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May 31st, 2013, 09:14 PM   #11
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Re: A new approach to Fermat's last theorem

Quote:
Originally Posted by Dacu
If and then necessarily are the sides of a triangle.
I feel obliged to point out that this is false.
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May 31st, 2013, 09:48 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 and then necessarily 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 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!
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June 1st, 2013, 02:54 AM   #13
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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?
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June 1st, 2013, 02:58 AM   #14
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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
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June 1st, 2013, 03:33 AM   #15
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Re: A new approach to Fermat's last theorem

What about FLT?
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June 1st, 2013, 03:45 AM   #16
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Re: A new approach to Fermat's last theorem

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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, ...}.
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June 4th, 2013, 10:11 AM   #17
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Re: A new approach to Fermat's last theorem

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Therefore:
(
and accoriding to relations (3) and the uniqueness of (a,b,c) we have proved,we have

which is false and thus,the asumption:

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
(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):

and therefore,in a totally analogous way we can conclude now:

and thus:
r\:N|b \rightarrow \:a=0\r\:b=0" />
absurdity
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June 6th, 2013, 08:46 PM   #18
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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 ?
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June 7th, 2013, 02:07 AM   #19
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Re: A new approach to Fermat's last theorem

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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
or equavalently:
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:

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:
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June 7th, 2013, 02:42 PM   #20
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Re: A new approach to Fermat's last theorem

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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
Without slightest intention to be offensive I would like to quote:

Quote:
FLT1
From Wikipedia, the free encyclopedia
Vascular endothelial growth factor receptor 1 is a protein that in humans is encoded by the FLT1 gene.
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
My question is whether and 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 coprime with then has no solution.

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