Fermat's Last Theorem; unacceptable proof, why?

[v4theory]

Wikipedia says Fermat's last theorem has the greatest number of failed proofs in history. I presume this simple "proof" is one of them. It must have been thought up before me. I first considered it years ago when I first heard of the problem. Figured it was so simple someone else must have thought of it before and disposed of it so I didn't really flesh it out then. My question is for anyone familiar with the history of the theorem is why this "proof" wouldn't be considered an acceptable proof.

First consideration is that all Pythagorean triples with a, b, or c greater than the square root of 2 are scale able by the same coefficient for a b and c that leaves c equal to the square root of 2. This disposes of all numbers greater than the square root of 2 for a b and c.

Second consideration is that when a=1 b=1 and c=sqrt2. As a approaches 0 b approaches sqrt2. Trivially when a=0 b=sqrt2 and c=sqrt2. As 0 isn't a positive number it isn't included in fermat's last theorem because Fermat says the theorem covers positive values for a b and c.

Third consideration is when a =1 then a to any exponent n =1 and b to n=1 and c (=sqrt2) to any n larger than 2 is going to be greater than 2 and therefore cannot be a to n+b to n=c to n when n>2.

Fourth consideration; as the first consideration says c=sqrt 2 divided by any scaling coefficient when c> sqrt2 then a to n + b to n must equal 2 else a squared + b squared would not equal c squared. So as a to n approaches 0 b to n approaches but is always smaller than c to n.

Therefore any a to n + b to n cannot equal c to n for any number a b and c or any n larger than 2. That's Fermats' last theorem in a nutshell.

So why is this "proof" not satisfactory.

If you have any questions about the above, something not clear, I could flesh it out some more for you. If you know if this "proof" has been previously published and disposed of I'd appreciate a direction to look into the disposition of this.

 

[PeroK]

First consideration is that all Pythagorean triples with a, b, or c greater than the square root of 2 are scale able by the same coefficient for a and b that leaves c equal to the square root of 2. This disposes of all numbers greater than the square root of 2 for a b and c.

What does this mean?

 

[v4theory]

What does this mean?

Which part? The scaling function? Or the fact that the scaling function allows all numbers for a b and c to be treated by the same formula as a b and c when c=sqrt2 divided by the scale coefficient for other a b and c Pythagorean triples therefore disposing of all Pythagorean triple sets with numbers for c larger or smaller than sqrt2?

 

[PeroK]

Which part? The scaling function? Or the fact that the scaling function allows all numbers for a b and c to be treated by the same formula as a b and c when c=sqrt2 divided by the scale coefficient for other a b and c Pythagorean triples therefore disposing of all Pythagorean triple sets with numbers for c larger or smaller than sqrt2?

Can you put that into mathematics?

 

[fresh_42]

You've gone from integers to reals but you didn't come back. There is no problem with real numbers. And what should

So as a to n approaches 0 b to n

mean? $n$ approaches infinity, not zero. If I remember correctly, FLT has been known up to $4,003$ before Wyles by more or less elementary proofs. The question was, what's above, not what's at zero. If $n \to 0$ we get $1+1=1$ which is wrong. I assume that your considerations rely on this simple fact. And the case $a^n,b^n \to 0$ with a fixed $c$ makes no sense. You cannot have any equation and let the LHS run to zero and keep the RHS fixed. With that you can prove whatever you want to.

 

[v4theory]

You've gone from integers to reals but you didn't come back. There is no problem with real numbers. And what should

mean? $n$ approaches infinity, not zero. If I remember correctly, FLT has been known up to $4,003$ before Wyles by more or less elementary proofs. The question was, what's above, not what's at zero. If $n \to 0$ we get $1+1=1$ which is wrong. I assume that your considerations rely on this simple fact. And the case $a^n,b^n \to 0$ with a fixed $c$ makes no sense. You cannot have any equation and let the LHS run to zero and keep the RHS fixed. With that you can prove whatever you want to.

You understand the issue of Fermat's Last theorem is about numbers a b and c that are in Pythagorean triples (P trips), right triangles? Not just any numbers a b and c pulled out of a hat? That's why the scaling function disposes of all numbers above and below sqrt2 that are in P trips. Multiplying one side of a triangle multiplies the other sides by the same number. For instance, as convention is c is the larger number of a P trip (the hypoteneuse of a triangle), then take c as 2 times sqrt2 (2x1.41421=2.82843) then a and b must be multiplied by 2. Then a squared + b squared must equal 8 because c squared is 8. So the formula that solves for c=sqrt2 also solves for any number in P trips larger or smaller than sqrt2. Get it?

