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G. I. OVCHINNIKOV
PROOF OF FERMAT'S THEOREM
1
ANNOTATION
This collection contains four articles. In the article "Proof of Fermat's theorem" showed that the equation of Fermat's theorem is a transcendental equation. This transcendental equation has no solution in integers. Therefore, Fermat's Last Theorem is true. In the article "Proof of Fermat's theorem for degree three" the theorem is proved by means of the auxiliary equations of the third degree. In the article "The proof of Fermat's theorem for a prime number by method of Decomposition" showed that Fermat's last theorem is true for two special cases. Degree of the equation of Fermat's theorem is an even number; one of the variables (the smallest) is a prime number, and two others are integers. Degree of the equation is an odd number; one of the variables (the smallest) is a prime number, and two others are complete squares of integers. In the article "The solution the equation of Pythagoras in integers" showed that by the method of decomposition allows solved the equation of Pythagoras in integers. These solutions are wider than the solutions (in integers) obtained by the formulas of Hindus.
2
CONTENTS
PROOF OF FERMAT'S THEOREM
4
PROOF OF FERMAT'S THEOREM FOR THE DEGREE THREE 30
THE PROOF OF FERMAT'S THEOREM FOR A PRIME NUMBER BY METHOD OF DECOMPOSITION 64
THE SOLUTION THE EQUATION OF PYTHAGORAS IN INTEGERS 71
3
PROOF OF FERMAT'S THEOREM
CONTENTS
1. Equation (V + y ) p  y p = A p
5
2. Transcendental equation 3. Auxiliary equation of the third degree
15 21
4
1. EQUATION (V + y ) p  y p = A p
1. We consider the equation:
xp + yp = zp ,
p 3.
(1)
Fermat's theorem: the equation (1) has no solutions in integers ( x , y , z ) , if integer p 3 , [1, 2]. If the number is not an integer, then it is either a rational fraction, or irrational value. Let the numbers ( x , y , z ) are the rational fractions, then the relation (1) we can lead to a common denominator. In this case, the equation (1) will have solutions in integers, contradicting Fermat's theorem. Fermat's theorem we state as follows: if an integer p 3 , then in the equation (1) at least one of the numbers ( x , y , z ) is an irrational number. We consider the equation (1), provided that the variables positive integers. 2. Assuming that x = y , the equation (1) becomes:
( x , y , z ) are
2 yp = zp ,
p 3.
(2)
From (2), we have the expression:
1
y 1p = , z 2
p 3.
(3)
For integers
( y,z )
, the equality (3) is not feasible, since the righthand
side is an irrational number by definition. Consequently, at least one of the numbers
( y,z )
is an irrational number.
5
3. Provided 0 < x < y < z , we write equation (1) as follows:
zp  yp = xp , z > y > x > 0.
p 3,
(4)
If x is an integer, then by Fermat's theorem, at least one of the numbers ( y , z ) is an irrational number. Let x = A , where a positive integer is A . In this case, the equation (4) we write as:
z p  y p = Ap , z > y > A> 0.
p 3,
(5)
Provided 0 < y < x < z ( y = A ), a problem also reduces to the equation (5) in the form:
z p  x p = Ap , z > x > A> 0.
p 3,
(6)
4. Any composite integer can be uniquely represented as a product of prime factors. We represent the integer A p , ( p 3) as an expansion in integer factors
( V ,U )
provided that U > V 1 :
A p = V U ,
U >V 1,
p3 ,
(7)
V U =
{ Vk U k },
k = (1 , N ) ,
(8)
V1 U1 = ... = Vk U k = ... = VN U N , U k > Vk , V1 = 1 , U1 = U .
6
For an odd number A , integer multipliers ( V , U ) are odd numbers. For an even number A , integer multipliers ( V , U ) are either even integers or integers of different parity. 5. Consider the equation (5). In view of (7), we represent the equation (5) in the form of decomposition on the multipliers:
( z  y ) ( z p 1 + z p  2 y + ... + z y p  2 + y p 1 ) = V U ,
(9)
p = 2 n , n 2 , p = 2n + 1 , n 1 .
Since ( y , z , V , U
)  integers,
z > y > 0 U > V > 0 then the equation (9) can
be represented as a system of equations:
z  y = V ,
(10)
z p 1 + z p  2 y + ... + z y p  2 + y p 1 = U
,
(11)
V Vk , U U k , y yk , z zk , k = ( 1 , N
).
In view of (10), the equation (5) takes the following form:
(V + y ) p  y p = A p ,
p3 ,
(12)
(V + y ) > A ,
y>A .
In the equation (12)
(V , A )
are integers of the samy parity, by definition.
6. For equation (5) we have the condition:
z < A + y.
(13)
7
In view of (10), the inequality (13) has the form:
V < A.
(14)
Therefore, under the condition V A , the equation (12) has no solution in the integers ( y, V , A ) .
7. We consider the special solutions of equation (1). Assuming that y = 1 , we have the solution:
1 + V = (1 + A p )1 / p ,
p 3.
(15)
Assuming that y = A , we have the solution:
V =
(
p
2 1 A ,
)
p 3.
(16)
Assuming that y = V , we have the solution:
A = V 2 p 1
(
)
1/ p
,
p 3.
(17)
Provided that V = A , we have the inequality:
( A + y )p > A p + y p ,
Since for the integers
p 2.
( y ,V , A )
and subject to p 3 , the equalities (15)
(17) are not feasible, then for the equation (12) we have the following inequalities:
A < y < V, 1 < y < A < V,
(18)
A < V < y, V < A < y,
(19)
8
V < y < A, 1 < y < V < A.
