# Fermat’s Last Theorem

Fermat’s last theorem is one of the best known mathematical puzzles ever posed. It is very easy to understand yet it eluded a proof for 350 years. Fermat stated in the margin of Arithmetica that he had the most marvellous proof of the conjecture, but it was too long to fit in the margin. It has always been known as Fermat’s last theorem even though it has only been a conjecture for 350 years. Pierre de Fermat stated that

it is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second, into two like powers. I have discovered a truly marvellous proof of this, which this margin is too narrow to contain.

In other words

$\displaystyle a^n + b^n = c^n$

does not have solutions for n > 2.

For n = 2 there exist infinitely many solutions and we have been dealing with them in problem 9 of Project Euler.

The problem was finally solved in 1995 by Andrew Wiles after he dedicated 8 years struggling to prove the theorem. In order to prove the theorem he had to prove several other conjectures and not least use results and methods in many branches of mathematics developed within the last 100 years. So there is no way that Fermat could have proven it the same way as Andrew Wiles did. Wikipedia has a section on Fermat’s Last Theorem where they briefly go through the history and the content of the proof.

## My story on the theorem

I was first introduced to Fermat’s last theorem when I went in high school a mere 4 years after the theorem of proven. Our math teacher (whom I owe a lot of thanks for sparking my curiosity) wanted us to watch the movie on the subject made by Simon Singh and John Lynch. Many of my fellow students giggled at Andrew Wiles and thought he was a complete nut job, but I saw something different. I saw a man with a burning passion for solving this problem, and by the end of the movie I was so touched that I was almost crying. To me it was a real story and a treasure hunt for the truth.

This movie sparked something in me, and inspired me in many ways. I don’t claim to be good at mathematics and I am not rigorous enough to prove many things. But my passion and curiosity for math was ignited and will burn forever after this movie. It has been aired on television in many countries and until Google video was closed it was available through that service. Today it might be available through other means on the internet, but I haven’t found a source for it. If you a legal source for the movie I would be very interested in hearing from you.

Simon Singh is also the author of a book on the subject called Fermat’s Last Theorem. Simon Singh is a great story teller and manages to take the reader through the story in a way that most people can follow. The book takes you all the way from Fermat’s life and achievement and through the long history of Fermat’s Theorem which many people have spend many hours trying to prove without success until Wiles finally did his brilliant work and finally proved it. So if you love a good story I will highly recommend you to read the book and watch the movie.

On a side note Simon Singh has written other great books on different subjects such as The Code Book.

## But did he prove it?

The big question is; did he prove it? Not Wiles of course but Fermat – did he prove it? Most sources believe not. Wolfram has the very good argument that he later on looked for proofs of n=4 and n=5, which would have been meaningless if he had already proven it.

Due to my personal pride I hope and doubt he in fact did not prove it, because that means he would have had an insight that has eluded the rest of humanity for 350 years even though the mathematics has evolved incredibly since then.

### Posted by Kristian

w.johnson

I think Fermat did come up with the proof. It was probably something like this: Fermat’s equation is simply the Pythagorean identity times some factor c to the n. Sine squared theta and cosine squared theta are irrational via Niven’s theorem. So a and b to the n are irrational so they can’t be integers.

James Tomson
Huen Yeong Kong

Why Wiles’ proof need 100 pages. I cannot even under the first page.
Here is a proof I could understand:
A^n+B^n=C^n
If A^n and B^n are general symmetrical twin-primes, then globally:
1/2*(A^n+B^n)=C^n
But both sides must be integeral. Therefore
C = ( 1/2*(A^n+B^n))^(1/n)
But the right-side is impossible since the nth root of 2 is irrational. Therefore FLT is proved globally. Q.E.D

Huen Yeong Kong

Corrections:

Why Wiles’ proof need 100 pages. I cannot even under the first page.
Here is a proof I could understand:
A^n+B^n=C^n
If A^n and B^n are general symmetrical twin-compositess, then globally:
1/2*(A^n+B^n)=C^n
But both sides must be integeral. Therefore
C = ( 1/2*(A^n+B^n))^(1/n)
But the right-side is impossible since the nth root of 2 is irrational. Therefore FLT is proved globally. Q.E.D

Kristian

Without knowing what you mean with general symmetrical twin-composites, I really doubt that the proof holds. In fact as far as I can see, your assumption that you can pull out 1/2, shows an assumption you cannot make. And therefore your proof does not hold.

