Simple solution of FLT?

[vantheman]

Am still trying to find a simple solution of FLT. I think I remember that someone proved that if a solution exist to FLT that either x, y or z must have a factor of n. Is this true and, if so, where can I find the proof of this?

 

[matt grime]

save your efforts, please.

 

[Zurtex]

I'm assuming you mean Fermats Last Theorem and not Fermats Little Theorem.

The theorem states that there exits no solution for:

$$\left. x^n + y^n = z^n \quad \forall \, x, y, z, n \in \mathbb{N} \right\backslash \{ 1 \}$$

And it has been proven true.

 

[vantheman]

Yes, I was referring to Fermat's Last Theorem. I know Andrew Wile has already solved it, but there is that remote possibility that Fermat's "truely marvelous proof" does exist.
It may seem like a waste of time to a serious mathematician, but amuse me if you will.
I don't understand your reply because I do not have the mathmetical credentials. Let me re-phrase the question. Prior to Wile's proof, did anyone prove that, in order for a solution to exist, either x, y or z must have a factor of n?

I don't feel my efforts have been wasted. I have developed an independent proof for n=3 using simple algebra and congruences. It has been reviewed by the Math Dept of a local university and sent to the College Mathmetical Journal for possible publication.
If the above proof exists, it will go a long way to extending the proof for n=3 to be a proof for all prime values of n. HELP.

 

[HallsofIvy]

I'm assuming you mean Fermats Last Theorem and not Fermats Little Theorem.

The theorem states that there exits no solution for:

$$\left. x^n + y^n = z^n \quad \forall \, x, y, z, n \in \mathbb{N} \right\backslash \{ 1 \}$$

And it has been proven true.

I presume you meant $$n \in \mathbb{N}\right\backslash\{1,2\}$$

 

[HallsofIvy]

Yes, I was referring to Fermat's Last Theorem. I know Andrew Wile has already solved it, but there is that remote possibility that Fermat's "truely marvelous proof" does exist.
It may seem like a waste of time to a serious mathematician, but amuse me if you will.
I don't understand your reply because I do not have the mathmetical credentials. Let me re-phrase the question. Prior to Wile's proof, did anyone prove that, in order for a solution to exist, either x, y or z must have a factor of n?

Do you mean this n to be the same as the power in xⁿ+ yⁿ= zⁿ?

I don't feel my efforts have been wasted. I have developed an independent proof for n=3 using simple algebra and congruences. It has been reviewed by the Math Dept of a local university and sent to the College Mathmetical Journal for possible publication.
If the above proof exists, it will go a long way to extending the proof for n=3 to be a proof for all prime values of n. HELP.

Are you sure of that? The cases n= 3, 4, 5 were quickly proved but the method used did not extend to n larger than 5.

 

[vantheman]

It's Greek to me.

Yes, my Greek is a little rusty.

 

[Icebreaker]

I'm sure Fermat actually wrote, "Wouldn't it be funny if I have a truly marvelous proof..." and the first part was torn away somehow. :rofl:

 

[vantheman]

Let me state it again:
Prior to Wile's proof, did anyone prove that in order for a solution to exist to the equation
x^n + y^n = z^n in non-zero integers
that either x, y or z must have a factor of n?

Also, since my proof for n=3 has the possibility, through binomial expansion techniques, to be a tool for proving the equation for all prime values of n, an existing proof of the question I posed will help me in extending my proof for n=3 to n = all prime numbers.

 

[Zurtex]

I presume you meant $$\left. n \in \mathbb{N}\right\backslash\{1,2\}$$

You are of course right, doh!

Let me state it again:
Prior to Wile's proof, did anyone prove that in order for a solution to exist to the equation
x^n + y^n = z^n in non-zero integers
that either x, y or z must have a factor of n?

Also, since my proof for n=3 has the possibility, through binomial expansion techniques, to be a tool for proving the equation for all prime values of n, an existing proof of the question I posed will help me in extending my proof for n=3 to n = all prime numbers.

I've never heard of that before. In fact that sounds weird, because the trouble with the methods of proving it for specific cases only struggled on certain prime numbers. I think composite numbers were really easy to deal with, before Wile's proof I think it was something like all n < 10'000'000 were proven.

 

[matt grime]

n=3 and n=4 are easy to do by elementary methods, 4 is easier than 3 if i recall correctly.

there are many things out there in number theory (not to mention other parts of mathematics) that are interesting and have open questions to work towards, and indeed studying the number theory of various things to do with FLT can be rewarding but don't bother trying to get an elementary proof of FLT as there is almost certainly no such. There have been many plausible attempts that failed. some for example because they assumed unique factorization in rings that turned out not to be UFD's. in fact there are probably sufficiently many "proofs" out there that Fermat's original "proof" has been discovered, and is incorrect. Fermat was a genius but like lots of us he got many things wrong you'd be surprised how many old well known results attributed to the greats were never properly proven by them. it took some 40 years for euler to anwer some simple questions of fermat's never mind the complicated ones (relatively speaking that is).


http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html

contains the information you want (amazing what google can do: google for fermat's last theorem known conditions)

 

[vantheman]

Thanks all. Have to run now. Will check back later.

 

[matt grime]

http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html

i case you didn't read my edited post has all the answers you need, i imagine. but why not just study number theory in general.

 

[shmoe]

Let me state it again:
Prior to Wile's proof, did anyone prove that in order for a solution to exist to the equation
x^n + y^n = z^n in non-zero integers
that either x, y or z must have a factor of n?

