Proof 2 Cases BTF
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There were questions, gave answers. There were questions on the merits. It seems to be able to give an explanation. They asked, asked, and closed the topic.
In the same place, he published the “Proof of 1 Case of BTF”.
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I hope to know the opinion on Habrahabr.
It is necessary to prove that equality
a n + b n = c n ; (1.1)
')
with integer a, b, c and n> 2, is impossible.
[1] M.M. Postnikov, "Introduction to the theory of algebraic numbers."
Currently, the BTF must be proved in an elementary way for the case when
n is a prime number, and one of the bases, for example b, contains the factor n.
(2 Case BTF).
[2] G. Edwards "Fermat's last theorem".
Express the basis of equality 1.1 through common arguments, for which we introduce the following notation:
(a + b) = D c i = (c i ) n ; (2.1)
(cb) = D a i = (a i ) n ; (2.2)
(ca) = D b i = (b i ) n ; (2.1)
Where:
a; b; c - integers.
Therefore, equality 1.1 can be represented as:
(a i ) n Ă— (a x ) n +
(b i ) n Ă— (b x ) n =
(c i ) n Ă— (c x ) n ; 1.2
All bases of degrees in the expression 1.2 are integers.
Proof of the 2 Case of BTF is based on a comparison of the grounds and degrees
mod 2n, using Binom Newton.
[3] M.Ya.Vygodsky “Reference book on elementary mathematics”.
This comparison and the use of the Newton binomial allows us to consider the difference of degrees as the difference of the sum of the terms, which, ultimately, allows for the analysis and comparison of the exact degrees and degrees of the assumed.
Since, always,
b x n ≡1 (mod 2n)
Consider a formalized expression of the degrees of this class of residues.
To do this, enter the notation:
F a n = (a n -1) / (2n) - with a power meter
a n mod 2n;
F a = (a-1) / (2n) - with base meter degree
a mod 2n;
The existing pattern for degrees belonging to the first class of residues in mod 2n .:
F a n ≡n × a 1 (mod 2n);
Therefore, it is possible that in the expression
F (b x ) n
assess the presence of a factor n;
Consider the option when
c≡a≡1 (mod 2n);
Further it will be shown that the consideration of this option covers all possible options that need to be considered.
Consideration of the chosen option is explained by the clarity of the determination of the presence of the factor n in the
F (b x ) n
.
The analysis is carried out on consideration of the difference of cubes, when c 3
and a 3 , that is, when the occurrence of an exact cube is expected in the difference of cubes, whose bases are numbers belonging to the first class of residues in mod 2n.
When considering it becomes clear that the analysis of the equality 1.2 for the cube provides proof of the 2 Case of the BTF for all degrees required for consideration.
We express through c 1 and a i the bases c and a,
and determine the difference of degrees c 3 -a 3 , as the difference of the sums:
b 3 = c 3 -a 3 =
(6 Ă— c 1 +1) 3 - (6 Ă— a 1 +1) 3 =
216 Ă— (c 1 ) 3 + 3 Ă— 36 Ă— (c 1 ) 2 +
3 Ă— 6 Ă— (c 1 ) 1 +1 - 216 Ă— (a 1 ) 3 +3 Ă— 36 Ă— (a 1 ) 2 +
3 Ă— 6 Ă— (a 1 ) 1 + 1 =
216 Ă— (c 1 -a 1 ) (c 1 2 +
c 1 1 Ă— a 1 1 +
c 1 2 ) +
3 Ă— 36 Ă— (c 1 -a 1 ) (c 1 + a 1 ) +
3 Ă— 6 Ă— (c 1 -a 1 ); 1.3
We define (b x ) 3 by dividing 1.3 by
3 Ă— 6 Ă— (c 1 -a 1 ):
(b x ) 3 =
12 Ă— (c 1 2 +
c 1 1 Ă— a 1 1 +
c 1 2 ) +
6 Ă— (c 1 + a 1 ) +1; 1.4
Define F (b x ) n :
F (b x ) n =
2 Ă— (c 1 2 +
c 1 1 Ă— a 1 1 +
(c 1 2 ) +
(c 1 + a 1 ); 1.5
We get the sum of two terms, the first of which contains the factor 3, and the second is not.
Mean, and the value of F (b x ) n , cannot contain the factor 3.