And if you really are correct, I think you should publish that rather in a peer-reviewed journal rather than here, because that would be a really feat to find such a simple proof for something that have puzzled mathematicians for 300 years.

Huen Yeong Kong

Here is my short proof of Fermat’s Lsst Theorem, please critize if
found wrong.

Volume 1 ( 2013 ) , Issue 2
Previous | Next | Back

A Short Proof of Fermat’s Last Theorem
Huen Yeong Kong
Pages 1-4
View Details Abstract References

The paper starts with a linear equation c = (a+b) where c, a and b are positive integers. Global integral equality is assured in this equation. Raising both sides of this equation to the nth power we get cn = (a+b)n which still retains global integral equality. If the right-side is expanded as Binomial Theorem, we get cn = an + unexpanded intermediate bionomial terms + bn. If the intermediate binomial terms could be reduced to zero, we get Fermat’s Last Theorem. But this is an impossibility since the right-side is uniformly additive with a and b as positive integers and n>2. This proves Fermat’s Last Theorem using only 17th centuary mathematics.

Kristian

You are right that if we get that all the intermediate terms can be removed we have Fermat’s last theorem. However, you have not shown that there isn’t a way to group all these terms into the form a^n + b^n. And therefore your proof does not hold in general.

Huen Yeong Kong

I think you have a point. I will think about it. Meanwhile if
anyone could improve on my paper, that would be welcome. Thanks

Huen Yeong Kong, Singapore

Huen Yeong Kong

Here is my attempt at an alternative proof of FLT:
For integral equality of FLT: z^n=x^n+y^n, let x=y then
z^n=2*x^n
Then z must be equal to 2^(1/n)*x for integral equality.
However 2^(1/n) is irrational, therefore 2^(1/n)*x is also irrational. Therefore there is no integral equality for FLT
based on FLT assertions.

Huen Yeopng Kong

Kristian

Hi Huen

Yep that would work for the special case where x = y, now you just need to show it for the infinity of cases where x != y.

I am sorry to sound rude, but I think you should stop these attempts. You wont find a proof for FLT like this. You should rather use your energy and skills on solving some other problems. I can’t guide you in the right direction here, you would need a mathematician for that.

Huen Yeong Kong

Yes, you are right. Should not spend too much energy on this topic now. But I did learn a lesson on Fermat’s strategy in setting up his conjecture. Start with a globally well-behaved equation, expand it,
and drop part of it. Then present the deformed formula to the world. It took 370+ year before Wiles solved his problem. Have a good day.
Huen Yeong Kong

Kristian

Yes, that is a very good point. Start from something you know for certain and then work towards what you want to prove. That is always a good approach.

Not the only approach, but a good one.

Huen Yeong Kong

Elementary number theory is simple. School leavers and amateur number theorist like me could comprehend statements of FLT and Goldbach’s Conjecture almost instantly. With modern number theory, things are not so simple. This group has grown inward cutting themselves off from school leavers and amateur number theorists. I am 82 years old. If I am younger I would like to be an activist to revive interest in elementary number theory. Even Wolfram gives only a scant one line remark on elementary number theory. There are still plenty of interesting ideas coming out from elementary number theory.
Huen Yeong Kong

I believe Fermat had a proof when he he made his famous statement that he had a marvelous proof. But he might found a mistake in it later.I believe his proof is based on a parametric solution of the equation of the theorem x^n+y^n=z^n.Still I could not derive what he had in his mind but the one ‘A simple and short analytical proof of Fermat’s last theorem’ which one can read on the internet(Published in CMNSEM)is close to it,to my mind.Within a short period of time I hope to derive a much shorter one which I believe the one that Fermat had in his mind.

Earl Bellinger

In other words

\displaystyle a^n b^n = c^n

does not have solutions for n > 2.

For n = 2 there exist finitely many solutions

I think you mean *infinitely* here.

Kristian

Yes I do, thanks for noting.

Kumaresan K

Thanks,I wish to share the following reg FLT. Consider any two positive
integers, I have taken 9 & 10, only a random choice.
9^1+10^1=19
9^2+10^2=181 -> (181)^1/2 a irrational number between 13 & 14,
9^3+10^3=1729 -> (1729)^1/3 a irrational number between 12 & 13,
9^4+10^4=16561 -> (16561)^1/4 a irrational number between 11 & 12,
9^5+10^5=159049 ->(159049)^1/5 a irrational number between 10 & 11,
Now what about nth root of 9^n+10^n for n>5?. The sum 9^n+10^n converges
to 10^n and nth root lies between 10 & 11, all are irrational Nos.
FLT for 9 & 10 verified.

