New approach to FLT Proof for prime powers of n

[robert80]

Dear all, here is the new approach how to prove the Fermats Last theorem for the prime powers of n. Thank you all that you have mentioned the Diophantine equations.
The proof has still one missing link. It should be proved that l is coprime to (c-b) and the same kind of proof should arise for (a+b) respectively. All in all its an interesting approach, cause it includes triangle rules and Diophantine equations. Please help me to improve this one.



ABSTRACT:

For the past few centuries, there was always doubt if the elegant solution to the Fermat¨s Last Theorem exists. Although proven in the 90s the proof itself did not answer the question whether there is more simple solution to this problem. The following proof is the proof to Fermat¨s Last theorem for the prime powers of n. It contains applying triangle rules, coprimety and they both result in 2 unsolvable Diophantine equations. 1 for general case and the other for c – b = 1.

THE PROOF:

Let us suppose the solution to the Equation

a^n + b^n = c^n (1) exists and a,b,c are positive integers, coprime to each other and n is a prime number.

Let us now rearange the equation(1):
a^n = (c-b)*(c^(n-1) + c^(n-2)*b...+ b^(n-1)) (2)

since from the solution when exist, we can form a Fermats triangle from the positive integers a,b,c, those triangle rules apply:

a > c - b, b > c - a and 2c > b + a
.
so we can easily see that c – b is coprime to one part of a AND is not coprime to the other part of a if we factorize a into prime factors in the case when c – b does not equal to 1. When c – b equals to 1, there will be a second prof applied. The same apply for b respectively. And since b is even, we see that c – a is never 1. Let us first prove the case when c – b does not equal to 1.


If we rearrange the (2) into:

a^n = (c-b)*((c-b)^(n-1) + n*c*b*l)) (3) l being the positive integer. Since b, c and l are coprimes to c – b the only 2 cases apply : n is coprime to (c – b) or n is not coprime to (c – b).

lets check the two cases, when n is not coprime to (c – b)

n = c – b (1*) or
n*m = c – b (2*) m being the positive integer, BUT the case when
(n = (c – b)* k k being the positive integer does not apply, since n is prime).




If we apply (1*) in (3) we see that (c - b)^2 part is square AND
(c – b)^2 does not divide a^n since n is prime. So we got a contradiction. The second (2*) case when applied would give the contradiction too. It follows that if (c – b) = n*m when we apply this in (3) we get that n^2 divides a^n. This does not hold true since n is prime so the only remaining possibility would be that n^(n-2) is hidden in the c*b*l but we know that l is coprime to (c – b) and so are c and b, so this could not be the case..
We see now that in order that the solution exists (c - b) has to be coprime to n and since (c - b) is coprime to l,


******?When n is prime, a + b divides (1) and the very same thinking with coprimety of a + b to what is left when a^n + b^n = c^n is divided by a + b we can see again that a + b = c1 ^n. This we will apply in the second part of the proof. Since 2c > a + b > c we see that a + b does not include all the factors of c. So when the solution exist c is not a prime number.********

IT FOLLOWS (c – b) MUST take the form of (a1)^n when (c – b) does not equal to 1.

c – b =(a1)^ n where a1 is one part of a and the another part of a is a2. And a2 is coprime to a2 respectivelly. This part a2 is hidden in the ((c-b)^(n-1) + n*a*b*l)) for the coprimety reason and if we see a as being factored to primes and a > c – b so a = a1*a2
In order to get the solutions a is odd, c is odd and respectively b is even.
The same conclusions come for (c – a) respectively:

b^n = (c-a) (c^(n-1) + c^(n-2)*a...+a^(n-1)) (4)...

The same conclusions come from c^n, since when n is prime, is divisible by a + b. (this we will apply in the special case of proof when c – b = 1



SO WE GET NOW 2 STRONG CONDITIONS IN ORDER TO GET THE SOLUTION(S) FOR THE a,b,c being positive integers a, b, c forming a Fermat¨s triangle, n being prime and (c-b) being coprime to n and (c-b) does not equal 1.

These 2 are:

(c – b) = (a1)^n and (c – a) = (b1)^n (5)

So, when these 2 combined we get:
(a1)^n + b = (b1)^n + a (6)
b = b1*b2 and a = a1*a2
b and a are factored to primes so that a1 and a2 are coprimes and b1 and b2 are coprimes. It follows that a1, a2, b1 and b2 are coprimes.
so
b1*b2 – a1*a2 = (b1)^n – (a1)^n (7)

b1*b2 – a1*a2 = (b1 – a1) * r (8) r being the positive integer. The resulting linear Diophantine equation never has the solution when b1, b2, a1 and a2 are coprimes to each other OR to say another way, we get the contradiction with coprimarity from the beginning of the proof.



