Sunday, May 22, 2005

Fermat's Last Theorem: Proof for n=3

Leonhard Euler came up with two proofs for Fermat's Last Theorem: n = 3.

One proof involved a very innovative method using irrational numbers. Unfortunately, Euler made a mistake in his proof. Despite this, his method revealed a very promising approach to Fermat's Last Theorem which was later taken up by Gauss, Dirichlet , and Kummer. I discuss the details of this method and Euler's mistake in another blog.

The other proof is less generalizable but still brilliant. This is the proof that I will present in today's blog.

The details of this proof are based largely on the work by H. M. Edwards in his book: Fermat's Last Theorem: A Genetic Introduction to Algebraic Number Theory.

Theorem: Euler's Proof for FLT: n = 3

x3 + y3 = z3 has integer solutions -> xyz = 0

(1) Let's assume that we have solutions x,y,z to the above equation.

(2) We can assume that x,y,z are coprime. [See here for the proof]

(3) First, we observe that there must exist p,q such that (see here for proof):
(a) gcd(p,q)=1
(b) p,q have opposite parities (one is odd; one is even)
(c) p,q are positive.
(d) 2p*(p2 + 3q2) is a cube.

(4) Second, we know that gcd(2p,p2+3q2) is either 1 or 3. (see here for proof).

(5) If gcd(2p,p2+3q2)=1, then there must be a smaller solution to Fermat's Last Theorem n=3. (see here for proof).

(6) Likewise, if gcd(2p,p2+3q2)=3, then there must be a smaller solution to Fermat's Last Theorem n=3. (see here for proof).

(7) But then there is necessarily a smaller solution and we could use the same argument on this smaller solution to show the existence of an even smaller solution. We have thus shown a condition of infinite descent.



jenny said...

Hi Larry,

Could you please teach me how to solve FLT for exponent three according to the information below?
Prove Fermat's Last theorem for n=3 : X^3 + Y^3 = Z^3
where X, Y, Z are rational integers, then X, Y, or Z is 0.

* Show that if X^3 + Y^3 = Epsilon* Z^3, where X, Y, Z are quadratic integers in Q[sqrt(-3)], and epsilon is a unit in Q[sqrt(-3)], then X, Y, or Z is 0. (recall how to find all solutions to X^2 + Y^2 =Z^2 since it similar.)

*Begin with if X^3 + Y^3 = Epsilon* Z^3, where X, Y, Z are quadratic integers in Q[sqrt(-3)], where epsilon is a unit, then lambda divides X, Y or Z, where lambda is (3+sqrt(-3))/2. Also show that (lambda)^2 is an associate of 3.

*It will be useful to show that if X is congruent to 1 mod lambda, then X^3 is congruent to 1 (mod lambda)^4). Work out a similar desciption for when X is congruent to -1 mod lambda.
Use these fact to show that if X^3 + Y^3 = Epsilon* Z^3, if X and Y are not multiples of lambda, but Z is , then Z is a multiple of (lambda)^2. Do this by reducing (mod lamda)^4.

note that X^3 + Y^3 = Epsilon* Z^3 can be factored:
(x+y)(x+wy)(x+w^2*y)= Epsilon* Z^3 ,
where w is an appropriately chosen quadratic integer.

Consider each of these factors as quadratic integers p, q, r. Express x and y in terms of p, q, and r. The fact that we have three equations and two unknowns indicates there will be some extra constraint on p, q, and r.

Consider how many time lambda occurs in the prime factorization of p, q, and r. Use unique factorization of Q[sqrt(-3)] to show that except for the factors of lambda that you computed , p, q, and r are cubes times units.

Use the extra constraint on p, q, and r to find another solution
X^3 + Y^3 = Epsilon* Z^3 ,
where Z has one less factor of lambda. Note that it may be necessary to exchange the roles of x, y, z, -x, -y, or -z.

Derive a contradiction along the lines of Fermat's method of descent.
Thank you very much.


Larry Freeman said...

Hi Jenny,

I think you will have to work out the details but I provide a proof using Eisenstein integers here.

As you probably know, Eisenstein integers are quadratic integers in Q[sqrt(-3)].


jenny said...

No, I don't . I will go through your link.

I have more questions which related to part of the FLT proof. I can't do them too. If I finish proving them, it may help me to figure out more for the proof of FLT for n=3.

I have some idea about how to prove FLT for n=3, but I must solve the below question first, then can link up the idea.

I need your help again. I type it here. Could you please teach me how to solve it?

Prove that if x^3 + y^3 = z^3, and x , y, z are quadratic integers in Q[sqrt(-3)]
, then alpha = (3+sqrt(-3))/2 is belonging to Q[sqrt(3)] must divide one of x, y, or z.

