Fermat's Last Theorem

[McPgr]

Hello everybody on the Forum.

 

I'm not a mathematician and as so many amateurs could not resist temptation to prove the FLT.

 

I found that for equation a^n+b^n=c^n to be true it is required: a=uwv+v^n; b=uwv+w^n; c=uwv+v^n+w^n. (with modifications).

 

Then it becomes possible to obtain polynomial equations P(u)=0 as well as F(a)=0; F(B)=0; F©=0. And by means of basic algebra there was eventually obtained a contradiction that may prove the theorem.

 

Understanding how unbelievable it is and that there may be flaws I was unable to see I would be thankful to everybody willing to examine this proof. I'm prepared the flaws even fatal may be discovered but will not be disappointed if this does not happen.

 

Sorry for very unprofessional language of the paper.

 

Please go to

http://sites.google.com/site/mcpogorspace

 

 

McPgr

[Qfwfq]

Understanding how unbelievable it is and that there may be flaws I was unable to see I would be thankful to everybody willing to examine this proof. I'm prepared the flaws even fatal may be discovered but will not be disappointed if this does not happen.

The step from eq. 2 to eq. 3 doesn't work for me; it doesn't seem to follow of consequence.

[McPgr]

The step from eq. 2 to eq. 3 doesn't work for me; it doesn't seem to follow of consequence.

By substitution of a, b, c in the eq 1 with right hand parts of eq 2 we obtain P(u)=0. Then b=a-v^n+w^n ; c=a+w^n. After substitution of these r. h. parts for b and c in eq 1 we have F(a)=0 etc. In all equations v and w are considered invariable. ;

[jakuta]

Hello everybody on the Forum.

 

I'm not a mathematician and as so many amateurs could not resist temptation to prove the FLT.

 

I found that for equation a^n+b^n=c^n to be true it is required: a=uwv+v^n; b=uwv+w^n; c=uwv+v^n+w^n. (with modifications).

 

Then it becomes possible to obtain polynomial equations P(u)=0 as well as F(a)=0; F(B)=0; F©=0. And by means of basic algebra there was eventually obtained a contradiction that may prove the theorem.

 

Understanding how unbelievable it is and that there may be flaws I was unable to see I would be thankful to everybody willing to examine this proof. I'm prepared the flaws even fatal may be discovered but will not be disappointed if this does not happen.

 

Sorry for very unprofessional language of the paper.

 

Please go to

http://sites.google....te/mcpogorspace

 

 

McPgr

 

 

 

Hey. This looks fun.

 

Could you please post a step by step procedure of what you are trying to do?

 

For example, start with some abstract of your overall objective and how you achieve it. For example, I can see you are proceeding by reductio ad absurdum, but the rest was hard for me to follow because I can't see your line of reasoning.

 

Then, after the abstract, give an intro with a step by step procedure of how you reach A to B to C .....

 

Anyway, looks like you put lots of work into it and I am look forward to digging through it.

[jakuta]

Hello everybody on the Forum.

 

I'm not a mathematician and as so many amateurs could not resist temptation to prove the FLT.

 

I found that for equation a^n+b^n=c^n to be true it is required: a=uwv+v^n; b=uwv+w^n; c=uwv+v^n+w^n. (with modifications).

 

Then it becomes possible to obtain polynomial equations P(u)=0 as well as F(a)=0; F(B)=0; F©=0. And by means of basic algebra there was eventually obtained a contradiction that may prove the theorem.

 

Understanding how unbelievable it is and that there may be flaws I was unable to see I would be thankful to everybody willing to examine this proof. I'm prepared the flaws even fatal may be discovered but will not be disappointed if this does not happen.

 

Sorry for very unprofessional language of the paper.

 

Please go to

http://sites.google....te/mcpogorspace

 

 

McPgr

 

OK, in your paper,

 

According to the Fermat's Last Theorem the equation (1)

 

aⁿ + bⁿ = cⁿ

cannot be true when a, b, c and n are positive integers and n > 2.

 

It is assumed in following discussion that three terms are coprime and the exponent is a prime number.

 

 

 

Assume the equation (1) is true.

 

Let us express

 

c = a + k = b + f (2)

 

 

You can't just say equation 2 you have to prove it from equation 1.