* The proof when c – b = 1*
c – b = 1
c – a = b1^n
2*c = a + b + 1 + b1^n

since a + b = c1^n
2*b + 2 = a + b + 1 + b1^n

2*b + 1 = c1^n + b1^n

since 2*b + 1 equals to c + b when c – b equals to 1 we get:

c + b = c1^n + b1^n

since c + b divides c1^n* c2^n + b1^n* b2^n it never divides c1^n + b1^n
so we again got a contradictory.

The end of the proof.

 

[robert80]

I mean I find it at least interesting. It could be the Fermats proof goes this way: applying triangle rules showing coprimarity and ending with 2 Diophantine equations. I have a good feeling the idea how to prove it is correct, but since I am the Math enthusiast only. I would ask the Mathematicians, to prove the coprimarity more precisely. I am not skilled in Math.

 

[robert80]

REVISED:


SO WE GET NOW 2 STRONG CONDITIONS IN ORDER TO GET THE SOLUTION(S) FOR THE a,b,c being positive integers a, b, c forming a Fermat¨s triangle, n being prime and (c-b), (a+b), (c-a) being coprime to n and (c-b) does not equal 1.

And the last proof: c + b = c1^n + b1^n

since c + b divides c1^n* c2^n + b1^n* b2^n and since c1^n + b1^n never divides c1^n* c2^n + b1^n* b2^n
we again got a contradictory.

I will put it into more readable form with some math application soon. Anybody who would like to help me with this proof could contact me through pm.

 

[ramsey2879]

REVISED:


SO WE GET NOW 2 STRONG CONDITIONS IN ORDER TO GET THE SOLUTION(S) FOR THE a,b,c being positive integers a, b, c forming a Fermat¨s triangle, n being prime and (c-b), (a+b), (c-a) being coprime to n and (c-b) does not equal 1.

And the last proof: c + b = c1^n + b1^n

since c + b divides c1^n* c2^n + b1^n* b2^n and since c1^n + b1^n never divides c1^n* c2^n + b1^n* b2^n
we again got a contradictory.

I will put it into more readable form with some math application soon. Anybody who would like to help me with this proof could contact me through pm.

I would be glad to help you but I am of the opinion that the "rules for Fermat's 'right' triangle" do not apply to Fermat's theorem in any way. So where is the contradiction?

 

[robert80]

I would be glad to help you but I am of the opinion that the "rules for Fermat's 'right' triangle" do not apply to Fermat's theorem in any way. So where is the contradiction?

Fermats triangle is not right angled. We know nothing about angles. What we know is that is not right angled for sure. Its just clear that if a solution(s) exist we can form a triangle from (a,b,c) being positive integrers and that's enough for the proof. I just called it Fermats triangle.

 

[robert80]

And Ramsey I have a good feeling about this approach. Could you please prove that (c-b) is coprime to l in equation (3)? I guess the (c-b) parts should be formed on even numbers, like in (3) in brackets, so finaly we get ((c-b)^2 +c*b*j) so when we are done its just necessary to prove that (c-b) and j are coprimes. I know its hard to take me seriously, concerning all the mistakes in former proofs, but before few months I knew very little about this problem. I am sorry for my impulsiveness concerning this forum, but most of the guys were really patient with me.

 

[ramsey2879]

And Ramsey I have a good feeling about this approach. Could you please prove that (c-b) is coprime to l in equation (3)? I guess the (c-b) parts should be formed on even numbers, like in (3) in brackets, so finaly we get ((c-b)^2 +c*b*j) so when we are done its just necessary to prove that (c-b) and j are coprimes. I know its hard to take me seriously, concerning all the mistakes in former proofs, but before few months I knew very little about this problem. I am sorry for my impulsiveness concerning this forum, but most of the guys were really patient with me.

Sorry, I was confused with another triangle named a Fermat triangle which was a right triangle. However, there is one error. c-b and a^n/(c-b) are the two parts which multiplied together give a^n. c-b thus can only comprise factors of "a" to certain powers. For (c-b) to be coprime to one part of a then "a" would have to include a factor not in c-b. But you have no proof of that. Merely that c-b does not divide the other part of "a" does not mean that they can not have a common factor.

 

[robert80]

Sorry, I was confused with another triangle named a Fermat triangle which was a right triangle. However, there is one error. c-b and a^n/(c-b) are the two parts which multiplied together give a^n. c-b thus can only comprise factors of "a" to certain powers. For (c-b) to be coprime to one part of a then "a" would have to include a factor not in c-b. But you have no proof of that. Merely that c-b does not divide the other part of "a" does not mean that they can not have a common factor.

