Re: FERMAT LAST THEOREM:ANOTHER PROOF

From: george ghiata (george_ghiata_at_hotmail.com)
Date: 01/14/05 

Date: Fri, 14 Jan 2005 18:25:09 +0000 (UTC)

HI Michael Orion,
 A MISTAKE is made in EQ4.
So I have to continuu the search.
george ghiata
 See my answers to your questions:
1)The algebra in this case applies only to n=odd.
when n=even that is a different story which Fermat gave the answer
 when he outline his proof by infinite descent that X^4+Y^4=Z^4
is impossible.
2)Yes. I assumed that that the EQF is possible and try to get to a
contradiction of this.
3)I think that people with college math can figure out by themselves
 that if m is not prime then m=[(n1)^a]*[(n2)^b]*....
where (n1),(n2)... are prime numbers.
So x^m =X^(n1).IT is a simple reasoning which college math students
should not miss to figure out.
4)The Little Theorem says that a^n-a is divisible by n if n
is prime number.So X^n-X is divisible by n
                   Y^n-Y is divisible by n
                   Z^n-Z is divisible by n
That is why B is divisible by n
5)The contradiction at which I wanted to get is that
assuming that EQF is possible then we have to accept that this
implies that U-q-p must be divisible by n
So the proof follows a line which tries to contradict that,and by
doing A MISTAKE(see EQ4) I contradict that by showing that
U-1,q-1,and p-1 are divisible by n.
6)You have to know What the words "simple proof"
  mean in the context of Last theorem.
It means A proof which uses the mathematical knowledge of
 Fermat's time .Well,today a college 4years graduate
in general does not have that !
Thank you for your reply!
I am keeping working on IT.IT is my hobby as were the othe problems
which i posted .
george ghiata
 
