Clearly, if x, y and z are integers, or ANY numbers, where x?y?z and xyz forms a triangle, then for sure z<x+y

June 1st, 2013, 03:33 AM  #15 
Senior Member Joined: May 2013 Posts: 118 Thanks: 10  Re: A new approach to Fermat's last theorem
What about FLT?

June 1st, 2013, 03:45 AM  #16  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A new approach to Fermat's last theorem Quote:
 
June 4th, 2013, 10:11 AM  #17  
Senior Member Joined: May 2013 Posts: 118 Thanks: 10  Re: A new approach to Fermat's last theorem Quote:
Comparing ( to the already known relation (9) we observe that in ( a has replaced c and ab has replaced a.By (9) we had cocnluded (see step 2,subcases ?a and ?b): and therefore,in a totally analogous way we can conclude now: and thus: r\:Nb \rightarrow \:a=0\r\:b=0" /> absurdity  
June 6th, 2013, 08:46 PM  #18 
Newbie Joined: Jun 2013 Posts: 6 Thanks: 0  Re: A new approach to Fermat's last theorem
I HAVE FOUND A COUNTEREXAMPLE TO WHAT YOU SAY. IF N=7 THEN THE TRIPLE (a,b,c)=(1,2,3) SATISFIES ALL YOYR CRITERIA SO MY QUESTION IS : IF X,Y AND Z DON'T EXIST, HOW CAN THEIR REMAINDERS EXIST ? 
June 7th, 2013, 02:07 AM  #19  
Senior Member Joined: May 2013 Posts: 118 Thanks: 10  Re: A new approach to Fermat's last theorem Quote:
For every prime N of the 6q+1 form,there are exactly N1 non trivial trios of remainders (a,b,c), satisfying the following relations 0<a<N, 0<b<N, 0<c<N and or equavalently: r\:N^2a^Nc^N(ac)^N\r\:N^2b^Nc^N(bc)^N" /> Here's a list for the first 2 such primes for ?=7 the trios (a.b,c) are: (2,4,6), (1,4,5), (1,2,3), (6,5,4), (6,3,2), (5,3,1) for N=13 the trios are3,9,12), (5,6,11), (1,9,10), (2,6,, (2,5,7), (1,3,4), (12,10,9), (12,4,3), (11,8,6), (11,7,5), (10,4,1), (8,7,2) A trio of the form (b,a,c) is trivial to (a,b,c) In half of these N1 trios we have a+b=c and in the other half we have a+b=N+c As the reader can see,we always have: Primes of the 6q1 form don't have trios at all Numbers X=AN+a, Y=BN+b and Z=CN+c don't exist,because you can't find quotients A,B,C to satisfy:  
June 7th, 2013, 02:42 PM  #20  
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20  Re: A new approach to Fermat's last theorem Quote:
Quote:
Quote:
It is well known and can be easily proved that to have solution the latter requires XYZ to be divisible by N. Then the inverse statement must be true as well: If coprime with then has no solution. It is an equivalent of this elaborated proof, isn't it.  

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