Yes Ramsey but when we apply triangle rule that a + b > c we can easily see that a > c - b. so (c-b) does not include all the factors of a (But there is a special case when (c-b) equals to 1.. As with the proof that when solutions a,b,c being a positive integrers exist. Than a +b divides c^n and since 2c > a + b > c and since a + b can not equal to 1. we can see that if the solutions exists, c couldnt be a prime number. With such a simple thing as triangle rules, we can prove that. I hope that helps.

 

[robert80]

And you are right in order the proof would be complete, there is this missing link I am talking about all the time. Experiences teach us that (c-b) is coprime to l in (3) Thats what I am kindly asking the Mathematicians to prove. And the same technique applies for (a + b), than the last equations are as they are writen. Thanks for your time Ramsey.

 

[ramsey2879]

Yes Ramsey but when we apply triangle rule that a + b > c we can easily see that a > c - b. so (c-b) does not include all the factors of a (But there is a special case when (c-b) equals to 1.. As with the proof that when solutions a,b,c being a positive integrers exist. Than a +b divides c^n and since 2c > a + b > c and since a + b can not equal to 1. we can see that if the solutions exists, c couldnt be a prime number. With such a simple thing as triangle rules, we can prove that. I hope that helps.

That is no proof since a -b + c can contain factors that are not in "a" without contradiction of the fact that b-c divides a^n. Also a number that divides a^n doesn't have can be less than a and still contain all the prime factors of a. As to c the question as to whether or not c is prime is immaterial since c only has to be coprime with b and a.

 

[robert80]

That is no proof since 2^7 is greater than 2^4. 2^4 does not divide 2^7 but yet they both only contain the prime factor 2. Thus there is no reason that c could not be prime. Also, c does not have to be prime, it only has to be coprime to both b and a.

if c is prime and a + b divides a^n + b^n a + b equals to c. this is not possible. so c is not a prime. a + b just take another factor of c and leaves one out.

 

[robert80]

and (a + b) of course couldn't be in form of c^2 when c is prime since c^2 is greater than 2c. I hope its ok.

 

[robert80]

All in all I believe the proof in general is ok, just the things I am asking on this forum should be done and maybe some improvement of Math formulation. I hope to find somebody who will take his or her time.

 

[robert80]

That is no proof since a -b + c can contain factors that are not in "a" without contradiction of the fact that b-c divides a^n.

I know what you mean, a could have prime factors on powers. the triangle rule solely its not enough. BUT
when we show (c-b) is coprime to l (the thing I am asking for several times) and a > c - b -----------> we get the 2 equations and factor coprimety as they are written at the end of the proof. I just wanted to show on the case of c how convinient the triangle rules are in the Fermats equation, leading us to the proof that c is not a prime number. Thanks for your time.

 

[ramsey2879]

I know what you mean, a could have prime factors on powers. the triangle rule solely its not enough. BUT
when we show (c-b) is coprime to l (the thing I am asking for several times) and a > c - b -----------> we get the 2 equations and factor coprimety as they are written at the end of the proof. I just wanted to show on the case of c how convinient the triangle rules are in the Fermats equation, leading us to the proof that c is not a prime number. Thanks for your time.

I am trying to figure how you got I as being an integer since (c-b)^(n-1) includes parts that are not divisible by n. Therefor, to say that (3) a^n = (c-b)*((c-b)^(n-1) + n*c*b*l)) is not sound.

 

[robert80]

I am trying to figure how you got I as being an integer since (c-b)^(n-1) includes parts that are not divisible by n. Therefor, to say that (3) a^n = (c-b)*((c-b)^(n-1) + n*c*b*l)) is not sound.

When you write a^n = (c-b)*(c^(n-1) +c^(n-2)b .....+ cb^(n-2) + b^(n-1)) (1**)
and (c-b)^(n-1) with binomial factors as , (c-b)^(n-1) = c^(n-1) -(n-1)*c^(n-2)*b....-(n-1)*b^(n-2)*c + b^(n-1) (2**)you see if you transform (2**) into c^(n-1) +c^(n-2)b .....+ cb^(n-2) + b^(n-1) you have to add n*c*b*l part to (c-b)^(n-1) so you finally get :
(c^(n-1) +c^(n-2)b .....+ cb^(n-2) + b^(n-1) = (c-b)^(n-1) + n*c*b*l

 

[ramsey2879]

When you write a^n = (c-b)*(c^(n-1) +c^(n-2)b .....+ cb^(n-2) + b^(n-1)) (1**)
and (c-b)^(n-1) with binomial factors as , (c-b)^(n-1) = c^(n-1) -(n-1)*c^(n-2)*b....-(n-1)*b^(n-2)*c + b^(n-1) (2**)you see if you transform (2**) into c^(n-1) +c^(n-2)b .....+ cb^(n-2) + b^(n-1) you have to add n*c*b*l part to (c-b)^(n-1) so you finally get :
(c^(n-1) +c^(n-2)b .....+ cb^(n-2) + b^(n-1) = (c-b)^(n-1) + n*c*b*l

Yes adjacent binomial factors of the expansion (c-b)^(n-1) sum to zero mod n. One is 1 and the other is -1 mod n. they are not all 1 mod n. But the odd powers of b are negative so by simply adding or subtracting a n*c*b*I part will no get you to the desired expression.
I forgot to make amends for the negative powers of b.