(20)
The inequalities (20) are in contradiction to equation (12), since by condition y > A .
8. Consider the equation (12) subject to (18). We represent the equation (12) in the form:
p p p V 1 + y  y = 1 , A V V
p 3.
(21)
For the equality (21) we have the following inequality:
y y 1 +  > 1 , V V
p
p
p 2.
So, for the integers
( y , p , V , A ) and under the condition V > A
, the
equality (21) is impossible: the left side is greater than 1. Therefore, under condition (18), the equation (12) has no solutions in integers. Consider the equation (12) subject to (19). We write the equation (12) as:
p p y V 1 +  1 = 1 , y A
p 3.
(22)
Since y > A , then from (22) we have the inequality:
V 1 + y
p
 1 < 1,
p 3.
(23)
9
From (23), we obtain the condition for integers ( y , p , V ) :
pV < p y
(
p
2 1
) 3(
3
2  1 < 1, p 3.
)
(24)
In the expression (22), we separate the factor ( p V ) :
y A
p 1
pV A
y f = 1, V
p 3,
(25)
y y f = pV V
V p 1 +  1 , f > 1 . y
(26)
So, for the integers ( y, V , A) and subject to A < V < y , the equality (25) is impossible: the lefthand side is greater than 1. Thus, under the condition y > V > A , equation (12) has no solutions in integers. 9. Consider the equation (25) provided y > A > V . If p V A , then the equality (25) is impossible: the lefthand side is greater than 1. Provided A > p V , from (11), we have the following inequality:
pV y f <1, A V
pV < A < y .
(27)
In vief of (26), from (27), we have the inequality (23) and the condition (24). Thus, for equation (12), we have the condition (24) and the following inequality:
y > A > p V , p 3.
(28)
10
10. We represent the equation (12) as follows:
v 1 + p
p
 1 = u ,
p 3,
(29)
A pV v = , u = . y y
p
We use the binomial formula:
v p + 1
p
=
v C pk p k =0
p
pk
.
(30)
k C p  binomial coefficient:
k Cp =
p! , k ! ( p  k )!
p k 0,
0 p p C p = C p = 1 , C1 = C p 1 = p , 0! = 1! = 1 . p
With (30), we write (29) as follows:
p  k 1 p2 1 + 1 Ck v v p p = u, v< p, p k =0
p 3.
(31)
Under the condition v u , equality (31) is impossible: the lefthand side of equality more the righthand, by definition. Consequently, for equation (12) we have the following condition 1 > u > v :
A 1> y
p
>
pV , y
y > A > pV ,
p3 .
(32)
11
11. Consider the equation (1.12) under the condition A = a , a  a prime number, ( a 2 ). In this case, equation (12) has the form:
(V
In equation (33),
+ y)p  y
p
= a
p
, a 2 , p 3.
(33)
(a ,V )
are integers of the same parity.
In view of (14), for the equation (33), we have the following conditions:
(V (V
+ y) > a,
y > a >V ,
(34)
+ y ) > a , a > y , a >V .
(35)
Imagine a whole number a p as an expansion in integer factorization ( V , U
):
a p = V U , U > V 1 ,
p 3.
(36)
Under the condition p = 2 n , n 2 , the factors
( V ,U )
are:
V = a ni ,
U = a n + i , i = (1 , n ) .
(37)
Under the condition p = 2 n + 1 , n 1 , the factors
( V ,U )
are:
V = a ni ,
U = a n + i +1 , i = ( 0 , n ) .
(38)
In view of (37) and (38), inequality a > V has the form:
a > a ni , i n .
(39)
The inequality (39) is feasible only if i = n .
12
In this case, V = 1 and the conditions (34)  (35), respectively, take the following form:
y > a  1, a 2 .
(40)
1 + y > a > y , a 2.
(41)
For integers
( a , y ) , the condition (41) is not feasible.
Consequently, the expansion (36) can be represented only in the form:
a p = V U , V = 1 , U = a p ,
p 3.
(42)
Thus, equation (33) has the form:
(1
+ y)p  y
p
= ap,
p 3,
(43)
y > a  1, a 2 .
Under the condition a = 2 , equation (33) is:
(1
+ y)p  y
p
= 2p , p 3 .
(44)
For an integer y , any parity, equality (44) can not be satisfied: the left hand side is odd. Consequently, for the equation (33) we have the condition: a prime number a 3 . Under the condition y = a , equation (33) has a solution:
(
p
2  1 a = 1 .
)
(45)
For an integer a , equality (45) is not feasible.
13
Thus, for the equation (33) we have the condition:
y > a , a 3.
(46)
Equation (33), we will present in the form:
1 1 + y
p
a 1 = y
p
,
a 3, p 3.
(47)
According to (24) and (32), provided that V = 1 , A = a , for the equation (47) we have the following inequalities:
p < p y
(
p
2  1 < 1, p 3 ,
)
(48)
a p < < 1, y y
p
y > a > p 3.
(49)
Equation (47) is a special case of (29), provided that V = 1 , A = a , a prime number a > 3 .
14
2. THE TRANSCENDENTAL EQUATION 1. We consider the equation (1.29):
v 1 + p
p
 1 = u ,
p3 ,
(1)
0 < v < u <
p
u < 1.
(2)
Transform equation (1) to mean:
1 +
v = p
(1
+ u
)1 / p ,
0 < u <1,
p3 .
(3)
We represent the right side of (3) as a power series:
( 1 + u )1 / p
= 1 +
uk 1 ( 1 ) k +1 Bk . p k =1 k
(4)
The coefficients Bk , k 1 are: B1 = 1 ,
Bk < 1 , k 2
Bk =
( p  1 ) ( ( k  1) p  1 )
p k 1 ( k  1)!