Hint: Reduce the equation modulo alpha^3.

Thank you very much.

Larry Freeman said...

The proof comes down to showing that for any integer beta in Q[sqrt(-3)], if gcd(alpha,beta)=1, then beta^3 = +/- 1 (mod alpha^3)

Then you can show that if alpha doesn't divide x,y,z say that:

x^3 + y^3 + z^3 = (+/- 1) + (+/- 1) + (+/- 1) (mod alpha^3)

which is impossible if

x^3 + y^3 + z^3 = 0


jenny said...

I see. Thank you very much Larry. Now , I can forcus on proving the FLT for n=3.


Unknown said...

Hi everyone!
This is Cheryl. Can any1 prove if n>3 then 2^n>n!? Thanks

Larry Freeman said...

Hi Cheryl,

In fact, the opposite is true. For n > 3, n! > 2^n

For n=3, 2^3 = 8 and 3! = 6

For n=4, 2^4 = 16 and 4! = 24

If n > 3 and n! > 2^n, then n*(n!) > 2*(2^n) so (n+1)! > 2^(n+1).


Joe said...

Hi Larry,

I found that your FLT blog is very intriguing. I have a question: How big will it be if someone finds an elementary proof to FLT?

Side note: Let me say this much, Pierre was right all along! Margin on his paper was not enough, it would take at least one full page.


Larry Freeman said...

Hi Joe,

Thanks for your question.

The proof by Andrew Wiles is over 100 pages long.

It is possible that there is a major insight that takes just 1 page but it is very unlikely.



Gangerolf said...

Hi Larry,

Please comment on the blogger Gangerolf's "proof".

Larry Freeman said...

Hi Gangerolf,

I posted a comment on your blog.

On this blog, I will only talk about the history of FLT.



neat_maths said...

Proof of Case I of Fermat’s Last Theorem for n = 3
The following is the most elegant solution imaginable.

(x+y+z)^3 can be neatly rearranged as follows;

(x+y+z)^3 = x^3 + y^3 + z^3 +

Therefore, assuming that x^3 + y^3 + z^3 = 0

(x+y+z)^3 = 3(z+x)(z+y)(x+y)

clearly, 3 only appears on the right hand side once as it cannot appear in (z+x) or (z+y) or (x+y) under case I.

This is because all prime factors of (x+y) must be contained in -z^3 and hence z.

Similarly, all prime factors of (x+z) must be contained in –y^3 and hence y

Also all prime factors of (y+z) must be contained in –x^3 and hence must be contained in x
and under this particular iteration of Case I, none of x, y and z is divisible by 3.

Therefore (z+x), (z+y) and (x+y) cannot be divisible by 3 if x, y and z cannot be divisible by 3

Therefore the right hand side can never be a cube, even if (z+x), (z+y) and (x+y) are all cubes

Therefore (x+y+z)^3 = 3(z+x)(z+y)(x+y) is impossible for case I
where 3 does not divide x, y, z

which means x^3 + y^3 + z^3 cannot equal 0

This proves Case I of Fermat’s Last Theorem for n=3

Unknown said...

Hi, this is Bisher, I wanted to know if A^3+B^3=C^3 is actually a possible equation with a positive integer answer.

Larry Freeman said...

By a well known proof, we know that the only integer answer is where abc=0.

So, that we could have for example:

a = 3, b = -3, c = 0 so that we have:

(3)^3 + (-3)^3 = 0^3

27 + -27 = 0

If abc != 0, then it follows that a,b,c cannot all be integers if the equation holds true.

Larry Freeman said...

A positive integer answer would have to be something where a=c such that:

a^3 + 0^3 = a^3


Joe said...

Hi Larry,

It's been 3 years since I made comments about Fermat might have it right about his comments.

I can prove it but that didn't mean it would be the same as Fermat had thought. Everybody were trying to find the root. But if the equation is invalid when tripletts are integers to begin with, then what is the point for looking for the root.

This year will be 375 years since he conjected. This number is nice be cause it is 3*5^3. I think it's time to reveal the riddle.

I believe that with your ability you probably could come up with a proof to FLT. Think outside the box: FLT is a mathematical false statement with integers.



Joe said...

Hi Larry,

Another hint, for x, y, z are all integers and power integer nth is greater than 2, then the equation

x^n + y^n = z^n

is as false as 1 = 2.


Joe Nilaad

Unknown said...

Sir i have a proof of Fermat's theorem for n=3 in new way and with the help of odd number techniques i prepare a table with the help of which we calculate the value of square of hypotenuse.

admin said...

nice blog :)

Unknown said...