 

[PeroK]

You understand the issue of Fermat's Last theorem is about numbers a b and c that are in Pythagorean triples (P trips), right triangles? Not just any numbers a b and c pulled out of a hat? That's why the scaling function disposes of all numbers above sqrt2 that are in P trips. Multiplying one side of a triangle multiplies the other sides by the same number. For instance, as convention is c is the larger number of a P trip (the hypoteneuse of a triangle), then take c as 2 times sqrt2 (2x1.41421=2.82843) then a and b must be multiplied by 2. Then a squared + b squared must equal 8 because c squared is 8. So the formula that solves for c=sqrt2 also solves for any number in P trips larger or smaller than sqrt2. Get it?

You have shown that if $a^2 + b^2 = c^2$ then $a^n + b^n \ne c^n$. But that certainly isn't Fermat's last theorem, which is for any whole numbers $a, b, c$ "pulled out of the hat", not just Pythagorean triples.

 

[v4theory]

You have shown that if $a^2 + b^2 = c^2$ then $a^n + b^n \ne c^n$. But that certainly isn't Fermat's last theorem, which is for any whole numbers $a, b, c$ "pulled out of the hat", not just Pythagorean triples.

No. Because any set of numbers pulled out of a hat not in Pythagorean triples won't satisfy the equation a squared + b squared = c squared. For instance 1 2 and 3 pulled out of a hat for a b and c. 1 squared =1, 2 squared=4 and 3 squared= 9. Obviously 1 squared + 2 squared does not equal 9.

 

[PeroK]

No. Because any set of numbers pulled out of a hat not in Pythagorean triples won't satisfy the equation a squared + b squared = c squared. For instance 1 2 and 3 pulled out of a hat for a b and c. 1 squared =1, 2 squared=4 and 3 squared= 9. Obviously 1 squared + 2 squared does not equal 9.

I knew you were going to say that! Well, I guess there's nothing I can say that will convince you that you have misunderstood FLT.

Aren't you at all suspicious that a theorem that remained unproved for over 300 years is so easy to prove?

Anyway, for the benefit of the rest of us, are you able to state FLT?

 

[QuantumQuest]

You understand the issue of Fermat's Last theorem is about numbers a b and c that are in Pythagorean triples (P trips), right triangles? Not just any numbers a b and c pulled out of a hat? That's why the scaling function disposes of all numbers above and below sqrt2 that are in P trips. Multiplying one side of a triangle multiplies the other sides by the same number. For instance, as convention is c is the larger number of a P trip (the hypoteneuse of a triangle), then take c as 2 times sqrt2 (2x1.41421=2.82843) then a and b must be multiplied by 2. Then a squared + b squared must equal 8 because c squared is 8. So the formula that solves for c=sqrt2 also solves for any number in P trips larger or smaller than sqrt2. Get it?

Fermat's equation $x^n + y^n = z^n$ is a Diophantine equation. Now, if $n = 2$ you have the quadratic Diophantine equation $x^2 + y^2 = z^2$ which has solutions given by the Pythagorean triples. When $n$ takes higher values you have different kinds of Diophantine equations. What have Pythagorean triples to do with these? Are you claiming that any kind of Diophantine equation has necessarily Pythagorean triples as solutions?

 

[v4theory]

I knew you were going to say that! Well, I guess there's nothing I can say that will convince you that you have misunderstood FLT.

Aren't you at all suspicious that a theorem that remained unproved for over 300 years is so easy to prove?

Anyway, for the benefit of the rest of us, are you able to state FLT?

"It is impossible (snip) in general for any number that is a power greater than the second to be the sum of two like powers".

That involves 3 numbers and a power. Since any 3 numbers define a triangle, and any triangle can be divided into 2 right triangles it has long been understood that solving FLT for right triangles solves for all triangles and therefore all numbers and "powers greater than the second".

By the way FLT doesn't say anything about whole numbers as you stated earlier.

 

[PeroK]

By the way FLT doesn't say anything about whole numbers as you stated earlier.

Let $a = 1, b = 1, c = 2^{1/3}$, then:

$a^3 + b^3 = c^3$

Of course FLT has to be talking about whole numbers!

 

[fresh_42]

That involves 3 numbers and a power. Since any 3 numbers define a triangle...

They do not.

... and any triangle can be divided into 2 right triangles it has long been understood that solving FLT for right triangles solves for all triangles and therefore all numbers and "powers greater than the second".

No.

By the way FLT doesn't say anything about whole numbers as you stated earlier.

Obviously whole numbers are something else to you than to the rest of us.

I knew you were going to say that! Well, I guess there's nothing I can say that will convince you that you have misunderstood FLT.

Agreed.

Anyway, for the benefit of the rest of us ...

... I'll close this thread now. We generally do not debunk obviously false statements, although the will was there, unfortunately not the language, not to say manners.