, k2 .
(5)
We use the definition of the function of the natural logarithm:
ln( 1 + ) =
(  1 ) k +1
k =1
k
k
, 0 < <1 .
(6)
Using (9), we represent (7) as:
( 1 + u )1 / p
= 1 +
1 1 ln ( 1 + u ) + g( u ) , 0 < u < 1 . p p
(7)
15
The function g ( u ) is the power series of the form:
g( u ) =
u (  1 ) k ( 1  Bk )
k =2
k
k
.
(8)
In view of (7), equation (3) is:
ln( 1 + u ) = v  g ( u ) , 0 < v < u < 1 .
For the equality (3), we have the relation:
v = ln ( 1 + u ) . p ln 1 + p
(9)
(10)
In view of (10), equation (9) is:
v 1 + p v  ln 1 + p 1 = 1 + g( u ) , p
p 3.
(11)
For the variable v , equation (14) is a transcendental equation of the natural logarithm. We write the equation (11) for the variables
( y , A,V )
:
V 1 + y
V  ln 1 + y
1 A = 1 + g , p y
(12)
y > A > pV ,
p3 .
For the variable y , the equation (12) is a transcendental equation of the natural logarithm. So, the algebraic equation (1.12) reduces to the transcendental equation However, the transcendental equations can be solved only approximately. In other words, for the equations (12), we can not find the solutions in integers. Consequently, for the equation (1.12) are also impossible to find the solutions in integers. This in fact means that the equation (1.1) has no solution in integers
( x, y,z) .
16
2. Consider the equation (11). In view of (6), the lefthand side of equation (11), we represent in the form:
v v 1 +  ln 1 + = 1 + p p
1 v (  1) k p . k =2
k
k
(13)
In view of (8) and (13), equation (14) takes the following form:
1 (  1) k k =2
k
k v uk = 0,  ( 1  Bk ) p p
p3 .
(14)
We introduce the notation:
Rk = ( 1  Bk ) p k 1 .
(15)
From (5) and (15), we have the expressions:
B1 = 1 ,
B2 =
p 1 , R1 = 0 , R2 = 1 , p
Bk +1 kp  1 , k 1, = Bk kp
p 3,
(16)
1  Bk +1 Rk +1 = 1  B p , k 2 , Rk k
p 3.
From (16), we have the following inequalities:
1 Bk > Bk +1 , k 1 ,
(17)
Rk +1 > Rk 1 , k 2 .
17
With (15), we write (14) for the variables
( y , A,V )
:
1 (  1) k p k k =2
k
p k k R A  pV = 0 , y k y
(18)
y > A > pV ,
p 3.
3. We write the equation (18) as follows:
( 1 )
k=2
k
bk = 0 ,
(19)
bk =
1 Rk u k  v k , k kp
(
)
bk > 0 .
(20)
Equation (19), we will present in the form of:
( b
k =1
2k
 b2 k +1 ) = 0 .
(21)
So, we have the equality:
b2 k = b2 k +1 , k 1 .
(22)
In view of (20), the equality (22) has the form:
R2 k 
2k u R2 k +1 u 2 k = 1  ( 2 k + 1 ) p
2 k ( 2 k + 1 ) p
2k ,
(23)
R 2 k +1 R 2k
1  B 2 k +1 , R 2 k +1 > R 2 k 1 , k 1 , = p 1  B 2k
(24)
18
B 2 k +1 B 2k
=
2k p  1 , 2k p
1 > B 2 k > B 2 k +1 , k 1 .
(25)
For the righthand side of (29), we have the condition:
2k < 1, 0 < <1, k 1, ( 2 k + 1 ) p
p 3.
(26)
According to (26), for equation (29), the following condition is necessary:
2k u R 2 k +1 < 1 , 0 < u < 1 , k 1 , ( 2 k + 1 ) p R2k
p 3.
(27)
For the condition (27) we have the inequality:
2k u R 2 k +1 < ( 2 k + 1 ) p R2 k
2k R 2 k +1 , 0 < u < 1 . ( 2 k + 1 ) p R2k
(28)
Consider the following inequality:
2k R 2 k +1 1 , k 1 , ( 2 k + 1 ) p R2k
p 3.
(29)
In view of (24) and (25), the inequality (29) has the form:
B2k
p , k 1, p +1
p 3.
(30)
Provided k = 1 , we have the expressions:
B2 =
p 1 , p
p 1 p < , p p +1
p 3.
Since B 2 k > B 2 k +1 , then condition (30) holds for all k 1 .
19
Thus, for the equation (23), we have the inequalities (26) and (27). The equality (23), we presented in the form:
u 2 k ( ( 2 k + 1 ) p R2 k  2 k u R2 k +1 ) =
(31)
= 2 k ( ( 2 k + 1) p  2 k ) , k 1 .
The equality (31) must hold for all values of k 1 . We represent the expression (31) as follows:
u
=
2k
Dk , k 1.
(32)
The function Dk is:
Dk =
( 2k
+ 1 ) p R 2 k  2 k u R 2 k +1
( 2k
+ 1 ) p  2 k
, k 1.
(33)
Under the condition k = 1 , we have the expressions:
R2 = 1 , R3 =
1 ( 3p 1 ) , 2
1
3 p  ( 3 p 1 ) u 2 , = u 3 p  2
p 3.
(34)
Analysis (34) we give below, in Paragraph 3. The lefthand side of (32) and (34), for the variables
( y , A,V )
, has of the form:
pV y = y A , u
p
p 3.