But you claim that "c-b" is coprime with I. I don't see where you get that. Please explain.

Then you say that "If we apply (1*) in (3) we see that (c - b)^2 part is square AND
(c – b)^2 does not divide a^n since n is prime. So we got a contradiction. The second (2*) case when applied would give the contradiction too. It follows that if (c – b) = n*m when we apply this in (3) we get that n^2 divides a^n. This does not hold true since n is prime so the only remaining possibility would be that n^(n-2) is hidden in the c*b*l but we know that l is coprime to (c – b) and so are c and b, so this could not be the case..
We see now that in order that the solution exists (c - b) has to be coprime to n and since (c - b) is coprime to l,"

I don't follow any of this since a^2 divides a^7 and 7 is prime. So (c-b)^2 could divide a^n

 

[robert80]

Yes adjacent binomial factors of the expansion (c-b)^(n-1) sum to zero mod n. One is 1 and the other is -1 mod n. they are not all 1 mod n. But the odd powers of b are negative so by simply adding or subtracting a n*c*b*I part will no get you to the desired expression.
I forgot to make amends for the negative powers of b.

But you claim that "c-b" is coprime with I. I don't see where you get that. Please explain.

Then you say that "If we apply (1*) in (3) we see that (c - b)^2 part is square AND
(c – b)^2 does not divide a^n since n is prime. So we got a contradiction. The second (2*) case when applied would give the contradiction too. It follows that if (c – b) = n*m when we apply this in (3) we get that n^2 divides a^n. This does not hold true since n is prime so the only remaining possibility would be that n^(n-2) is hidden in the c*b*l but we know that l is coprime to (c – b) and so are c and b, so this could not be the case..
We see now that in order that the solution exists (c - b) has to be coprime to n and since (c - b) is coprime to l,"

I don't follow any of this since a^2 divides a^7 and 7 is prime. So (c-b)^2 could divide a^n

Yes but what you get when you divide a^7 with a^2 is not coprime to a. So when you set (c-b) = n you can move out of the brackets in (#) another (c-b) part. Since when we prove l is coprime to (c-b) what is left in the brackets is coprime to (c-b) so because of that and triangle inequality, (c-b)^2 divides one whole part of a on the n power, and this never has a solution.I will take now some days off. In summer I will try to prove something else, when I have more time.I am tired of this FLT problem. Have you seen Ramsey the other one? That an odd perfect number does not exist? I have a good intuition sometimes and I was always good in abstract logic, But in Math formulation I am not. I am preety sure the last proof goes this way. There could be some mistakes. It would be the best for me if I could connect with someone who is good at math and when I have an idea, he proves it or not within few minutes.

I know nothing about modularity sorry. All I know is that the equation (3) satisfy the condition. When (c-b) =n it follows that (c-b)^2 divides c^n - b^n OR when (a+b) = n (a+b)^2 divides a^n + b^n alternating binomial factors form (c-b)^(n-1) when transformed to (2) they give rise to equation (3) no doubt about that. Now I will take a break for few days. I have enough of FLT. Ramsey are there any other Math puzzles in Number theory? The ones you don't need complicated mathematical tools to understand them? I mean I would like to start something, which is formulated very simple...Is there any list?

Thanks and we hear in few days.

I claimed that (c-b) is coprime to l. This I haven't proved, it arise from the experiences.Thats that's the thing I am asking MATH EXPERTS several times. Is it so hard to prove that (c-b) is coprime to l? I mean when this is not working simply leave it.

 

[robert80]

And Ramsey, please move the questions to my private mail, I have sent it to you days ago. But till Wednesday, I am occupied with my professional work . Greetings

 

[robert80]

I have found the missing link and assistant on Uni found another special case which I proved. I will put this elementary proof into more readable form in few weeks. Thank you for all your patience.

 

[robert80]

Its possible to show (c - b) = (a1)^n when we apply in equation c = d + b, it follows directly, when we rearrange equations. Its possible to show with the same trick (c - a) = (b1)^n but its impossible to show, that (a + b) = (c1)^n, so I am closing this thread. This exact way is not a Fermats proof. Thread closed.

 

[micromass]

Thread closed.

Good idea.