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.
QED
29 comments:
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)].
-Larry
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
-Larry
Hi everyone!
This is Cheryl. Can any1 prove if n>3 then 2^n>n!? Thanks
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).
-Larry
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.
Joe
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.
Cheers,
-Larry
Hi Larry,
Please comment on the blogger Gangerolf's "proof".
http://gangerolf-gangerolf.blogspot.com/
Hi Gangerolf,
I posted a comment on your blog.
On this blog, I will only talk about the history of FLT.
Cheers,
-Larry
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 +
3(x+y)(y+z)(x+z)
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
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.
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.
A positive integer answer would have to be something where a=c such that:
a^3 + 0^3 = a^3
-Larry
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.
Cheers,
Joe
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.
Cheers,
Joe Nilaad
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.
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
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.
Hi Larry,
If I had a proof for FLT where would I submit it?
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
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.
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.
I have proved fermat last theorem as ferma proved which is very easy method. Where should I public this ferma proof.
I have proof of fermet last theorem as ferma proved shortly.
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.
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
(x^3)^k+(y^3)^k=(z^3)^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?
It really seems an elementary proof of Fermat's Last Theorem has finally been found !: https://www.quora.com/Can-you-verify-this-proposed-elementary-proof-of-Fermats-Last-Theorem-in-the-commentary-section-below-this-question?srid=1zDk
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:
$C^n= 2* \sum_{1}^{A} (X^n-(X-1)^n) + \sum_{A+1}^{B} (X^n-(X-1)^n$
and/or
$C^n= 2* \sum_{1}^{B} (X^n-(X-1)^n) - \sum_{A+1}^{B} (X^n-(X-1)^n$
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
x^n+y^n=z^n ..?
have solution !
if n=1,2,3,....
how much solution ?
Hi there-The PROOF OF PROOFS for n=3 - FERMAT-MURGU QUADRUPLETS
ARE A MATH BEAUTY IN CONCURRENCE NOW WITH PYTHAGOREAN TRIPLETS.
IT A REAL HUNTING FOR AND CONTAIN A COMPLETE MODULAR METHOD BY DEFINITION.
BUT ANYWAY WE SOLVED FERMAT'S LAST THEOREM WITH ABSOLUTE ACCURACY AND FOR
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).
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