(35)
20
For integers
( y , A , V ) , the lefthand side of (32) is a rational number, and the
righthand side is a radical degree 2 k of rational numbers Dk , k 1 . However, for equality (32), a radical degree 2 k must be a rational number for all values of k 1 . This requirement is actually not feasible for an infinite sequence k = 1 , 2 , 3 , . . . . In other words, the radical in the righthand side of (32), at least for one of k 1 , is an irrational quantity. So, for integers
( y , A , V ) , equality (32) is impossible: the lefthand side is a rational ( y ,V, A)
is an
number, and the righthand is an irrational quantity. In this case, according to (35), at least one of the variables irrational quantity. Since, by hypothesis, irrational number. So, equation (1) has no solution in integers ( y , A , V ) . Thus, equation (1.1) also has no solution in integers ( x , y , z ) . If in equation (1.1) x is integer, then at least one of the variables ( y , z ) is an irrational quantity. Therefore, Fermat's Last Theorem is true.
(V, A)
 integers, then the variable y is an
3. AUXILIARY EQUATION OF THE THIRD DEGREE 1. Consider the equality (2.34):
1
3 p  ( 3 p 1 ) u 2 , = u 3p  2v
p 3,
(1)
0 < v < u < 1.
The variables ( v , u ) are:
pV v = , y
A u = , y
p
p3
(2)
21
Taking into account (2), we represent equation (1) in the form:
pV pV A 3p y  2 y = 3 p y
2
3
2p
A  2 ( 3 p  1 ) y
3p
.
(3)
We write equation (3) with respect to the variable y :
3 p 3 V 2 y 3 p  2  2 p 3 V 3 y 3 p 3  3 p A2 p y p + ( 3 p  1 ) A3 p = 0 . (4)
Equation (4) is a transcendental equation of degree change y .
(3p  2 )
with respect to the
According to (4), the equation for the number of A is:
( 3 p  1 ) A3 p  3 p y p A2 p + 3 p 3 y 3 p  2 V 2 1  2V
= 0 . (5) 3y
By (1.7), from (5) we obtain the equation for the number of U :
( 3 p  1 ) V U 3  3 p y p U 2 + 3 p 3 y 3 p  2 1  2V
= 0. 3y
(6)
Equation (6) is a cubic equation for the number of U .
2. Consider the equation (6). Equation (6), we will present in the form of a cubic equation:
t3 + 3h t + 2 q = 0 ,
(7)
U = t + ,
p yp = , >1 ( 3 p  1 ) V
(8)
h =  2 ,
q =  (1  ) 3 ,
(9)
22
3 = 2
2V ( 3 p  1 ) V 1  3y y
.
2
(10)
For the solution of (7), we apply the method of auxiliary variables ( , ) , (Chapter 2.4., [3]). The numbers of real roots of equation (7) depends on the sign of the discriminate D :
D = q 2 + h3 =  ( 2  ) 6 .
(11)
Auxiliary value is equal to:
= ±
h
= ± .
(12)
Sign must coincide with the sign of q . Auxiliary value is determined depending on the signs of ( h , D ) . According to (9), we have h < 0 . Provided that D 0 , the value of and the solutions of equation (7) are:
cos = ,
=
q
3
,
(13)
t1 =  2 cos , 3
(14a)
ti = + 2 cos 3
, i = 2, 3 . 3
(14b)
Under the condition D > 0 , we have the expressions:
ch = ,
=
q
3
,
(15)
23
t1 =  2 ch . 3
(16)
The solutions
( t2 , t3 ) are imaginary.
, we have the following values:
A. Provided that > 2
q > 0 , D > 0 , = + , =  1 .
According to (15), the value
is:
= ln  1 +
(
(  2
) ), > 2 , > 0 .
(17)
According to (8), the real solution of equation (6) has the form:
U =  2 ch  1 , > 0 , > 0 . 3
(18)
So, provided > 2 , the equation (6) has no positive solutions. According to (10), condition > 2 can be written as:
2 3 pV < < 1. 9 y
B. Provided that = 2 , we have the following values:
q = 3 , D = 0 , = + ,
=1,
=0 .
In this case, the solutions (14a)  (14b) are:
t1 =  2 ,
t 2 = t3 = .
24
In view of (8), the solutions of equation (6) are:
U1 =  , U 2 = U 3 = 2 .
Thus, equation (6) has one positive solution:
U = 2 .
(19)
With (8), the solution (19), we will present in the form of:
2p yp U = ( 3 p  1 )V
.
(20)
Using (1.7) and (20), we obtain an expression for A p :
Ap =
2p yp , (3p 1)
p3 .
(21)
We will present (21) in the form:
1
2p p A = 3 p 1 , y
p3 .
(22)
The righthand side of (22) is an irrational quantity, by definition. Consequently, at least one of the variables
( y , A)
is an irrational number.
So, provided = 2 , the equation (5) has no solution in integers According to (10), condition = 2 can be written as:
( y , A) .
3 3 p 1 4 p
2
pV 2 pV y 1  3p y = 1 .
2
25
C. Provided that = 1 , we have the following values:
q = 0 , D < 0 , = + ,
=0, =
2
.
In this case, the solutions (14a)  (14b) are:
t1 = 
3 ,
t2 = + 3 ,
t3 = 0 .
(23)
In view of (8) and (23), the solutions of equation (6) are:
U1 = 
( (
3  1 , 3 + 1 ,
)
U2 = + U3 = .
)
(24)
According to (24), equation (6) has only one rational solution:
U = .
