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June 7th, 2013, 09:41 PM   #21
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Re: A new approach to Fermat's last theorem

Quote:
Originally Posted by McPogor
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.
If we assume there are integers solutions,then the two equations are not equivalent,as the $x^p+y^p+z^p$ accepts (actually needs) negative solutions,while the other one does not.But if we suppose or prove the lack of integer solutions, then the two are equivalent
I'm eager to see the easy proof you mentioned.Ceck this http://en.wikipedia.org/wiki/First_case ... st_theorem
For the $N|XYZ$ case,see my may 31 12:23 remark

 June 8th, 2013, 01:36 AM #22 Newbie   Joined: May 2013 From: india,odisha,jatni,harirajpur Posts: 1 Thanks: 0 numbers The smallest value of n,for which 2n +1 is not a prime number is...................................
June 8th, 2013, 03:46 PM   #23
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Re: numbers

Quote:
 Originally Posted by sanjay_jena13 The smallest value of n,for which 2n +1 is not a prime number is...................................
Ooooh! I know! 4!

Why?

June 8th, 2013, 06:25 PM   #24
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Re: numbers

Quote:
 Originally Posted by johnr Ooooh! I know! 4! Why?
n=0

If we do not consider prime number the "1"

Regards.

 June 8th, 2013, 11:39 PM #25 Senior Member   Joined: May 2013 Posts: 118 Thanks: 10 Re: A new approach to Fermat's last theorem Poor FLT
June 9th, 2013, 02:17 AM   #26
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by bruno59 Poor FLT
You have reason, I sit it.
I ask him to excuse myself.

Regards.

June 9th, 2013, 05:00 AM   #27
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Re: numbers

Quote:
Originally Posted by mente oscura
Quote:
 Originally Posted by johnr Ooooh! I know! 4! Why?
n=0

If we do not consider prime number the "1"

Regards.
Zzzzzzziinnggg! Point taken!

June 11th, 2013, 02:38 PM   #28
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Re: A new approach to Fermat's last theorem

Quote:
 For the $N|XYZ$ case,see my may 31 12:23 remark
LET ME SEE IF I GET THE POINT
IF N DIVIDES ONE OF X,Y,Z, SAY Y (b=0), THEN FROM $N|a^2+ab+b^2$ WE GET N|a
SO a=0, ABSURD

June 11th, 2013, 10:29 PM   #29
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by akenaton LET ME SEE IF I GET THE POINT IF N DIVIDES ONE OF X,Y,Z, SAY Y (b=0), THEN FROM $N|a^2+ab+b^2$ WE GET N|a SO a=0, ABSURD
That's right

June 14th, 2013, 08:07 AM   #30
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Re: A new approach to Fermat's last theorem

Quote:
 Originally Posted by bruno59 ] Step 1. Basic relations between X,Y,Z,N Let $P_{xy}=\sum_{i=1}^{N}X^{N-i}Y^{i-1}$ We first show that $(X+Y,P_{xy})=1$ Suppose this was'nt true.Then there would be a prime Q common in these two numbers: $X+Y=GQ^g\:and\: HPQ^h=P_{XY}=\sum_{i=1}^{N}X^{N-i}Y^{i-1}=$$\sum_{i=1}^{N}X^{N-i}(GQ^g-X)^{i-1}=NX^{N-1}+GQ^gJ\rightarrow NX^{N-1}=UQ^u$ and if $(Q^u,X^{N-1})\neq1\rightarrow(Q,X)\neq1=j$ and as Q is a prime,we get j=Q and so $Q|X$ but since $Q|X+Y$ we conclude that $Q|Y$ which is false,because X and Y are coprimes.So: $(Q^u,X^{N-1})=1$ and from (1) we see that $Q^u$ is all "hidden" in N: $N=wQ^u$ and as N is a prime: $w=u=1\:and\: Q^u=Q=N\rightarrow N|X+Y\rightarrow (sentence\:1b)\:N|Z$ which is false.Therefore: $(X+Y,P_{xy})=1$ and since: $(X+Y)P_{xy}=Z^N$
i am just posing doubt don't misunderstand...
is the last equation correct...

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