FLT-Solution for prime values of n .

[vantheman]

FLT-Solution for prime values of "n".

The proof is posted in my journal. It has been blessed by two Math academics. Take a look. By the way, Victor was very close.

 

[HallsofIvy]

1. I didn't know "Math academics" blessed anything! I thought that was left to priests.

2. I don't have access to your journal.

3. I have no idea who "Victor" is.

 

[Hurkyl]

Your abstract alone makes me quite skeptical -- you don't even claim to have a proof, you merely claim that your goal was a proof.

The introduction makes me even more skeptical -- it doesn't even mention the possibility that you are providing a proof, and the main thing it seems to attempt is only mentioned as a "possibility".

Let us assume in equation (5) that x has three relatively prime factors,
a, b and c, so that
`
x^n = (abc)^n``````````````````````````````````````````` ````(7)

Sure, though it isn't necessarily of any use; for instance, a = 1, b = 1, and c = x satisfies this assumption, although it is entirely uninteresting.

Since we know that x^n is divisible by k, let us further assume that
`
k = [b^(n-s)][c^(n-t)]````````````````````````````````````````(8)
Where s < n, t < n; therefore

But I see no reason why we could make this assumption.

Anyways, continuing on for the sake of argument...

nz[ f(z,k) ] = (a^n)(b^s)(c^t) - [b(n-s) c(n-t)]^(n-1)````````````(10)
`
Now, we can see that the second and third terms of equation (10) are both divisible by (b^s)(c^t), therefore nz[ f(z,k) ] would also have to be
divisible by this term. Unless s and t are equal to zero, all three terms of
the equation are divisible by at least bc.

What if s is zero but t is not?

However, bc, which is a factor of x and k, cannot be a factor of z or [ f(z,k) ] without violating the condition that equation (10) can have no common factors. So, in this case, the only way to eliminate the problem of common factors is for s and t to be equal to zero. Therefore,

Thus far, you've given no reason to think that bc cannot divide f(z, k).

Let's set x = (x')(x"), where x' = a and x" = b

This directly contradicts your previous assumption that sets x = abc. (unless c = 1)

2.4 Inserting the key
Since we know that (x + y) = (z")^n, and z" is a factor of z, we can say that

Why would we know that?

 

[vantheman]

1. I didn't know "Math academics" blessed anything! I thought that was left to priests.
2. I don't have access to your journal.
3. I have no idea who "Victor" is.

Victor is username "victor.sorokine" who last month posted numerous attempts at proofs of Fermat's Last Theorem.

To have access to my journal, just click on "read my journal" under my username.

You are right. Science and religion aren't interchangable.

Unfortunately, the journal format does not support LaTeX typesetting. If you do take the time to look at the contents of my journal and want a copy in PDF format, with proper mathematical notation, it will be sent.

 

[vantheman]

Your abstract alone makes me quite skeptical -- you don't even claim to have a proof, you merely claim that your goal was a proof.
The introduction makes me even more skeptical -- it doesn't even mention the possibility that you are providing a proof, and the main thing it seems to attempt is only mentioned as a "possibility".
Sure, though it isn't necessarily of any use; for instance, a = 1, b = 1, and c = x satisfies this assumption, although it is entirely uninteresting.
But I see no reason why we could make this assumption.
Anyways, continuing on for the sake of argument...
What if s is zero but t is not?
Thus far, you've given no reason to think that bc cannot divide f(z, k).
This directly contradicts your previous assumption that sets x = abc. (unless c = 1)
Why would we know that?

Since I don't claim to be a mathematician, it is quite possible that I have made errors that invalidate the proof that I am attempting. That's why I'm seeking review by mentors such as you.

Since k must divide x, x cannot be prime and therefore must have at least two factors greater than 1. - If you can follow my logic to the end, s and t must both be zero or the three terms of the equation will have a common factor. - bc cannot divide f(z,k) because bc and k have common factors, but bc and z cannot have common factors or the initial condition that x, y and z have no common factors is violated.
Have to sign off now but will return tomorrow. Thanks for the comments.