(25)
Using (1.7) and (8), we will present the solution (25) in the form:
1
p p A = 3 p 1 , y
p3 .
(26)
The righthand side of (26) is an irrational quantity, by definition. Consequently, at least one of the variables
( y , A)
is an irrational number.
So, provided = 1 , the equation (5) has no solution in integers According to (10), condition = 1 can be written as:
( y , A) .
3 3 p 1 2 p
2
pV 2 pV y 1  3p y = 1 .
2
26
D. Provided that 1 < < 2 , we have the following values:
q > 0 , D < 0 , = + ,
=  1 ,
= arc cos (  1 ) , 0 <  1 < 1 .
Solutions (14a) (14b) take the form:
t1 =  2 cos , 3
(27)
(28a)
ti = + cos ± 3
3 sin
, i = 2, 3 . 3
(28b)
In view of (27), for the parameter of , we have the inequality:
0 <
3
<
6
,
(29)
3 < cos < 1, 2 3 0 < 3 sin
3
<
3 . 2
Using (8) and (28), we will present the solutions of equation (6) in the form:
t U = 1 + i , i = 1, 2 , 3 .
(30)
By (29), the functions cos , sin , under the condition 0 < < , take 3 3 3 6
only the irrational values. Since is a rational number, then the solutions ( t1 , t 2 , t3 ) are the irrational values. Hence, according to (30), U is also an irrational number.
27
In view of (1.7), (8) and (30), we obtain the following expression:
1
p p A ti p = 3 p 1 1 + , y
1
p 3 , i =1, 2 , 3 .
(31)
The righthand side of (31) is an irrational quantity, by definition. Consequently, at least one of the variables
( y , A)
is an irrational number.
So, provided 1 < < 2 , the equation (5) has no solution in integers According to (10), condition 1 < < 2 can be written as:
( y , A) .
6 2 3 pV < < 9 y 9
.
E. Provided that 0 < < 1 , we have the following values:
q < 0, D < 0, =  ,
=1 ,
= arc cos ( 1  ) , 0 < 1  < 1 .
Solutions (14a) (14b) take the form:
t1 = + 2 cos , 3
(32)
(33a)
ti =  cos ± 3
3 sin
, i = 2, 3 . 3
(33b)
In view of (32), for the parameter of , we have the inequalities (29). In view of (8) and (33), solutions of equation (6) have the form (30).
A The parameter is defined by expression (31). y
So, provided 0 < < 1 , the equation (5) also has no solution in integers
( y , A) .
28
According to (10), condition 0 < < 1 can be written as:
0 <
pV < y
6 , 9
p 3.
3. According to (2), the lefthand side of equation (1) there is a proper fraction (2.35):
pV y = y A , u
p
p 3 , 0 < < u < 1.
Since equation (5) has no solution in integers
( y , A)
, then for integers
( y , A,V ) ( y , A,V )
the equality (1) is not feasible, the lefthand side there is a proper fraction,
while the righthand side is an irrational quantity, by definition. Consequently, for integers and under the condition k 1 , the equality (2.32) also is impossible: the left
hand side there is a rational number, and the right side is an irrational quantity. So, equation (1.12) has no integer solutions
( y , A , V ) . Thus, equation (1.1)
also has no solution in integers ( x , y , z ) . If in the equation (1.1) x is integer, then at least one of the variables ( y , z ) is an irrational quantity. Therefore, Fermat's Last Theorem is true.
REFERENCES 1. . . . .. 1934. 2. . . . . 1983. 3. . ., .. . . 1986. June 2011. Tartu
29
PROOF OF FERMAT'S THEOREM FOR THE DEGREE THREE
CONTENTS
1. DERIVAITION OF THE EQUATION
31
2. EQUATION (1 + y ) 3  y 3 = a 3
34
3. EQUATION ( V + y )3  y 3 = A3
41
4. EQUATION 1 +  1 = u 3
3
46
APPENDIX 1 APPENDIX 2 APPENDIX 3 REFERENCES
55 59 62 63
30
1. DERIVAITION OF THE EQUATION
1. We consider the equation:
x3 + y 3 = z 3 .
(1)
Fermat's theorem: the equation (1) has no solution in integers ( x, y, z ) . Consider the equation (1) under the conditions: ( x, y, z )  integers. Under the condition x = y , we represent the equation (1) as follows:
z =
3
2y.
If
( y , z )  integers, then this equality is not feasible, by definition.
Thus, we consider the equation (2) provided:
0 < x < y < z.
(2)
Let x = A , where A  an integer (a prime, odd, even number). We represent the integer A3 as an expansion on the multipliers of integers:
A3 = V U , U > V 1.
(3)
Where ( V , U )  the integers, on which can be decomposed an integer A3 , provided that U > V 1 . In view of (2), we represent the equation (1) as follows:
z 3  y 3 = A3 , z > y > A.
(4)
In view of (3), we will present the equation (4) in the form of an expansion:
( z  y ) ( z 2 + z y + y 2 ) = V U .
(5)
31
Since ( y , z , V , A )  integers, 0 < y < z , 0 < V < U then the equation (5) can be represented as a system of two equations:
z  y =V , z2 + z y + y2 = U .
(6) (7)
In view of (6), the equation (3) becomes:
( V + y )3  y 3 = A3 .
(8)
For equation (3) we have the condition:
z < A + y.
(9)
In view of (6), the inequality (9) has the form:
V < A.
(10)
Therefore, under the condition V A , the equation (8) has no solution in the integers ( y , A , V ) . So, for equation (3) we have the following condition:
(V
+ y) > A,
y >A > V.
