- #1

- 44

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter vantheman
- Start date

- #1

- 44

- 0

- #2

Science Advisor

Homework Helper

- 9,426

- 5

save your efforts, please.

- #3

Science Advisor

Homework Helper

- 1,120

- 1

The theorem states that there exits no solution for:

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

And it has been proven true.

- #4

- 44

- 0

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.

- #5

Science Advisor

Homework Helper

- 43,010

- 973

Zurtex said:

The theorem states that there exits no solution for:

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

And it has been proven true.

I presume you meant [tex] n \in \mathbb{N}\right\backslash\{1,2\}[/tex]

- #6

Science Advisor

Homework Helper

- 43,010

- 973

Do you mean this n to be the same as the power in xvantheman said: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?

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.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.

- #7

- 44

- 0

Yes, my Greek is a little rusty.

- #8

- #9

- 44

- 0

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.

- #10

Science Advisor

Homework Helper

- 1,120

- 1

You are of course right, doh!HallsofIvy said:I presume you meant [tex]\left. n \in \mathbb{N}\right\backslash\{1,2\}[/tex]

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.vantheman said:

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.

- #11

Science Advisor

Homework Helper

- 9,426

- 5

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)

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)

Last edited:

- #12

- 44

- 0

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

- #13

Science Advisor

Homework Helper

- 9,426

- 5

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.

- #14

Science Advisor

Homework Helper

- 1,994

- 1

vantheman said: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.

- #15

- 19

- 0

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

- #16

Science Advisor

Homework Helper

- 1,994

- 1

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

...in a simple way

- #17

- 1,059

- 1

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,

Last edited:

- #18

- 1,059

- 1

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]

Last edited by a moderator:

- #19

- 44

- 0

- #20

Science Advisor

Homework Helper

- 11,464

- 1,717

- #21

Science Advisor

Homework Helper

- 43,010

- 973

- #22

- 12

- 0

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

- #23

- 129

- 0

WeeDie said:

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:

- #24

- 20

- 0

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

- #25

- 1,059

- 1

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.

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.

Last edited by a moderator:

- #26

- 1,059

- 1

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.

Last edited:

- #27

Science Advisor

Homework Helper

- 43,010

- 973

keebs said: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!

- #28

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,106

- 136

Well, he didn't do it one and a half years ago, so I suspect he is unwilling to enlighten us now.HallsofIvy said:And you are just going to LEAVE it at that? How about telling us what the "abc conjecture" is!

- #29

Science Advisor

Homework Helper

- 1,994

- 1

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

Share:

- Replies
- 9

- Views
- 147

- Replies
- 14

- Views
- 1K

- Replies
- 13

- Views
- 741

- Replies
- 4

- Views
- 396

- Replies
- 2

- Views
- 563

- Replies
- 1

- Views
- 460

- Replies
- 15

- Views
- 986

MHB
Condition for a unique solution of matrix A that have infinite solutions and optimisation of tr(A)

- Replies
- 0

- Views
- 1K

- Replies
- 6

- Views
- 2K

- Replies
- 2

- Views
- 788