 

[ramsey2879]

Quote:

2.4 Inserting the key
Since we know that (x + y) = (z")^n, and z" is a factor of z, we can say that

Why would we know that?

The above argument has been made by first showing that (x+y) does not divide Z^n/(x+y). Posters then conclude that the GCD of (x+y) and (Z^n)/(x+y) must equal 1, but this is not so. For instance, the factor (x+y) could equal (z'')^i where n/2 < i < n, and that the other factor include (z'')^(n-i) as a factor. Thus the assumption is not valid.

 

[vantheman]

The above argument has been made by first showing that (x+y) does not divide Z^n/(x+y). Posters then conclude that the GCD of (x+y) and (Z^n)/(x+y) must equal 1, but this is not so. For instance, the factor (x+y) could equal (z'')^i where n/2 < i < n, and that the other factor include (z'')^(n-i) as a factor. Thus the assumption is not valid.

I have studied the above for two days and fail to see your point. Perhaps you couild re-phrase it. I fail to see that no matter how you factor (x+y) you cannot prove that (x+y) and (z^n)/(x+y) have common factors, since the latter (z")^n = [(z')^n (z")^n)]/(x+y) and since (x+y)=(z")^n it reduces to (z")^n. We go the full circle.

Hurkey's comment on Para. 2.4 - "Why would we know that?"
I will cover that this PM

 

[vantheman]

Your abstract alone makes me quite skeptical -- you don't even claim to have a proof, you merely claim that your goal was a proof.
The introduction makes me even more skeptical -- it doesn't even mention the possibility that you are providing a proof, and the main thing it seems to attempt is only mentioned as a "possibility".
Sure, though it isn't necessarily of any use; for instance, a = 1, b = 1, and c = x satisfies this assumption, although it is entirely uninteresting.
But I see no reason why we could make this assumption.
Anyways, continuing on for the sake of argument...
What if s is zero but t is not?
Thus far, you've given no reason to think that bc cannot divide f(z, k).
This directly contradicts your previous assumption that sets x = abc. (unless c = 1)
Why would we know that?

"Why would we know that?"
If we can agree on the validity of the derivation of the equation
(z-y) = k = (x")^n, then, based on symmetry, (z-x) = l = (y")^n and
(x+y) = m = (z")^n
If we can't agree that (z-y) = k = (x")^n then let's consider the case where n = 3.
3zyk = x^3 - k^3 Since x^3 and k must have common factors let's assume x has three factors such that x^3 = (abc)^3
We can assume that x has only two factors or multiple factors, but for illustrative purposes let's just assume three.
Let's further assume that a does not have a common factor with k but b and c do have common factors with k. Let's assume further that k = bc.
Then, factoring out k, 3zy = (a^3)(bc)^2 - (bc)^2
Since x, y and z can have no common factors, bc, which is a factor of x, cannot be a factor of z or y; therefore the assumption that k = bc is invalid. If we assume k = (bc)^2 or bc^2 or cb^2, we can see that these assumptions are all invalid. The only assumption that does not violate the initial condition that x, y and z have no common factors is if k = (bc)^n

Since x>k, x must have at least two factors, one of which is the product of all the factors of x that are not common factors of k (let's call this x') and the other is the product of all the factors of x that are common factors of k (let's call this x"). In the example (x")^n = (bc)^n = k.

"What if s is zero and t is not?"
Let's use the n=3 example, then k = (b^3)c or (b^3)(c^2). In the latter case 3zy(b^3)(c^2) = (abc)^3 - [(b^3)(c^2)]^3
If we factor out k
3zy = (a^3)c - [(b^3)(c^2)]^2
Here again, c is a factor of the right hand side of the equation and a factor of x, so it should be a factor of the left side, but this is not possible without violating the initial condition prohibiting common factors.

 

[ramsey2879]

I have studied the above for two days and fail to see your point. Perhaps you could re-phrase it. I fail to see that no matter how you factor (x+y) you cannot prove that (x+y) and (z^n)/(x+y) have common factors...