Pham Duc Sinh

1. There is another explanation of a simple proof of Fermat’s last theorem as follows:

X^p + Y^p ?= Z^p (X,Y,Z are integers, p: any prime >2) (1)

2. Let‘s divide (1) by (Z-X)^p, we shall get:

(X/(Z-X))^p +(Y/(Z-X))^p ?= (Z/(Z-X))^p (2)

3. That means we shall have:

X’^p + Y’^p ?= Z’^p and Z’ = X’+1 , with X’ =(X/(Z-X)), Y’ =(Y/(Z-X)), Z’ =(Z/(Z-X)) (3)

4. From (3), we shall have these equivalent forms (4) and (5):

Y’^p ?= pX’^(p-1) + …+pX’ +1 (4)
Y’^p ?= p(-Z’)^(p-1) + …+p(-Z’) +1 (5)

5. Similarly, let’s divide (1) by (Z-Y)^p, we shall get:

(X/(Z-Y))^p +(Y/(Z-Y))^p ?= (Z/(Z-Y))^p (6)

That means we shall have these equivalent forms (7), (8) and (9):

X”^p + Y”^p ?= Z”^p and Z” = Y”+1 , with X” =(X/(Z-Y)), Y” =(Y/(Z-Y)), Z” =(Z/(Z-Y)) (7)

From (7), we shall have:

X”^p ?= pY”^(p-1) + …+pY” +1 (8)
X”^p ?= p(-Z”)^(p-1) + …+p(-Z”) +1 (9)

Since p is a prime that is greater than 2, p is an odd number. Then, in (4), for any X’ we should have only one Y’ (that corresponds with X’) as a solution of (1), (3), (4), (5), if X’ could generate any solution of Fermat’s last theorem in (4).

By the equivalence between X’^p + Y’^p ?= Z’^p (3) and X”^p + Y”^p ?= Z”^p (7), we can deduce a result, that for any X” in (8), we should have only one Y” (that corresponds with X’’ ) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.

X” cannot generate any solution of Fermat’s last theorem, because we have illogical mathematical deductions, for examples, as follows:

i)In (8), (9), if an X”1 could generate any solution of Fermat’s last theorem, there had to be at least two values Y”1 and Y”2 or at most (p-1) values Y”1, Y”2,…, Y”(p-1),
that were solutions generated by X”, of Fermat’s last theorem. (Please note the even number (p-1) of pY”^(p-1) in (8)). But we already have a condition stated above, that for any X” we should have only one Y” (that corresponds with X”) as a solution of (1),(7),(8),(9), if X” could generate any solution of Fermat’s last theorem.
Fermat’s last theorem is simply proved!

ii)With X”^p + Y”^p ?= Z”^p, if an X”1 could generate any solution of Fermat’s last theorem, there had to be correspondingly one Y” and one Z” that were solutions generated by X”, of Fermat’s last theorem. But let’s look at (8) and (9), we must have Y” = -Z”. This is impossible by further logical reasoning such as, for example:

We should have : X”^p + Y”^p ?= Z”^p , then X”^p ?= 2Z”^p or (X”/Z”)^p ?= 2. The equal sign, in (X”/Z”)^p ?= 2, is impossible.
Fermat’s last theorem is simply again proved, with the connection to the concept of (X”/Z”)^p ?= 2. Is it interesting?

Kristian

I must admit that I don’t follow you all the way through, but just the fact that you assume that p is a prime means that you have not proven Fermat’s last theorem. You might have proven something interesting, but FLT it is not.

Pham Duc Sinh

With X^n+Y^n ?= Z^n, while n>2, n =a.p, p is a prime >2, we can have:

(X^a)^p+ (Y^a)^p ?= (Z^a)^p.

The case n=4 has been proved (by Fermat), then we can have the same reasoning for any n in which 4 divides n.

Pham Duc Sinh

With X^n+Y^n ?= Z^n,n>2, while n=a.p, we can have:

(X^a)^p +(Y^a)^p ?= (Z^a)^p.

Fermat has proven the case n= 4, then we can applied Fermat’s last theorem for any n in which 4 divides n.