In other words, you have to prove there exists k and f that satisfy equations 1 and 2.

[McPgr]

OK, in your paper,

 

According to the Fermat's Last Theorem the equation (1)

 

aⁿ + bⁿ = cⁿ

cannot be true when a, b, c and n are positive integers and n > 2.

 

It is assumed in following discussion that three terms are coprime and the exponent is a prime number.

 

 

 

Assume the equation (1) is true.

 

Let us express

 

c = a + k = b + f (2)

 

 

You can't just say equation 2 you have to prove it from equation 1.

In other words, you have to prove there exists k and f that satisfy equations 1 and 2.

 

If there are three positive integers a, b, c and from eq (1) we know c>a and c>b there must exist integers k=c-a and f=c-b. How they can satisfy the (1) can be determined only after substitution of a+k and b+f for c in (1).

[McPgr]

I'm surprised it is possible to reach my paper through damaged link. For reason unknown to me I was unable to post it undamaged.

[Qfwfq]

Your link isn not damaged Michael, it simply shows that way when parsed automatically by the forum app, in order to keep it short in the text, but under the bonnet the full link is stil there.

 

By substitution of a, b, c in the eq 1 with right hand parts of eq 2

Gosh I've been gathering dust the past few days, things are so dreary over here. I had tried getting (3) as a consequence of only (2) without considering (1). :doh: For clarity, perhaps you should edit the line after (2) so, instead of just "Then" it could read "Then (1) can be written:"

 

P(u)=0. Then b=a-v^n+w^n ; c=a+w^n

It would be a good idea to use LaTeX tags:

 

[math]b=a-v^n+w^n; \; c=a+w^n[/math]

 

Oops I hit the submit button by mistake.

 

I find various lack of clarity in your paper, I think I get what you mean by lemma 1, 2 and 3 but it took some figuring especially for 2; for one thing the requisite of being coprime should be part of the statement rather than mentioned only in the proof. From lemma 4 onward I think I'll need a lot more time to follow you.

[jakuta]

If there are three positive integers a, b, c and from eq (1) we know c>a and c>b there must exist integers k=c-a and f=c-b. How they can satisfy the (1) can be determined only after substitution of a+k and b+f for c in (1).

 

Yes, that is correct.

 

But, you are trying to prove assuming aⁿ + bⁿ = cⁿ is true

 

that is the case that there are three positive integers a, b, c or you arrive at a contradiction.

 

So you cannot write a + k = c with k integer as k might be a real decimal number in order to satisfy aⁿ + bⁿ = cⁿ

You must prove k can only be an integer.

[McPgr]

Your link isn not damaged Michael, it simply shows that way when parsed automatically by the forum app, in order to keep it short in the text, but under the bonnet the full link is stil there.

 

Gosh I've been gathering dust the past few days, things are so dreary over here. I had tried getting (3) as a consequence of only (2) without considering (1). :doh: For clarity, perhaps you should edit the line after (2) so, instead of just "Then" it could read "Then (1) can be written:"

 

It would be a good idea to use LaTeX tags:

 

[math]b=a-v^n+w^n; \; c=a+w^n[/math]

 

Oops I hit the submit button by mistake.

 

I find various lack of clarity in your paper, I think I get what you mean by lemma 1, 2 and 3 but it took some figuring especially for 2; for one thing the requisite of being coprime should be part of the statement rather than mentioned only in the proof. From lemma 4 onward I think I'll need a lot more time to follow you.

 

Sorry again for unprofessional style and language. The paper definitely needs a lot of editing but maybe only in case if no fatal flaws are found in it.

I have absolutely no training in LaTeX.

The statement: Lemma-2 The sum with all terms but one containing the same factor... implies that this one term and the factor are coprime.

[McPgr]

Yes, that is correct.

 

But, you are trying to prove assuming aⁿ + bⁿ = cⁿ is true

 

that is the case that there are three positive integers a, b, c or you arrive at a contradiction.

 

So you cannot write a + k = c with k integer as k might be a real decimal number in order to satisfy aⁿ + bⁿ = cⁿ

You must prove k can only be an integer.