One point was that just because two numbers don't divide each other dosn't mean that they have no common prime divisor.
As to factoring (x+y), that can't be done. Many binominals have no algebraic factors but that does not mean that they are prime for all values of x and y. Likewise just because two polynominals don't divide algebraically doesn't mean that for certain values of x and y that there is no common prime factor.

But more importantly, you seem to be missing my point, which is that you have not proven that given x^n + y^n = z^n, that for all values of x and y that solve the above equation where x and y are coprime, that

(x+y) and \frac{z^n}{x+y} can have no common prime factor.

It is not for me to prove that they can. It is for you to prove that they can't if your proof is to have any merit.

Cheers

 

[JasonRox]

The world of mathematics is not like the business world.

You can't just have an idea and get others to prove it, then get credit for it.

 

[ComputerGeek]

The proof is posted in my journal. It has been blessed by two Math academics. Take a look. By the way, Victor was very close.

Why are you even bothering? IT has been proven... Ever hear of Wiles.

 

[ComputerGeek]

The proof is posted in my journal. It has been blessed by two Math academics. Take a look. By the way, Victor was very close.

Oh.. and do us all a favor, LEARN LATEX!

 

[shmoe]

Why are you even bothering? IT has been proven... Ever hear of Wiles.

Finding a proof of a theorem is no reason to stop looking for alternate proofs, They may provide more insight to the problem. The pythagorean theorem has many different proofs, Gauss proved quadratic reciprocity some 8 different ways, Riemann proved the functional equation two different ways in the same paper, etc.

 

[ComputerGeek]

Finding a proof of a theorem is no reason to stop looking for alternate proofs, They may provide more insight to the problem. The pythagorean theorem has many different proofs, Gauss proved quadratic reciprocity some 8 different ways, Riemann proved the functional equation two different ways in the same paper, etc.

quite true, but he is not even starting with the wiles proof as a starting off point.

would it not be more productive to start with wiles and simplify parts until the proof is simplified?

starting from nothing is pointless, especially when you are retreading number theory methods that have been thought through many times.

 

[JasonRox]

quite true, but he is not even starting with the wiles proof as a starting off point.
would it not be more productive to start with wiles and simplify parts until the proof is simplified?
starting from nothing is pointless, especially when you are retreading number theory methods that have been thought through many times.

Who's to say you can't prove it using entirely different areas of Number Theory?

There are many proofs for certain theorems that are proven with completely different methods.

 

[shmoe]

quite true, but he is not even starting with the wiles proof as a starting off point.

So?

would it not be more productive to start with wiles and simplify parts until the proof is simplified?

Simplifying an existing proof vs. coming up with a new one are quite different things.

starting from nothing is pointless, especially when you are retreading number theory methods that have been thought through many times.

While I think it's very unlikely that anyone will find an "easy" proof of fermat's last theorem, this doesn't mean it's impossible. As long as it's done with an open mind, discovering the many things that don't work can be instructive and hardly what I'd call pointless.

 

[matt grime]

Proofs in special cases can beinteresting. The proof for cubes for instance (or even quartics) of FLT are of interest and only a foolw wouold dismiss them or try to invoke Wiles's proof in these cases. Not that you should think i imagine this proof has interest or is even remotely correct.

 

[vantheman]

Why are you even bothering? IT has been proven... Ever hear of Wiles.

Wiles' proof is not Fermat's proof, if you believe Fermat's claim to have a proof. Wouldn't it be nice to find a proof that you could teach to a high school algebra class?

I tried to use LATEX in my journal. It didn't work. I have been told that the journal will not support LATEX. True or false?

 

[matt grime]

But no one believes Fermat had a proof (that was correct). He even felt the need to later give a proof for the special case of n=4, so it is unlikely he had a general correct solution. Plausible proofs abound if one assumes all quadratic extensions have unique factorization. Indeed, given the use of the word proof in that era of mathematics, there is no reason to suppose he is claiming to know how to prove it in the modern sense. For example Fermat claimed a theorem by induction that the primes represented by certain binary forms were as we now know they are. The 'proof' was a series of (non-exhaustive) examples.