(11)
In the equation (8), ( V , A ) are the integers of the same parity, by definition. 2. Consider the special case of equation (8), subject to V = 1 and subject to V = 2 . Under the condition A  an odd number and subject to V = 1 equation (8) is:
(1
+ y)
3
 y 3 = A3 ,
y > A 3.
(12)
Under the condition A = B , ( B  an even number), and subject to V = 2 equation (8) is:
(2
+ y)
3
 y 3 = B3 .
(13)
32
We will present an even number B in the general form:
B = 2N A ,
N 1,
A 1,
(14)
A  An odd number.
In view of (14) and subject to y = 2m , m 1 equation (13) is:
(1
+ m )  m 3 = 23 ( N 1) A 3 .
3
(15)
Under the condition N = 1 , equation (15) takes the form (12), ( m = y , m > A 3 ). Under the condition N 2 , the equality (15) is impossible: the left hand side is odd. In view of (14) and subject to y = 2m  1 , m 1 , equation (13) is:
( 2m
+ 1)
3

( 2m
 1)
3
= 23 N A 3 .
(16)
We write equation (16) as follows:
12 m 2 + 1 = 23 N 1 A3 .
(17)
For integers ( m , A ) , equality (17) is impossible: the left hand side is odd. 3. We consider the equation (8) under condition A = a , (a prime number a 2 ). We define the factors ( V , U ) for the number of a 3 :
V1 = 1 , U1 = a 3 ,
(18)
V2 = a , U 2 = a 2 .
(19)
For factors (19), equation (8) has the form:
3a y ( a + y ) = 0 .
(20)
33
If ( y , a ) are the positive integers, then equality (20) is impossible: to the left we have a sum of positive integers. For factors (18), the equation (8) and the condition (11) have the form:
(1 + y ) 3  y 3 = a 3 ,
y > a 3.
(21)
For an odd number A 3 and condition V = 1 , the equation (21) is analogous to equation (12).
2. EQUATION (1 + y ) 3  y 3 = a 3
1. Consider the equation (21). We represent equation (21) as follows:
3 y2 + 3y 
(a
3
 1 = 0.
)
(22)
The quadratic equation (22) has one positive solution:
y =
1 6
(
12 a 3  3  3 , a 3 .
)
(23)
Consider the following equation:
12 a 3  3 = 9b 2 , b = 2 k + 1 , k 1 .
(24)
In the equality (24), ( a , b ) are odd numbers. In view of (24), the solution (23) is integer:
y = k , k 1.
(25)
By hypothesis, a is odd:
a = 2m + 1 , m 1.
(26)
34
In view of (26), the equality (24) is:
4 ( 2 m + 1 ) = 3( 2 k + 1 ) + 1
3 2
(27)
We introduce the notation n = 2 m and represent equation (27) as follows:
n3 + 3 n 2  k 2
(
) + 3( n  k ) = 0
(28)
Provided to n k , the equality (28) is impossible. Consequently, for (28) we have the condition k > n . Transform (28) to mean:
3 ( k  n ) ( k + n + 1 ) = n3 , k > n .
(29)
For integers ( k , n ) , the inequality k > n can be written as:
k = n + l , l 1.
(30)
In view of (24) and (26), the equality (30) is:
b = 2a +
( 2l
 1 ) , a 3 , l 1.
In view of (30), equation (29) is:
n3  6 l n  3 l ( l + 1 ) = 0 .
(31)
Equation (31) is a cubic equation for the reduced form:
n3 + 3 p n + 2 q = 0 ,
(32)
p =  2l ,
q = 
3 l (l + 1). 2
(33)
35
The number of real roots of equation (32) depends on the sign of the discriminate D , [1]:
D = q 2 + p3 .
In view of (33), the parameter D is:
l 2 D = 9 ( l + 1 )  32 l . 2
2
(
)
According to (33), provided to l 1 , the parameter D > 0 . Under the condition D > 0 , the equation (32) has one real solution and two imaginary (imaginary solutions we do not consider). To solve the equation (32) applies the formula of Cardan [1]. As a result, the actual solution of (32) is:
1
l 3 n = 2
(c + h
)3
1
+ (c  h
)3 ,
1
(34)
h =
9 l 2 + 1  14 l
(
)
,
l 1, h > 0 ,
(35)
c = 3( l + 1 ) , l 1 , c > h .
(36)
In (34) c is an integer, and h is an integer or irrational value, c > h > 0 . In view of (25), (30) and (34), the solution of (22) for y is as follows:
1
l 3 y = l + ( c + h 2
)3
1
+
(c
 h
)3
1
.
(37)
According to (37), for an integer l = 1 , the solution y is an irrational number:
y = 1 +
3
2 1 +
(
3
2
)
.
36
2. Consider (37), provided that l 2 . Provided that l = 2n + 1 , n 1 , the solution (37) is:
y =
( 2n + 1)
+
3
c1 +
( 2n + 1) h1
+
3
c 1  ( 2n + 1) h1 ,
(38)
c 1 = 3 ( n + 1) ( 2 n + 1) ,
h1 =
9 n2 + 2 n + 1 .
(39)
Provided that l = 2n , n 1 , the solution (37) is:
y = 2n +
3
c2 + n h 2 +
3
c2  n h 2 ,
(40)
c2 = 3 n ( 2n + 1) ,
h2 =
36 n 2  28 n + 9 .
(41)
A. Consider the following equation
2 9 n 2 + 2 n  h1  1 = 0 .
(
)
(42)
We define integer n , where h1 is an integer. Equation (42) has the solution:
n =
1 9
(3h )
1
2
 8  1 , h1 1 .
