August 3, 2016 at 13:26BTF Case Proof 2
The work was shown on the mathematical forum d x d y .
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.
There he also published “Proof of Case 1 of BTF.”
There are enough readers, no reviews.
I hope to get an opinion on Habrahabr.
It is necessary to prove that the equality
a n + b n = c n ; (1.1)
for integers a, b, c and n> 2, it is impossible.
[1] M.M. Postnikov "Introduction to the theory of algebraic numbers."
At present, 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 of BTF).
[2] G. Edwards "Fermat's last theorem."
We express the grounds for equality 1.1 in terms of uniform 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 are integers.
Therefore, equality 1.1 can be represented as:
(a i ) n × (ax ) n +
(b i ) n × (b x ) n =
(c i ) n × (c x ) n ; 1.2
All degrees bases in the expression 1.2 are integers.
The proof of Case 2 of BTF is based on a comparison of the bases and degrees
of the 2n modes when using the Newton binomial.
[3] M.Ya. Vygodsky “Handbook of elementary mathematics”.
This comparison and use of Newton’s Binom allows us to consider the difference in degrees as the difference in the sum of the terms, which, ultimately, allows us to analyze and compare the exact degrees and the degrees assumed.
Since, always,
b x n ≡1 (mod 2n),
we consider the formalized expression of the degrees of this class of residues.
To do this, we introduce the following notation:
F a n = (a n -1) / (2n) - with a power meter
a n in mod 2n;
F a = (a-1) / (2n) - a base meter of degree
a in mod 2n;
The existing regularity for degrees related to the first class of residues by mode 2n .:
F a n ≡n × a 1 (mod 2n);
Therefore, it becomes possible that in the expression
F (b x ) n
evaluate the presence of the factor n;
Consider the case when
c≡a≡1 (mod 2n);
In the future, it will be shown that consideration of this option covers all possible options that require consideration.
The consideration of the chosen option is explained by the clarity of determining the presence of the factor n in the quantity
F (b x ) n
.
The analysis is carried out on the consideration of the difference of cubes, when c 3
and a 3 , that is, when the appearance of an exact cube is expected in the difference of cubes, the bases of which are numbers belonging to the first class of residues by mode 2n.
Upon consideration, it becomes clear that the analysis of equality 1.2 for the cube provides the proof of Case 2 of the BTF for all degrees necessary for consideration.
We express through c 1 and a i the bases of 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 × c 1 +1) 3 - (6 × a 1 +1) 3 =
216 × (c 1 ) 3 + 3 × 36 × (c 1 ) 2 +
3 × 6 × (c 1 ) 1+1 - 216 × (a 1 ) 3 +3 × 36 × (a 1 ) 2 +
3 × 6 × (a 1 ) 1 + 1 =
216 × (c 1 -a 1 ) (c 1 2 +
c 1 1 × a 1 1 +
c 1 2 ) +
3 × 36 × (c 1 -a 1 ) (c 1 + a 1 ) +
3 × 6 × (c 1 -a 1 ); 1.3
Define (b x ) 3 by dividing 1.3 by
3 × 6 × (c1 -a 1 ):
(b x ) 3 =
12 × (c 1 2 +
c 1 1 × a 1 1 +
c 1 2 ) +
6 × (c 1 + a 1 ) + 1; 1.4
Define F (b x ) n :
F (b x ) n =
2 × (c 1 2 +
c 1 1 × a 1 1 +
(c 1 2 ) +
(c1 + a 1 ); 1.5
We obtain the sum of two terms, the first of which contains a factor of 3, and the second does not.
Therefore, the quantity F (b x ) n cannot contain the factor 3. The
assumption that it is possible to choose the bases c and a when c 1
and a 1
belong to different classes of residues by mode 2n is 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 to various degrees.
This contradiction is inherent in any degree that requires consideration.
The shown regularity is preserved for any exponent, the smallest term in the quantity
F (b x ) n is
always represented (c 1 + a 1 ),
and the occurrence of the factor n in the quantity
F (b x ) n
can occur only if c 1 and a 1 contain factors n.
This option requires further consideration.
However, it should be noted that it can be argued that the quantity
(b x -1) / (2n) contains the factor (2n) (2p) , where
p is the number of common factors n in c1 and a 1 .
To complete the proof of Case 2 of the BTF case (according to the considered option), it remains to show that the bases c and a belonging to any odd class of residues by mode 2n can be transferred to the first class of residues, by this module, without distorting the assumption that the quantity
( b x ) n
may be an exact degree.
We ask ourselves:
What should be the factor k, through which the bases c and a can be transferred to the first class of residues.
To prove the existence of the laws of translation, it is necessary to turn to power-law values.
The regularity of the transfer of bases a and c to the first class of deductions for mode 2n is ensured by:
k = r {n-1} , where
n is the exponent under consideration
r is the residue class of bases c and a in mod 2n.
Since this degree always refers to the first class of deductions for the module under consideration. When r and n are coprime numbers.
(Fermat's Little Theorem).
And we, only such classes of residues are of interest, otherwise, the same factors appear in each of the degrees.
But for the possibility of considering the difference of degrees, in order to analyze the value of
(b x ) n ,
as the expected exact degree, it is necessary to multiply the bases c and a by the degree.
Therefore, the factor for translating the bases c and a should be equal to:
K = kn = (r {n-1} ) n ;
This consideration is only necessary in order to make sure that the possibility of transferring the bases relating to any class of residues for any module 2n to the first class of residues for mode 2n exists.
