It looks like the Russian article is the only one that gives the details of the proof. So, I am relying on this translation.

Let me start by saying that it is very important to be skeptical of anyone claiming to have a solution. In May, the Manilla Times stated that a local man found a mistake in the proof by Andrew Wiles. This of course turned out to be a hoax.

There are two tell-tales sign that the proof is most likely flawed. First, the professor who claims to have found the proof is not an expert in number theory. Second, the proof has not been reviewed by anyone who is an expert of number theory. From the mathematician Gabriel Lame to the very famous Leonhard Euler, it is very easy to make mistakes when it comes to number theory. Even Fermat made a mistake when it came to his conjecture about what are today known as Fermat primes.

Now, onto his proposed proof. After reviewing an English translation of the article, I found (no surprise) that the proof is not a valid demonstration of Fermat's Last Theorem.

First, let me summarize the proof:

(1) Assume that Fermat's Last Theorem is true.

(2) Then, there exists x

^{n}+ y

^{n}= z

^{n}with n ≥ 3.

(3) For any x,y, we can create a right triangle and so we can suppose a value r such that:

r

^{2}= x

^{2}+ y

^{2}.

(4) Since n ≥ 3, we know that z is less than r. [I will add the details for this later]

(5) We can construct a triangle based on z,x,y. Since z is less than r, we know that the angle opposite z (let's call it angle B) must be less than 90 degrees (and of course, greater than 0 degrees).

(6) Now, using the Law of Cosines, we know that:

z

^{2}= x

^{2}+ y

^{2}- 2xycosB.

(11) Since B is greater than 0 and less than 90, we know that cosB cannot be a whole number. [See here for a Cosine look up table]

QED

OK, so if the steps are valid, is Fermat's Last Theorem proven? Not at all. Just because cos B is not a whole number doesn't prove that z is not a whole number. That is ultimately what must be proven.

The professor must either prove that cos B is irrational or he must show that based on 2xycosB, z cannot be a whole number.

Since cos B is a continuous function, it necessarily covers rational values and therefore it is quite possible that 2xycosB is a whole number.

Thanks go out to David (see the comments section) for providing his thoughts.

## 9 comments:

About the Omsk mathematician's proof.

Me and my colleague read the proof carefully, and tried to

re-make it by our own:

----------------------------------------------------------------

We have X^n + Y^n = Z^n.

Hypothesis: X ingeger, Y integer, n > 2

Let's consider now a triangle formed of X, Y and Z...

and a square triangle X,Y,R where we know that the

angle between X and Y (we call from now B) is 90° and X^2+Y^2=R^2

We have that:

for n=2 B = 90° and Z=R

for n > 2 results that B < 90° and Z < R

B shrinks until the limit n-> (+inf)

in this case we'll have

Lim Z = MAX(X,Y).

n-> +inf

In the case X=Y will be Z=X=Y, meaning B = 60°,

but this is a limit. So we can safely proof 60° < B < 90°

Applying the basic formula for triangles then the proof says:

Z^2 = X^2 + Y^2 - 2XYcos(B)

where we have 0 < cos(B) < 0.5

Then the proof says: the last term isn't an integer, so Z^2 can't be integer too.

-----------------------------------------------------------------

BUT this is false because even if cos(B) isn't an integer, it can be a rational number: cos() is continuous, so it assumes all values in the range

0..0.5 Let's assume: X=3, Y=5, cos(B) = 0.1. Will be: 2XYcos(B) = 2*3*5*0.1=3

that means Z^2 can be integer and thus Z can possibly be integer too...

in a few words the proof is correct until the last formula:

Z^2 = X^2 + Y^2 - 2XYcos(B) where 60° < B < 90°.

Simply this formula seems to prove nothing... maybe there's another

piece of proof we're missing!

Cheers, David

Hi David,

You are right that the proposed proof does not take into account rational numbers.

If Z < R, then of course R * fraction = Z. Likewise, if the fraction is rational and can be canceled out by R, then Z is a whole number.

Even if the logic of his proof were to hold, it doesn't end up proving anything.

-Larry

Title: A simple proof of Fermat’s last theorem

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

Xp + Yp ?= Zp (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?

Author: Sinh Pham . Contact: sinh.pham65@yahoo.com

Title: A simple proof of Fermat’s last theorem

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

Xp + Yp ?= Zp (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?

Author: Sinh Pham . Contact: sinh.pham65@yahoo.com

The Relations of Barlow and FLT

Hi Guys

I have found The Relations of Barlow when I read Ribenboim's Book. "Playing" with it, I found another sistem of equations, and then, I used modular congruences and I found what is summarized below (we can work just with natural numbers to analize the fermat's integer soluctions,because if one or 2 of the numbers are negatives, them we can change the place in the equation in order to have just positive numbers. If al of them are negative, we can multiply the equation by -1):

I didn't make an English version*, because I am having dificulties in the mahts expression (loses its format when I paste here), but if you follow the steps below, then you will achiev the same results :

* I published the entire demonstration in a Blog I created (in this blog is better to read the maths equations) and in other forums in Brazil and others countries :

http://filosofbeer.blogspot.com.br/2015/04/o-ultimo-teorema-de-fermat-as-relacoes.html

From The Barlow Relations (See Fermat Last Theorem for Amateurs, pg 99 - but is very easy to show this) we have, in both Case 1 (n dont divide z) and Case 2 (n divide z) of FLT :

z-x = y0^n

z-y = x0^n

From this , we conclude thar

y-x = y0^n-x0^n = (y0-x0)(y0^(n-1)+....+x0^(n-1))

gdc(y-x, y0^(n-1)+....+x0^(n-1)) = y0^(n-1)+....+x0^(n-1)=p

y0^(n-1)+....+x0^(n-1) "=" 0 mod (p)

y-x "=" 0 mod (p)

Replacing x0 = (z-y)^(1/n) and y0 = (z-x)^(1/n)

[(z-y)^(n-1)]^(1/n)+....+[(z-x)^(n-1)](1/n) "=" 0 mod(p)

Replacing y "=" x mod (p)

n[(z-x)^(n-1)](1/n) "=" 0 mod(p)

ny0^(n-1)"=" 0 mod(p)

p = 1 or p divides y0^(n-1) or p=n

As x and y are both > 0, then y0^(n-1)+....+x0^(n-1) is always > 1. So p can not be 1

If p divides y0, then it will divide x. But gdc(x,y,z)=1, so p can not divide y0 or x0

If p=n, then

y0^(n-1)+....+x0^(n-1)= n

But this sum has n terms, and this is possible just if y0=1 and x0 = 1 , because both are > 0. And if one of them is larger then 1, then the sum wil be larger than n.

If x0=1 and y0=1, then using the Relations of Barlow, we conclude that x=y, wich contradicts te conditions z>y>x and gdc (x,y)=1.

The same analysis could be used in the case 2 (even if n divedes x ou y).

So, there's no integers soluction for the equation x^n+y^n = z^n

An English Version :

http://filosofbeer.blogspot.com.br/2015/04/the-relations-of-barlow-and-fermat-las.html

An English Version :

http://filosofbeer.blogspot.com.br/2015/04/the-relations-of-barlow-and-fermat-las.html

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