Germain showed this is the case if n and 2n+1 are prime. Actually she proved something a little more general have a look for Sophie Germain's theorem or Sophie Germain primes. I've never seen a full generalization that holds for all primes n.

 

[keebs]

A simple solution for FLT would arise if you could prove the abc conjecture...

 

[shmoe]

A simple solution for FLT would arise if you could prove the abc conjecture...

...in a simple way

 

[robert Ihnot]

VAntheman: I think I remember that someone proved that if a solution exist to FLT that either x, y or z must have a factor of n. Is this true and, if so, where can I find the proof of this?

Here is where it's at: We only need consider prime cases, 4 was handled by Fermat, and we look at:

X^P+Y^P=(X+Y)(X^(P-1) -X^(P-2)Y+-...Y^(P-1)) = Z^P

If X+Y is divisible by a factor of Z, call it M; then X==-Y Mod M, so that:

X^(P-1) -X^(P-2)(-X) +X^(P-3)(-X)^2 -+-==PX^(P-1), Mod M. But X, Y and Z are presumed to have no common factor, so the only factor that could be in common with the factorization is P. Other than that we have both sides of the factorization are perfect Pth powers. That is how we have classically divided the two cases, Case 1: P divides one of the terms, or Case 2: P does not.

I don't think the above proves it is necessary for P to divide one of X,Y, or Z. HOWEVER, case 1 historically is much easier to eliminate for a given P, which would say that for a solution, P would have to divide one of the terms. BUT THERE ARE NO SOLUTIONS, so yes! P would have to divide one of the terms for there to be a solution.

 

[robert Ihnot]

HallsofIvy: The cases n= 3, 4, 5 were quickly proved but the method used did not extend to n larger than 5.

Hold on here! N =4 was presented by Fermat himself, one of the few things he did prove, and by the method of infinite descent. Euler is credited with the proof for n=3 in 1700, but Dirichlet did not prove the case for n=5 until 1828. Next case n=7 was 7 years later by Gabriel Lamé and Henri Lebesque.
http://www.bath.ac.uk/~ma1jam/Fermat/ [Broken]

 

[vantheman]

Thanks, but this is a special case where x + y is factorable. If it is prime then it doesn't work. I was hoping for a general solution. I guess it doesn't exist.

 

[mathwonk]

isn't it more likely that fermat was mistaken, than that 350 years of mathematicians have failed to find his elementary proof?

 

[HallsofIvy]

Since Fermat published his proofs of special cases AFTER he wrote that he had a way of proving it for all n, that is almost certain. I suspect what happened is what happens to all mathematicians- he thought he had a proof but when he looked more closely he found it didn't generalize. It was, after all, simply a note written on the spur of the moment in the margin of a book.

 

[WeeDie]

The proof lies in understanding the connection between the prime 2 and 2nd orders of multitude. 2 being the only prime devisable by 2.
I think it would be easier to see the connection using 3 dimensional models with movement in time related to phi - using base 6, as there's only 5 platonic solids in a 3 dimensional universe; 5 dimensional universe if you include time and observer.
I hope this clears things up

 

[hello3719]

The proof lies in understanding the connection between the prime 2 and 2nd orders of multitude. 2 being the only prime devisable by 2.
I think it would be easier to see the connection using 3 dimensional models with movement in time related to phi - using base 6, as there's only 5 platonic solids in a 3 dimensional universe; 5 dimensional universe if you include time and observer.
I hope this clears things up

Wow, this clears things up, tx :rofl:

 

[omega16]

Can anyone show me the proof of FLT for n=3? Thanks

 

[robert Ihnot]

Case 1, is that P does not divide one of the Pth power terms, and is historically the much easier case of the two. Germain, as noted above did important work on this matter using elementary methods. A Germain prime is one of the form 2P+1, where P is also prime. In eliminating these cases, Germain showed that there is no solutions for the cubic Case 1, since 7 is also prime. Also the method could be generalized to some extent for cases like 4P+1, 6P+1, and so forth so that Germain was able to supply a prime and show that that no Case 1 exisits for P under 100. (Lagendre continued this to 197.)This was a substantial advance at the time.
http://www.agnesscott.edu/Lriddle/WOMEN/germain-FLT/SGandFLT.htm

As another writer points out http://www.mathpages.com/home/kmath367.htm [Broken], it is easy to eliminate Case 1 for P=3, by considering the Modulus 9. In this case we have X^6==1 Mod 9, so that X,Y,Z can only be congruent to +/- 1 Modulo 9, which is not possible.

Case 2 where P does divide one of the terms proved to be much harder, but, even so, as far as I know, Case 1 for all P still proved to present difficulities, all the way up to Wile.

 

[robert Ihnot]

mathwonk: isn't it more likely that fermat was mistaken, than that 350 years of mathematicians have failed to find his elementary proof?

That probably has something to do with whether you are a professional
mathematician or an amateur. The same problem is also is found in Physics and the matter of whether you believe in the Special Theory of Relativity. It amazes me that some have thought the theory bunk and openly admit they can not follow a mathematical equation.

 

[HallsofIvy]

A simple solution for FLT would arise if you could prove the abc conjecture...

And you are just going to LEAVE it at that? How about telling us what the "abc conjecture" is!

 

[arildno]

And you are just going to LEAVE it at that? How about telling us what the "abc conjecture" is!

Well, he didn't do it one and a half years ago, so I suspect he is unwilling to enlighten us now.

 

[shmoe]

The abc conjecture is quite well known, google will turn up many, many hits. Mathworld's,

http://mathworld.wolfram.com/abcConjecture.html