 

Assumption that k may be not integer contradicts the statement of the FLT. There may be any two integers c and b with c>b. Obviously c^n - b^n =A always will be some integer that if assumed A = a^n produces some real number (not integer) a. This is where your real not integral number k appears. But the theorem requires a to be an integer i.e. for k to be an integer is a must.

[Qfwfq]

I have absolutely no training in LaTeX.

Well you have a good chance here to practice, it isn't all that hard if you follow examples.

 

The statement: Lemma-2 The sum with all terms but one containing the same factor... implies that this one term and the factor are coprime.

I had got it in the end, only the statement seemed a bit confusing at first. It can really be reduced to the fact that, if [imath]a[/imath] is divisible by [imath]n[/imath] and [imath]b[/imath] isn't, then the sum [imath]a + b[/imath] isn't (by the distributive property).

 

Anyway, I found time yesterday to look at Lemma 4 and, by the way you apply Lemma 3, you are exploiting that [imath]f[/imath] and [imath]k[/imath] in eq,s 4a and 4b are each coprime with the other factors. My aging brain only now realized that this follows by Lemma 2, yeah, I s'pose I'm getting a bit of the dust out of it. By trying hard enough, I made it through to eq. (9) and it's clever work :) but I haven't shaken enough dust out yet to see why [imath]v[/imath] and [imath]w[/imath] must be coprime. Any hint?

[jakuta]

<br />Assumption that <i>k</i> may be not integer contradicts the statement of the FLT. There may be any two integers<i> c</i> and <i>b</i> with <i>c>b</i>. Obviously<i> c^n - b^n =A</i> always will be some integer that  if assumed <i>A = a^n</i> produces some real number (not integer) <i>a</i>. This is where your real not integral number <i>k</i> appears. But the theorem requires <i>a</i> to be an integer i.e. for<i> k</i> to be an integer is a must.<br />

<br /><br /><br />

 

OK, can you prove such a k exists. The theorem is saying one thing and you are saying another.

 

Simply prove the k exists. If you are correct, this will be no problem.

[McPgr]

Well you have a good chance here to practice, it isn't all that hard if you follow examples.

I'm too old for new tricks.

 

 

why [imath]v[/imath] and [imath]w[/imath] must be coprime. Any hint?

The divisors of coprime a and b correspondingly v and w must be coprime too. All three terms of eq (1) are coprime. If two of them have common divisor >1 the third must have it too and all three are divided by it.

[Qfwfq]

The divisors of coprime a and b correspondingly v and w must be coprime too. All three terms of eq (1) are coprime. If two of them have common divisor >1 the third must have it too and all three are divided by it.

Jease I wuz fergittin' 'bout that! :doh:

 

Well, that gets me through to Version A being unacceptable but, since I'm also getting too old for new tricks myself, I think I'll need a bit of a rest before examining Version B. I notice now that at the start of your paper, you also specify n will be considered prime; this explains why the first lemma might be useful but are you sure it doesn't lose generality?

 

But, of course, if n is prime then it must also be coprime with any other integer, including the two as per Version A, so the premise of prime n restricts you to that anyway. Actually, looking back at the premise of Version A it hits me I can't recall it being exploited; did I miss something? Can you sort it all out a little bit?

[McPgr]

[ name=Qfwfq' timestamp='1288264455' post='301638]

,

you also specify n will be considered prime; this explains why the first lemma might be useful but are you sure it doesn't lose generality?

 

The eq (74) deals with it.

the premise of prime n restricts you to that anyway. Actually, looking back at the premise of Version A it hits me I can't recall it being exploited; did I miss something? Can you sort it all out a little bit?

 

Since in version A n divides neither f nor k and according to lemma-2 is not divisor in (6a) and (6b) it plays no role in this version (except some specific case not used here)

[McPgr]

name='Qfwfq' timestamp='1288264455' post='301638']

you also specify n will be considered prime; this explains why the first lemma might be useful but are you sure it doesn't lose generality?

 

Equation (74) deals with it.

 

But, of course, if n is prime then it must also be coprime with any other integer, including the two as per Version A, so the premise of prime n restricts you to that anyway. Actually, looking back at the premise of Version A it hits me I can't recall it being exploited; did I miss something? Can you sort it all out a little bit?

 

Since in version A n divides neither f nor k and according to lemma-2 is not divisor in (6a) and (6b) it plays no role in this version (except some features not used here).