(43)
Consider the following equality
(3 h1 ) 2  d12 = 8 .
(44)
In the equation (44),
(h
1
, d1 ) are integers, by definition.
37
Solutions of the equation (44), in whole numbers, we will present in the form [4]:
3 h1 =
U +V , 2
d1 =
U V . 2
(45)
In (45) ( V , U ) are the integers, on which can be decomposed an even number 2 3 , subject to U > V 2 . The factors ( V , U ) , respectively, are V = 2 , U = 4 . As a result, we have the solution h1 = 1 , d 1 = 1 , n = 0 . So, for integer of n 1 , the parameter h1 is not an integer. Consequently, by (39), the value of h1 is an irrational number. In particular, under the condition that n = 1 , h1 = 2 3 .
B. Consider the following equation
36 n 2  28 n  h 2  9 = 0 . 2
(
)
(46)
We define integer n , where h 2 is an integer. Equation (46) has the solution:
n =
1 7 ± 18
(3h )
2
2
 32 , h 2 2 .
(47)
Provided that 9 h 2 = 32 , from (47) we have n = 7 / 18 . 2 If 3 h 2 2 , then equation (46) has a solution
n =
1 7  18
(3h )
2
2
 32 , 3 h 2 2 .
(47a)
If h 2 = 2 , then from (47a) we have n = 5 / 18 . If h 2 = 3 , then from (47a) we have n = 0 .
38
Provided that h 2 3 , equation (46) has a solution:
n =
1 7 + 18
(3h )
2
2
 32 , h 2 3 .
(47b)
Consider the following equation
2 (3 h 2 ) 2  d 2 = 32 .
(48)
In the equation (48),
(h
2
, d 2 ) are integers, by definition. Solutions of the
equation (48), in whole numbers, we will present in the form (45):
3h2 =
U +V , 2
d2 =
U V . 2
(49)
In (45) ( V , U ) are the integers, on which can be decomposed an even number 2 5 , subject to U > V 2 . The factors ( V , U ) , respectively, are V1 = 2 , U1 = 16 ; V2 = 4 , U 2 = 8 . For factors (V1 , U1 ) , we have the solution h 2 = 3 , d 2 = 7 , n = 7 / 9 . For factors (V2 , U 2 ) , we have the solution h 2 = 2 , d 2 = 2 , n = 1 / 2 . So, for integer of n 1 , the parameter h 2 is not an integer. Consequently, by (41), the value of h 2 is an irrational number. In particular, under the condition that n = 1 , h 2 = 17 .
3. Consider the solutions (38) and (40). We introduce the notations:
N1 = 9 n 2 + 2 n + 1 ,
N 2 = 36 n 2  28 n + 9 .
(50)
By (42) and (46) for integer of n 1 , the integers ( N1 , N 2 ) are not perfect squares.
39
Integers
( N1 , N 2 )
can be uniquely represented in the form:
N1 = m 2 a , m 1 1 ; 1
N 2 = m 2 b , m 2 1. 2
(51)
(m
1
, m 2 )  Integers.
( a , b )  The prime numbers. (a , b )
is the product of prime numbers that are not equal
In general, the value of to each other:
a = a1
ai
aK , ai +1 > ai , i = ( 1 , K ) ,
(52)
b = b1 bj bM , b j +1 > bi , j = (1 , M ) .
(a
definition:
i
, bj
)
 The prime numbers. a1 2 , b1 2 .
In view of (39), (41) (51), parameters
(h
1
, h2
)
are irrational numbers, by
h1 =
N1
= m1
a ;
h2 =
N2
= m2 b .
(53)
We introduce the functions ( 1 , 2 ) :
1 =
3
c1 + p1
N1 +
3
c1  p1
N1 ,
p1 = 2 n + 1 , n 1 ,
(54a)
2 =
3
c2 + p2
N2 +
3
c2  p2
N2 ,
p2 = n , n 1 .
(54b)
In Appendix 1, we have showed that functions ( 1 , 2 ) there are irrational expressions. Thus, for all values of integer of l 1 , the right part of the solution (37) there is an irrational number. Consequently, equation (21) has no solutions in integers ( a , y ) . Since, by hypothesis, a is odd integer, then the variable y is an irrational number.
40
So, if in equation (1) x = a , (a prime number a 3 ), then at least one of the variables
(y,z)
is an irrational value. In the equation (21) a is a prime number (in the
general case, a is odd integer). Therefore, a similar conclusion holds for the equation (12).
3. EQUATION ( V + y )3  y 3 = A3
1. Consider the equation (8), if V 3 . In view of (3), we represent the equation (8) in the form:
V
2
+ 3 y V  U  3y 2
(
)=
0,
y > A >V 3
(55)
Under the condition U > 3y 2 , ( A > 3V positive solution:
)
the quadratic equation (55) has one
2V =
(
4U  3 y 2
 3y
)
.
(56)
Consider the following equation
4U  3 y 2 = 9 d 2 ,
(57)
d  Integer.
In view of (57), the solution (56) is
2V = 3 ( d  y ) , d > y .
(58)
According to (58),
(d , y )
 integers of the same parity.
Under the condition d > y , for integers ( d , y ) , holds the relation:
d = y + 2k , k 1
(59)
41
In view of (59), solution (58) takes the form:
V = 3k , k 1 ,
(60)
(k ,V )
 The integers of the same parity.
In view of (59), the equality (57) is:
U = 3 d 2  k d + k 2 , d > k 1.
(
)
(61)
With regard to (3) and (60), we obtain the following equation:
9 k 3  d k 2 + d 2 k
(
)
= A 3, d > k 1.
