# Thread: New approach to Fermats Last theorem for the prime powers of n

1. ## New approach to Fermats Last theorem for the prime powers of n

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. Anyway its an interesting approach I believe. I improved my techniques for proving last weeks. If this is of any help to somebody, I am happy about it.

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 begining 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.

Snicerely

Robert KM, Slovenia

2. I have sent it to Science 10 days ago and got a reply I should send it to Math Journals. But I believe they are not willing to invest their time in order to improve it, so Ive decided, I will post it to Forums. If anyone is willing to take it seriously, I will be happy about it. I understand on the other case the journals, since they have very little time and they get dozens of "proofs" a week. Ive been thinking about the proof extensively for the past few weeks, few hours a day and I simply do not have energy to spend additional time.

3. Originally Posted by 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. Anyway its an interesting approach I believe. I improved my techniques for proving last weeks. If this is of any help to somebody, I am happy about it.

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 begining 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.

Snicerely

Robert KM, Slovenia

I won't invest my time in reading the above in the format in which it is written. If you want your work to be checked by mathematicians you better write it down
using LaTeX or any other mathematical format.

Tonio

4. Originally Posted by 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. Anyway its an interesting approach I believe. I improved my techniques for proving last weeks. If this is of any help to somebody, I am happy about it.

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 begining 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.

Snicerely

Robert KM, Slovenia
MHF is not the place to be seeking a review of your proof. Do what you were told and submit to a journal.