Thus, the proof of Case 2 of BTF is provided for any degree according to the considered option.
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.
There he also published “Proof of Case 1 of BTF.”
There are enough readers, no reviews.
I hope to get an opinion on Habrahabr.
It is necessary to prove that the equality
a n + b n = c n ; (1.1)
for integers a, b, c and n> 2, it is impossible.
[1] M.M. Postnikov "Introduction to the theory of algebraic numbers."
At present, 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 of BTF).
[2] G. Edwards "Fermat's last theorem."
We express the grounds for equality 1.1 in terms of uniform 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 are integers.
Therefore, equality 1.1 can be represented as:
(a i ) n × (ax ) n +
(b i ) n × (b x ) n =
(c i ) n × (c x ) n ; 1.2
All degrees bases in the expression 1.2 are integers.
The proof of Case 2 of BTF is based on a comparison of the bases and degrees
of the 2n modes when using the Newton binomial.
[3] M.Ya. Vygodsky “Handbook of elementary mathematics”.
This comparison and use of Newton’s Binom allows us to consider the difference in degrees as the difference in the sum of the terms, which, ultimately, allows us to analyze and compare the exact degrees and the degrees assumed.
Since, always,
b x n ≡1 (mod 2n),
we consider the formalized expression of the degrees of this class of residues.
To do this, we introduce the following notation:
F a n = (a n -1) / (2n) - with a power meter
a n in mod 2n;
F a = (a-1) / (2n) - a base meter of degree
a in mod 2n;
The existing regularity for degrees related to the first class of residues by mode 2n .:
F a n ≡n × a 1 (mod 2n);
Therefore, it becomes possible that in the expression
F (b x ) n
evaluate the presence of the factor n;
Consider the case when
c≡a≡1 (mod 2n);
In the future, it will be shown that consideration of this option covers all possible options that require consideration.
The consideration of the chosen option is explained by the clarity of determining the presence of the factor n in the quantity
F (b x ) n
.
The analysis is carried out on the consideration of the difference of cubes, when c 3
and a 3 , that is, when the appearance of an exact cube is expected in the difference of cubes, the bases of which are numbers belonging to the first class of residues by mode 2n.
Upon consideration, it becomes clear that the analysis of equality 1.2 for the cube provides the proof of Case 2 of the BTF for all degrees necessary for consideration.
We express through c 1 and a i the bases of 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 × c 1 +1) 3 - (6 × a 1 +1) 3 =
216 × (c 1 ) 3 + 3 × 36 × (c 1 ) 2 +
3 × 6 × (c 1 ) 1+1 - 216 × (a 1 ) 3 +3 × 36 × (a 1 ) 2 +
3 × 6 × (a 1 ) 1 + 1 =
216 × (c 1 -a 1 ) (c 1 2 +
c 1 1 × a 1 1 +
c 1 2 ) +
3 × 36 × (c 1 -a 1 ) (c 1 + a 1 ) +
3 × 6 × (c 1 -a 1 ); 1.3
Define (b x ) 3 by dividing 1.3 by
3 × 6 × (c1 -a 1 ):
(b x ) 3 =
12 × (c 1 2 +
c 1 1 × a 1 1 +
c 1 2 ) +
6 × (c 1 + a 1 ) + 1; 1.4
Define F (b x ) n :
F (b x ) n =
2 × (c 1 2 +
c 1 1 × a 1 1 +
(c 1 2 ) +
(c1 + a 1 ); 1.5
We obtain the sum of two terms, the first of which contains a factor of 3, and the second does not.
Therefore, the quantity F (b x ) n cannot contain the factor 3. The
assumption that it is possible to choose the bases c and a when c 1
and a 1
belong to different classes of residues by mode 2n is 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 to various degrees.
This contradiction is inherent in any degree that requires consideration.
The shown regularity is preserved for any exponent, the smallest term in the quantity
F (b x ) n is
always represented (c 1 + a 1 ),
and the occurrence of the factor n in the quantity
F (b x ) n
can occur only if c 1 and a 1 contain factors n.
This option requires further consideration.
However, it should be noted that it can be argued that the quantity
(b x -1) / (2n) contains the factor (2n) (2p) , where
p is the number of common factors n in c1 and a 1 .
To complete the proof of Case 2 of the BTF case (according to the considered option), it remains to show that the bases c and a belonging to any odd class of residues by mode 2n can be transferred to the first class of residues, by this module, without distorting the assumption that the quantity
( b x ) n
may be an exact degree.
We ask ourselves:
What should be the factor k, through which the bases c and a can be transferred to the first class of residues.
To prove the existence of the laws of translation, it is necessary to turn to power-law values.
The regularity of the transfer of bases a and c to the first class of deductions for mode 2n is ensured by:
k = r {n-1} , where
n is the exponent under consideration
r is the residue class of bases c and a in mod 2n.
Since this degree always refers to the first class of deductions for the module under consideration. When r and n are coprime numbers.
(Fermat's Little Theorem).
And we, only such classes of residues are of interest, otherwise, the same factors appear in each of the degrees.
But for the possibility of considering the difference of degrees, in order to analyze the value of
(b x ) n ,
as the expected exact degree, it is necessary to multiply the bases c and a by the degree.
Therefore, the factor for translating the bases c and a should be equal to:
K = kn = (r {n-1} ) n ;
This consideration is only necessary in order to make sure that the possibility of transferring the bases relating to any class of residues for any module 2n to the first class of residues for mode 2n exists.
Thus, the proof of Case 2 of BTF is provided for any degree according to the considered option.