The assumption that you can pick up the grounds with and a when c 1
and a 1
belonging to different classes of residues in mod 2n, erroneous.
Since, in this case, when determining
F (b x ) n
it is impossible to obtain an integer quotient, since each term of the expression 1.3. contains factors of 3 in various degrees.
This contradiction is inherent in any degree that requires consideration.
The regularity shown is preserved at any exponent, the smallest term in the value
F (b x ) n
always represented (c 1 + a 1 ),
and the occurrence of the factor n in the value
F (b x ) n
can occur only under the condition when c 1 and a 1 contain factors n.
This option requires additional consideration.
However, it should be noted that it can be argued that
(b x -1) / (2n) contains the factor (2n) (2p) , where
p is the number of common factors n in c 1 and a 1 .
To complete the proof of Case 2 of the BTF (according to the considered variant), it remains to show that the bases c and a, belonging to any odd class of class 2n residue, can be transferred to the first class of residue, by this module, without distorting the assumption that
(b x ) n
may be an exact degree.
Ask a question:
What should be the factor k, by means of which the transfer of bases c and a to the first class of residues can be made.
To prove the existence of a pattern of translation, it is necessary to refer to the power values.
The pattern of translation of the bases a and c to the first class of residue on mod 2n is provided by:
k = r {n-1} , where
n - considered exponent
r is the class of residues of bases c and a mod 2n.
Since such a degree always belongs to the first class of residues for the module under consideration. When r and n are mutually simple numbers.
(Small Fermat theorem).
And us, only such classes of deductions are of interest, otherwise, in each of the degrees the same factors appear.
But to be able to consider the difference of degrees, in order to analyze the magnitude
(b x ) n ,
as a supposed exact degree, multiplication of bases c and a by degree is necessary.
Therefore, the factor for the translation of bases c and a must be equal to:
K = k n = (r {n-1} ) n ;
This consideration is only necessary in order to make sure that the possibility of transferring grounds related to any class of residue by any module 2n to the first class of residue according to mod 2n exists.
Thus, evidence is provided for 2 Cases of BTF for any degree in the considered variant.
There were questions, gave answers. There were questions on the merits. It seems to be able to give an explanation. They asked, asked, and closed the topic.
In the same place, he published the “Proof of 1 Case of BTF”.
There are enough readers, no feedback.
I hope to know the opinion on Habrahabr.
It is necessary to prove that equality
a n + b n = c n ; (1.1)
')
with integer a, b, c and n> 2, is impossible.
[1] M.M. Postnikov, "Introduction to the theory of algebraic numbers."
Currently, the BTF must be proved in an elementary way for the case when
n is a prime number, and one of the bases, for example b, contains the factor n.
(2 Case BTF).
[2] G. Edwards "Fermat's last theorem".
Express the basis of equality 1.1 through common arguments, for which we introduce the following notation:
(a + b) = D c i = (c i ) n ; (2.1)
(cb) = D a i = (a i ) n ; (2.2)
(ca) = D b i = (b i ) n ; (2.1)
Where:
a; b; c - integers.
Therefore, equality 1.1 can be represented as:
(a i ) n Ă— (a x ) n +
(b i ) n Ă— (b x ) n =
(c i ) n Ă— (c x ) n ; 1.2
All bases of degrees in the expression 1.2 are integers.
Proof of the 2 Case of BTF is based on a comparison of the grounds and degrees
mod 2n, using Binom Newton.
[3] M.Ya.Vygodsky “Reference book on elementary mathematics”.
This comparison and the use of the Newton binomial allows us to consider the difference of degrees as the difference of the sum of the terms, which, ultimately, allows for the analysis and comparison of the exact degrees and degrees of the assumed.
Since, always,
b x n ≡1 (mod 2n)
Consider a formalized expression of the degrees of this class of residues.
To do this, enter the notation:
F a n = (a n -1) / (2n) - with a power meter
a n mod 2n;
F a = (a-1) / (2n) - with base meter degree
a mod 2n;
The existing pattern for degrees belonging to the first class of residues in mod 2n .:
F a n ≡n × a 1 (mod 2n);
Therefore, it is possible that in the expression
F (b x ) n
assess the presence of a factor n;
Consider the option when
c≡a≡1 (mod 2n);
Further it will be shown that the consideration of this option covers all possible options that need to be considered.