Jan 05 10:06:47 -0500 (EST), Michael Orion wrote:
>Well, this proof may be simple in comparison to Wiles', but it is
>beyond me. But then I admit that I am not particularly well educated
>in number theory. All the same, with my limited knowledge I have a
>couple of questions/critisms of the proof.
>
>The biggest problem I see with the proof is that I see nothing that
>precludes n = 2. I.e., it seems to prove that there is no non-zero
>integer solution to X^2 + Y^2 = Z^2, which of course is nonsense.
>This nonsense would seem to imply to me that the proof is not valid
>for any n.
>
>I have a couple of other comments/critisms. They are given below
>interspersed in the proof.
>
>- MO
>
>On 12 Jan 05 20:30:05 -0500 (EST), george ghiata wrote:
>>
>>This is a proof of Fermat Last Theorem by Gheorghe Ghiata
>>EQF: X^+Y^n=Z^n (X,Y,Z Are integers and n>=3) is impossible
>>PROOF:
>> X^n-X+Y^n-Y=Z^n-Z+Z-X-Y
>
>It seems that you are using the ~proof by contradiction~ method,
since
>you are starting by assuming that X^n + Y^n = Z^n for n >= 3 and
>some non-zero integers X, Y, and Z. Is this correct?
>
>> X+Y-Z=B; Little theorem makes B divisible n
>
>A comment and a question here.
>
>Comment: Fermat's Little Theorem applies only for n prime, which is
>all that is necessary to prove Fermat's Last Theorem. However, since
>this is supposed to be a simple proof, intended I presume for people
>with college math but not necessarily number theory, it would be
>helpful to show why you need only prove the theorem for n odd prime.
>
>Question: Why does Fermat's Little Theorem imply that B is divisible
>by n? I don't get it. You need to explain yourself more completely
>here.
>
>>Case1:X*Y*Z not divisible by n
>> X=B+Q
>> Y=B+P
>> Z=B+Q+P
>> EQ1 :(B+Q)^n+(B+P)^n=(B+Q+P)^n
>> From EQ1 we get:
>> X+Y=2*B+Q+P=U^n=W
>> Z=U*V
>> Q=q^n and P=p^n
>
>Heh? You define U^n = W = X+Y, V = Z/U, q^n = Q, and p^n = P
>but none of these definitions require EQ1. Do you also mean to
>say that U, V, q, and p are non-zero integers or some such thing?
>If so, state it. Then, how does EQ1 imply that U, V, q, and p have
>the form you assume?
>
>>Therefore 2*B=U^n-q^n-p^n
>>Little Theorem makes (U-q-p) divisible by n:
>
>Again, I do not see how Fermat's Little Theorem applies. Please
>explain yourself more completely.
>
>>Now we write:
>> X^n-Q^n+Y^n-P^n-Z^n +W^n=(-Q^n+Q)+(-P^n+P)+(W^n-W)+
>> +(W-Q-P)
>> Therefore (W-Q-P)=2*B is divisible by n^2
>>Therefore B is divisible by n^2
>
>At this point, I am completely lost. Your proof may be simple,
>but it is beyond me. It looks like nothing more than a mumble
>jumble of algebra manipulations. Perhaps I am merely exposing my
>incompentence. Or perhaps your proof really is just some mumble
>jumble.
>
>In spite of the fact that I am lost at this point, I have a couple
>more comments below (which sort of says how jumbled your proof seems
>to me that I can find further fault with it even though I can no
>longer follow it).
>
>>Now EQ1 is divided by (2*B+Q+P) and get
>> EQ2:
>>[B^(n-1)+(a1)*B^(n-2)+(a2)*B^(n-3).........+c*B]+[Q^n+P^n]/(Q+P)=V^n
>> From EQ2 we get that (V-1) is divisible by n and that
>> {V^n - [Q^n+P^N]/(Q+P)} is divisible by n^2
>>If we divide EQF by (X+Y) we get :
>> EQ3: X^(n-1)-Y*X^(n-2)+[Y^2}*X^(n-3)..........+Y^(n-1)=V^n
>>After we multiply EQ3 With Z we write It in two ways:
>>X+Y=Z+B=W
>> 1.
>>Z*{W^(n-1)-n*[W^(n-2)]*Y+(1/2)*n*(n-1)*[W^(n-3)]*Y^2........+n*Y^(n-1)=Z*V^n
>>
>> 2. The same as 1. exept we substitute Y with X
>> Now we Add -up (1.+2.) and get:
>> EQ4: B*G +(Z-1)*n*[Y^(n-1)+X^(n-1)] +Q^n+P^n+X^n+Y^n=2*Z*V^n
>> Now we substitute {X^n+Y^n] with Z^n and get to the left side of
>>EQ4
>>the value 2*Z*V^n.Since B is divisible by n^2 we get that (Z-1) is
>>divisible by n.
>>But Z=U*V and (V-1) is divisible by n.
>>Therefore (U-1) is divisible by n.
>>Repeating the proof twice we get That (q-1) and (p-1) are divisible
>by
>>n too.
>
>Heh? What do you mean by "repeating the proof twice"? If you mean
>that what you have so far reduces the original assumption that
>X^n + Y^n = Z^n to some reduced form, such as X^(n-1) + Y^(n-1)
>= Z^(n-1), then state the reduced form and how reducing it again
>gets you to your ascertian that n divides (q-1) and (p-1). If this
>is not what you mean, then what do you mean???
>
>>But We know that (u-q-p) must be divisible by n too.
>>That proves Fermat last theorem for Case1.
>
>How does this prove Fermat's Last Theorem for Case 1? Is the fact
>that n divides (u-p-q) contradict your assumption that Fermat's Last
>Theorem is false? Remember, you started by assuming that X^n + Y^n
>= Z^n for n > 3 and some non-zero integer X, Y, and Z, which is
>contrary to Fermat's Last Theorem. So I am looking for where this
>assumption leads to a contradiction. Please point out the
>contradiction.
>
>>Case 2; Z is divisible by n
>>In this case ((q+p) is divisible by n.
>>As in Case 1. we proof that (q-1) and (p-1) are divisible by n.
>>Therefore this proves Case2 of Fermat Last theorem
>>Therfore Fermat Last theorem Is true.
>
>Yes, Fermat's Last Theorem is true, but I am not so sure your proof
>proves its truth.
>
>>Created by Gheorghe Ghiata

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