Pham Duc Sinh

With X^n +Y^n ?= Z^n, while n=ap, n>3, we can have
(X^a)^p+(Y^a)^p ?= (Z^a)^p p: prime >3.

The case n=4 has been proven (by Fermat), then FLT holds true for all n, in which 4 divides n.

Bert

Hi Kristian,

The big question regarding FLT is indeed, did Fermat prove it himself? You mentioned that Wolfram had a good argument that he didn’t, because otherwise why would Fermat later on in life try to proof n=4 and n=5? In general if someone asks a retorical question it means that he doesn’t want an answer, he just wants to settle the matter and having it his way by stopping the discussion. For me that’s not good enough because there’s a reasonable answer to it.

First let me point out the Fermat said that his proof was ‘mirabilem’. This is often translated as ‘marvellous’ (like you did) or something similar like ‘magnificent’. However, the correct translation of ‘mirabilem’ is ‘singular’ or ‘unique’. In other words, the proof which Fermat said he found was one of a kind and not like something one would normally expect. So deductive reasoning such as saying that generally speaking it would be impossible to proof much of anything based on the mathematical knowlegde of Fermat’s time, so Fermat didn’t proof FLT, conflicts with what Fermat claimed in the first place: his proof was out of the ordinary.

So if for a moment we assume that Fermat’s own alleged proof was correct, then we still have to ask why he later tried to proof n=4 and n=5. My answer to that would be (a) simply because of the fun of it, because after all he loved to solve riddles, and (b) to have a little confirmation for himself from another perspective that he had it right the first time.

There’s more that can be said about this, for instance that it is in fact childs play to proof that FLT is at least very plausible, but maybe it’s best to await your reaction first on what I’ve said so far. And by the way, there’s nothing wrong with some personal pride, but don’t overdo it 🙂

Kind regards,

Bert Brouwer
(Netherlands)

Allen Conti

My interest in the Fermat Conjecture, (FC,) began as an interest in the Pythagorean theorem. I wasn’t looking for other integral solutions to the n greater than 2 problem. I was more interested in the fact that odd integral values of ‘a’ resulted in ‘b’ being equal to (c-1.) Then, I began looking at the FC, and his statement, in the margin, that he had a solution, but it was too big to fit in the margin. I realized that with the mathematical tools in existence at that time, it was probably algebraic or geometric in expression. The solution proposed in 1995 that exceeded 120 pages, and used mathematics far in excess of what was available to Fermat was probably much more complex that needed. I began by using simple substitution to eliminate variables. It was determined that b= [(a to the 2)-1]/2, and c=[(a to the 2)+1]/2. Using an integral ‘n’, the problem became, “Is there a positive integer ‘n’, such that the equation: [(a to the n)-1]/n is an integer?” Although (a to the n) / n would result in an integral solution, because 1 is subtracted from the numerator prior to the division, there are NO positive integers for ‘a’ and ‘n’ greater than 2 that would result in b being positively non-fractional. COMMENTS? A.Conti, Allen@TheWalnutGrove.com or OWInc@att.net

Simplest proof of Fermat’s last theorem
Proof of the theorem
Fermat’s last theorem for n(>2) can be stated thus: There are non-trivial integers x,y,z satisfying the equation
z^n=y^n+x^n,(x,y)=1,n>2 (1)
Rearranging and relabeling the integers, we can assume without loss of generality that z>y>x>0. From (1), we obtain
g^n=h^n+1 (2)
where g=z/x h= y/x. From the equation(2), we obtain
〖(g-h)[g^(n-1)+g^(n-2) h+⋯……..+gh^(n-2)+h^(n-1)]=g〗^n-h^n=1, and if g-h=d , we have g^(n-1)+g^(n-2) h+⋯……..+gh^(n-2)+h^(n-1)=1/d>0 and therefore d>0.
d([(d+h)^(n-1)+h(d+h)^(n-2)+⋯…….+(d+h) h^(n-2)+h^(n-1)])=1 Hence,
d^n+nhd^(n-1)+⋯……+(n(n-1)/2) h^(n-2) d^2+nh^(n-1) d-1=0 (3)
Since h,d>0,
(n(n-1)/2) h^(n-2) d^2+nh^(n-1) d-10 (4)
(4) is a quadratic expression in d and is positive and therefore the discrininant should be negative ,
i.e n^2 h^(2n-2)+2n(n-1) h^(n-2)<0 (5)
which never holds and we conclude that (1) is not satisfied by non trivial integer triples x,y,z.