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May 31st, 2013, 10:14 PM   #11
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by Dacu If $x,y,z\in Z*$ and $n>1$ then necessarily $x, y, z$ are the sides of a triangle.
I feel obliged to point out that this is false.

May 31st, 2013, 10:48 PM   #12
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Re: A new approach to Fermat's last theorem

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Originally Posted by CRGreathouse
Quote:
 Originally Posted by Dacu If $x,y,z\in Z*$ and $n>1$ then necessarily $x, y, z$ are the sides of a triangle.
I feel obliged to point out that this is false.
Hello!
You're right!Thousands of excuses!
Adjustment:If $x,y,z\in N*$ and the natural numbers $n>1$ then necessarily $x, y, z$ are the sides of a triangle.In fact even for $x,y,z\in R^+$ my assertion is valid and bviously $x,y,z\neq 0$.
Thank you very much!

 June 1st, 2013, 03:54 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, 03:58 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
 June 1st, 2013, 04:33 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, 04:45 AM   #16
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by johnr I'm not sure what the '*' is doing in 'N*'.
I'm not either. Usually X* denotes the group of units (= invertible elements) of X, so Z* would mean {-1, 1}. In this case I suspect the intended meaning is either Z \ {0} or else the positive integers {1, 2, ...}.

June 4th, 2013, 11:11 AM   #17
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by bruno59 Therefore: $N^2| (a-b)^N-a^N+b^N$ ( and accoriding to relations (3) and the uniqueness of (a,b,c) we have proved,we have $b=c\rightarrow N|c-b\rightarrow N|a$ which is false and thus,the asumption: $N\dagger XYZ$ is true no more
This is how I ended my topic.I feel the uniquness argument can't apply here,so I try another way:
Comparing ( to the already known relation
$N^2|a^N+b^N-c^N$ (9)
we observe that in ( a has replaced c and a-b has replaced a.By (9) we had cocnluded (see step 2,subcases ?a and ?b):
$N|a^2+ab+b^2$
and therefore,in a totally analogous way we can conclude now:
$N| (a-b)^2+(a-b)b+b^2= a^2-ab+b^2$
and thus:
$N|2ab \rightarrow (sentence\:5)\: N|a\r\:N|b \rightarrow \:a=0\r\:b=0" />
absurdity

 June 6th, 2013, 09:46 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, 03:07 AM   #19
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by akenaton 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 ?
Of course they can !
For every prime N of the 6q+1 form,there are exactly N-1 non trivial trios of remainders (a,b,c), satisfying the following relations
0<a<N, 0<b<N, 0<c<N and
$N^2|a^N+b^N-c^N$ or equavalently:
$N^2|a^N+b^N-(a+b)^N\r\:N^2|a^N-c^N-(a-c)^N\r\:N^2|b^N-c^N-(b-c)^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 N-1 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:
$N|a^2+ab+b^2,\:N|a^2-ac+c^2,\:N|b^2-bc+c^2$
Primes of the 6q-1 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:
$X^N+Y^N=Z^N$

June 7th, 2013, 03:42 PM   #20
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by bruno59 In this topic I will try to prove the first case of Fermat's last theorem (FLT1 as it is often written) So the goal is to prove that if N divides none of X,Y,Z and N>2,then we can't have $X^N+Y^N=Z^N$
Without slightest intention to be offensive I would like to quote:

Quote:
Quote:
 First case of Fermat's last theorem From Wikipedia, the free encyclopedia The first case of Fermat's last theorem says that for three integers x, y and z and a prime number p, where p does not divide the product xyz, there are no solutions to the equation $xp + yp + zp= 0$
My question is whether $x^p+y^p+z^p=0$ and $X^N+Y^N=Z^N$ are equivalent equations.
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 $XYZ$ coprime with $N$ then $X^N+Y^N=Z^N$ has no solution.

It is an equivalent of this elaborated proof, isn't it.

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