Fermat famously claimed to have a proof of his last theorem that he didn’t have room to write down. Mathematicians have speculated ever since what this proof must have been, though everyone is convinced the proof must have been wrong.

The usual argument for Fermat being wrong is that since it took over 350 years, and some very sophisticated mathematics, to prove the theorem, it’s highly unlikely that Fermat had a simple proof. That’s a reasonable argument, but somewhat unsatisfying because it’s risky business to speculate on what a proof must require. Who knows how complex the proof of FLT in The Book is?

André Weil offers what I find to be a more satisfying argument that Fermat did not have a proof, based on our knowledge of Fermat himself. Dana Mackinzie summarizes Weil’s argument as follows.

Fermat repeatedly bragged about the

n= 3 andn= 4 cases and posed them as challenges to other mathematicians … But the never mentioned the general case,n= 5 and higher, in any of his letters. Why such restraint? Most likely, Weil argues, because Fermat had realized that his “truly wonderful proof” did not work in those cases.

Dana comments:

Every mathematician has had days like this. You think you have a great insight, but then you go out for a walk, or you come back to the problem the next day, and you realize that your great idea has a flaw. Sometimes you can go back and fix it. And sometimes you can’t.

The quotes above come from The Universe in Zero Words. I met Dana Mackinzie in Heidelberg a few weeks ago, and when I came home I looked for this book and his book on the formation of the moon, The Big Splat.

**Related posts**:

Fermat’s unfinished business

ABC vs FLT

Feynman, Picard, and Fermat

I always thought the “didn’t have room to write down” was just a joke…

If only they had Post-It notes in Fermat’s time. He could’ve jotted his proof on one and stuck it to the page instead of fooling around with margins!

Yes. André Weil’s speculation feels certainly true.

Kermit

See this paper finding numbers not satisfying FLT.

http://iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf

The paper you link to does not give counterexamples to FLT. I haven’t read it closely, but the typos suggest it was not carefully reviewed.

I am trying to prove the impossibility of that counter example also.

A little Fermat Last Theorem proof here

http://one-zero.eu/resources/Riemann.pdf

Dear John D. Cook

Here is one of some simple proofs of Fermat’s last theorem:

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), 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), 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 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?

Take a look at this thread:

http://www.mymathforum.com/viewtopic.php?f=40&t=40857

You can go directly to the August 3th post with the improved version

http://iosrjournals.org/iosr-jm/papers/Vol7-issue4/A0740103.pdf

How could we get the wrong proof of FLT by given by Lame in 1847?