Of course, a proof for (odd)primes is a proof of the full FLT, hence I am happy to think you haven't got an elementary proof.

 

[vantheman]

One point was that just because two numbers don't divide each other dosn't mean that they have no common prime divisor.
As to factoring (x+y), that can't be done. Many binominals have no algebraic factors but that does not mean that they are prime for all values of x and y. Likewise just because two polynominals don't divide algebraically doesn't mean that for certain values of x and y that there is no common prime factor.
But more importantly, you seem to be missing my point, which is that you have not proven that given x^n + y^n = z^n, that for all values of x and y that solve the above equation where x and y are coprime, that
(x+y) and \frac{z^n}{x+y} can have no common prime factor.
It is not for me to prove that they can. It is for you to prove that they can't if your proof is to have any merit.
Cheers

I agree. You are right.
I had two approaches to a solution to the problem. I chose one that had a fatal flaw. I am re-writing from Para. 2.4 on. Expect to post it next week. I believe the intermediate proof that is in my journal that
(z-y)^(1/n), (z-x)^(1/n) and (x+y)^(1/n) are all integers is the key to the final proof.

 

[ramsey2879]

I agree. You are right.
I had two approaches to a solution to the problem. I chose one that had a fatal flaw. I am re-writing from Para. 2.4 on. Expect to post it next week. I believe the intermediate proof that is in my journal that
(z-y)^(1/n), (z-x)^(1/n) and (x+y)^(1/n) are all integers is the key to the final proof.

I have been trying to extend your proof for n=3 to all n myself but havn't had time to do so. I am happy to know that you agree with me and await your results.
P.S. in Tex format you should enclose your subscripts and superscripts within "{" "}" as {\left(\frac{1}{n}\right)}. The \left( ... \right) combination also sizes the parentheses to match the contents within but I am not familiar enough with tex yet to know if they are required for the parentheses here.

 

[vantheman]

I have been trying to extend your proof for n=3 to all n myself but havn't had time to do so. I am happy to know that you agree with me and await your results.
P.S. in Tex format you should enclose your subscripts and superscripts within "{" "}" as {\left(\frac{1}{n}\right)}. The \left( ... \right) combination also sizes the parentheses to match the contents within but I am not familiar enough with tex yet to know if they are required for the parentheses here.

I am not there yet, but let me review where I am and where I'm going.
So far I think my paper shows that, if an integer solution to FLT exists, then the nth roots of (z-y), (z-x) and (x+y) are all integers; that neither x, y or z have a factor of n and therefore the classical 'Case 2' of FLT is impossible.
In equation (5), x must be congruent to k mod n but neither can be congruent to n mod n. If we set x = nd + e and k = nf + e, where e determines the congruence of x and k, and then we substitute into (5) and subtract, the right side of the equation will be congruent to n^2 mod n. The only way the left side can be congruent to n^2 mod n is if f(z,k) is congruent to n mod n, since neither z nor k can be congruent to n mod n. Due to the symmetry of equations (5), (15) and (19), f(z,l) and f(x,m) must also be congruent to n mod n.-------Now, if we look at (13), Fermat's Little Theorem tells us that k^(n-1) is congruent to 1 mod n, therefore x' must be congruent to 1 mod n. Again by symmetry, y' and z' must also be congruent to 1 mod n. What this says is that
(z^n - y^n)/(z-y) , (z^n - x^n)/(z-x) and (x^n + y^n)/(x+y) are always congruent to 1 mod n. Interesting?

That's all for now. Thought you might want to take it further.

PS - I sent you my n=4 proof by email attachment.

 

[vantheman]

To continue, if n=3 then f(z,k) = (z-k) = y. We have established that f(z,k) must be divisible by n, but we have also established that y is not divisible by n - thus we have a contradiction. Is this not a proof of FLT for n=3?