Hi i am in a math research class and my topic is doing this theorem and i am reading this and i don't know where variables pq and gcd come from into the xyz

Larry Freeman said...

Hi Andy,

Click the link at that step to find out about p,q. It's explained in the lemma on the page of the link.

Unknown said...

Hi Larry,
If I had a proof for FLT where would I submit it?

Unknown said...

An interesting feature of x^n + y^n = z^n is when n=1, 0 and -1 given the right formula.

It will give values as if looking correspondingly along x, y and z and relative values of the other lines.

i.e. with x=3 and y=4 (giving z=5 then:
n=1 gives x=1.8 y=3.2 z=5
n=0 gives x=0.36 y=0.64 z=1
n=-1 gives x=0.072 y=0.128 z=0.2

Unknown said...

I admit that there is a basic premise of which I am unaware. Perhaps one being that the unknown must be a positive integer? X=1, Y= -1, makes Z = 0 and also solves for XYZ = 0. But I don't know why that is relevant. (2,-2), (3,-3), etc solve the equation.

Unknown said...

I admit that there is a basic premise of which I am unaware. Perhaps one being that the unknown must be a positive integer? X=1, Y= -1, makes Z = 0 and also solves for XYZ = 0. But I don't know why that is relevant. (2,-2), (3,-3), etc solve the equation.

GJ said...

I have proved fermat last theorem as ferma proved which is very easy method. Where should I public this ferma proof.

GJ said...

I have proof of fermet last theorem as ferma proved shortly.

olaleye oluwatosin said...

Hi, Larry, I want to ask: what is the conclusion of the infinite descent on n=3, is it that since we are talking about natural numbers, then infinite descent would not be applicable and thus we can conclude that there does not exist a solution to the FLT when n=3? Also in the case n=4, is the smaller solution n=2? If yes n=2 has a solution, so how does it work?
Thanks for your time.

MaNAtalia said...

Hi Larry
I need help with a problem. Knowing the result for FLT for n=3 I need to prove that if a positive integer n is divisible by 3 , then there are no x.y, z such that x^n+y^n=z^n.
What is did so far was say that if 3/n then 3=k.n, for some k positive integer. Then if (x^3, y^3,z^3) is a solution for the exponent k
and because 3.k=k.3 in integers
(x^k)^3+(y^k)^3= (z^k)^3

where x^k=a ,y^k,=b z^k=c are positive integers. Then there are no , a,b,c positive integers such that a^3+b^3=Z^3 .
Could that be remotely correct?

TK. said...

It really seems an elementary proof of Fermat's Last Theorem has finally been found !:

Ganpa4061 said...

Hallo evrybody,

Here is from where a simple proof will follow:

(1) $A^n= \sum_{1}^{A} (X^n-(X-1)^n)$

This by Telescoping Sum property, and it's a way to square any curve (the first, but also the follwing derivate) of the type $y'=nx^n$.

It can be aslo be seen in set Theory as an Ordinal Number.

This means that FLT can be rewritten as:

$C^n= \sum_{1}^{A} (X^n-(X-1)^n) + \sum_{1}^{B} (X^n-(X-1)^n$

but grouping the same terms of $A^n$ or $B^n$ will gives the Symmetric condition:

$C^n= 2* \sum_{1}^{A} (X^n-(X-1)^n) + \sum_{A+1}^{B} (X^n-(X-1)^n$


$C^n= 2* \sum_{1}^{B} (X^n-(X-1)^n) - \sum_{A+1}^{B} (X^n-(X-1)^n$

This condition imply also any $C^n/K$ is in bijection with other two rationals.

But since for the (1) follows that for $n>2$ the derivate is a curve, so the "balancing point" where the missin area bellow the curve it's equal to the exceding one, BOTH in X than in Y the balancing point IS NOT in the MIDDLE, follows Fermat is right.

To be more clear for the Tricotomic law, since we prove the distance of the balancing points always differs from the middle point for $n>2$, there cannot exist another Ordinal that state that such distance are equals.

Stefano Maruelli

Unknown said...

x^n+y^n=z^n ..?
have solution !
if n=1,2,3,....

how much solution ?

mionica said...

ALL n VIA Fermat-Murgu Impossible Equations
1. SENT all Fermat Equations Solutions in Irrational Field , without any doubts,
even for Z=Integers X,Y must to be Irrational.
2. Fermat-Murgu n Media Impassable for Fermat Triplets.
We Get it by Analyzing with Ion Murgu Math Millennium Equations all n neighbors
around of a supposed by absurd solutions (X,Y,Z).