Consideration of the chosen option is explained by the clarity of the determination of the presence of the factor n in the
F (b x ) n
.
The analysis is carried out on consideration of the difference of cubes, when c 3
and a 3 , that is, when the occurrence of an exact cube is expected in the difference of cubes, whose bases are numbers belonging to the first class of residues in mod 2n.
When considering it becomes clear that the analysis of the equality 1.2 for the cube provides proof of the 2 Case of the BTF for all degrees required for consideration.
We express through c 1 and a i the bases c and a,
and determine the difference of degrees c 3 -a 3 , as the difference of the sums:
b 3 = c 3 -a 3 =
(6 Ă— c 1 +1) 3 - (6 Ă— a 1 +1) 3 =
216 Ă— (c 1 ) 3 + 3 Ă— 36 Ă— (c 1 ) 2 +
3 Ă— 6 Ă— (c 1 ) 1 +1 - 216 Ă— (a 1 ) 3 +3 Ă— 36 Ă— (a 1 ) 2 +
3 Ă— 6 Ă— (a 1 ) 1 + 1 =
216 Ă— (c 1 -a 1 ) (c 1 2 +
c 1 1 Ă— a 1 1 +
c 1 2 ) +
3 Ă— 36 Ă— (c 1 -a 1 ) (c 1 + a 1 ) +
3 Ă— 6 Ă— (c 1 -a 1 ); 1.3
We define (b x ) 3 by dividing 1.3 by
3 Ă— 6 Ă— (c 1 -a 1 ):
(b x ) 3 =
12 Ă— (c 1 2 +
c 1 1 Ă— a 1 1 +
c 1 2 ) +
6 Ă— (c 1 + a 1 ) +1; 1.4
Define F (b x ) n :
F (b x ) n =
2 Ă— (c 1 2 +
c 1 1 Ă— a 1 1 +
(c 1 2 ) +
(c 1 + a 1 ); 1.5
We get the sum of two terms, the first of which contains the factor 3, and the second is not.
Mean, and the value of F (b x ) n , cannot contain the factor 3.
The assumption that you can pick up the grounds with and a when c 1
and a 1
belonging to different classes of residues in mod 2n, erroneous.
Since, in this case, when determining
F (b x ) n
it is impossible to obtain an integer quotient, since each term of the expression 1.3. contains factors of 3 in various degrees.
This contradiction is inherent in any degree that requires consideration.
The regularity shown is preserved at any exponent, the smallest term in the value
F (b x ) n
always represented (c 1 + a 1 ),
and the occurrence of the factor n in the value
F (b x ) n
can occur only under the condition when c 1 and a 1 contain factors n.
This option requires additional consideration.
However, it should be noted that it can be argued that
(b x -1) / (2n) contains the factor (2n) (2p) , where
p is the number of common factors n in c 1 and a 1 .
To complete the proof of Case 2 of the BTF (according to the considered variant), it remains to show that the bases c and a, belonging to any odd class of class 2n residue, can be transferred to the first class of residue, by this module, without distorting the assumption that
(b x ) n
may be an exact degree.
Ask a question:
What should be the factor k, by means of which the transfer of bases c and a to the first class of residues can be made.
To prove the existence of a pattern of translation, it is necessary to refer to the power values.
The pattern of translation of the bases a and c to the first class of residue on mod 2n is provided by:
k = r {n-1} , where
n - considered exponent
r is the class of residues of bases c and a mod 2n.
Since such a degree always belongs to the first class of residues for the module under consideration. When r and n are mutually simple numbers.
(Small Fermat theorem).
And us, only such classes of deductions are of interest, otherwise, in each of the degrees the same factors appear.
But to be able to consider the difference of degrees, in order to analyze the magnitude
(b x ) n ,
as a supposed exact degree, multiplication of bases c and a by degree is necessary.
Therefore, the factor for the translation of bases c and a must be equal to:
K = k n = (r {n-1} ) n ;
This consideration is only necessary in order to make sure that the possibility of transferring grounds related to any class of residue by any module 2n to the first class of residue according to mod 2n exists.
Thus, evidence is provided for 2 Cases of BTF for any degree in the considered variant.
Source: https://habr.com/ru/post/307010/
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