(62)
We introduce the following notation:
n = 3k , m = 3d .
(63)
In view of (63), equation (62) becomes:
n 3  mn 2 + m 2 n  3 A3 = 0 .
(64)
In equation (64), we introduce a new variable
t = n 
m . 3
(65)
In view of (65), equation (64) takes the form (32):
t 3 + 3 p t + 2q = 0 ,
(66)
m p = 2 , 3
2
q =
3 1 m 7  3A 3 2 3
.
(67)
42
In view of (63), expression (65) and (67) are as follows:
t = 3k  d ,
(68)
p = 2d 2 ,
q =
1 7d 3  3 A 3 . 2
(
)
(69)
According to (11) and (59), we have the inequality d > y > A . So, for the parameters
(p,q)
satisfies conditions p > 0 , q > 0 .
The number of real roots of equation (66) depends on the sign of the discriminate D = q 2 + p 3 , [1]. Since D > 0 , then the equation (66) has one real solution and two imaginary (imaginary solutions we do not consider). To solve the equation (66) applies the formula of Cardan [1]. As a result, the actual solution of (66) is:
1
1 3 t = ( r  s 2
)3
1
 (r + s
)3 ,
1
(70)
r =
s 2 + 32 d 6 ,
(71)
s = 7 d 3  3 A3 , d > A .
(72)
In view of (59), (68) and (70), the solution of equation of (8), for variable y , has the form:
3y = d +
3
4 ( r + s
)3
1

(r
 s
)3 ,
1
r >s.
(73)
According to (71)  (72), for integers integer or irrational value.
( d , A)
, the parameter r there is either an
43
2. In view of (72), we write the expression (71) in the form of an equation:
(3 A )
3 2
 14d 3 3 A3  r 2  32d 6
(
) (
)=
0.
(74)
Given the conditions d > A , equation (74) has one positive solution:
3 A3 = 7 d 3 
r 2  32d 6 .
(75)
Provided that r 2 = 32d 6 , from (75) we have Under the condition 81d 6 > r 2 > 32 d 6 Consider equality:
3
3 A =
3
7 d .
the solution (75) is positive.
s 2 + 32 d 6 = r 2 .
(76)
If
(s , 4d
3
,r
) (
are the reciprocal prime numbers, and also ( s , r ) are the odd
integers, then the equation (76) has no solution in positive integers [2]:
s = ± k 2  2l 2 , 2 d 3 = k l , r = k 2 + 2l 2 ,
)
(77)
( k , l )  The reciprocal prime integers,
integer. If
k  an odd integer, s  a positive
( d , A)
are integers of different parity, then according to (74) and (76),
(r,s)
are the odd numbers. According to (77), there are the definitions:
l = 2m ,
d3 = k m .
(78)
With (77) and (78), the solution (75) for the number of A is:
3
3 A =
3
7 k m ± 8 m2  k 2
(
).
(79)
44
For integers If
( k ,m, A)
, the equality (79) is not feasible (Appendix 3).
( d , A)
are integers of the same parity, then according to (74) and (76),
(r,s)
are the even numbers. If s = 2 p ,
r = 2 q , then the equation (76) becomes:
p2 + 8 d 6 = q2 .
(80)
If
( p , 2d
3
,q
) (
are the reciprocal prime numbers, and also
( p , q ) are the odd
integers, then the equation (76) has no solution in positive integers (77):
p = ± k 2  2l 2 ,
)
d 3 = k l , q = k 2 + 2l2 ,
(81)
In view of (81), the solution (73) for the number of y is:
3y =
3
k l ± 23 2
(
3
k2 
3
2l 2 .
)
(82)
Since k is an odd number, then the righthand side of (82) is an irrational expression. So, if
( d , A)
are the integers of the same parity, then according to (82), y
is an irrational number. If the solutions (77) and (81) are not feasible, then square integer. From the definition (51), we have the relation:
(r
2
 32 d 6
)
is not a perfect
(r
2
 32d 6
)=
C2 a , C 1.
(83)
C  Integer, a  the prime number of (52).
In view of (83), the solution (75) is:
3
3 A =
3
7d 3  C
a .
(84)
For integers
( a,C ,d , A)
, the equality (84) is not feasible.
45
By hypothesis,
( d , A)
are integers, hence, by (74), r is not an integer. In this
2
case, by (71), the expression
(s
+ 32 d 6
)
is not a perfect square integer.
From the definition (51), we have an expression:
s 2 + 32 d 6 = R 2 b , R 1 ,
(85)
R  Integer, b  the prime number of (52).
So, by (85), the solution (73) is:
3y = d +
3
4 3 R b + s 
3
R b  s .
(86)
In Appendix 2, we have showed that the righthand side (86) there is irrational expression. So, for integers
( y , d , A ) , the equality (73) is impossible: the righthand side of
is an
y is an
equality is an irrational quantity. Consequently, equation (8) has no solution in integers
( y , V , A ) . In other words, at least one of the variables ( y , V , A ) irrational number. By hypothesis, ( V , A ) are integers, hence, the variable
If in equation (1) x is integer, then at least one of the variables
irrational value.
(y, z)
is an
irrational number. Thus, for the degree three, Fermat's Last Theorem is true.
4. EQUATION 1 +  1 = u 3
3
1. We write equation (8) as follows:
1 +  1 = u , 3
3
(87)
3V = , 0 < <1; y
A u = , 0 < u < 1. y
3
(88)
46
For equation (87) satisfies conditions [5]:
3V A < < 1, V 1, y y
3V < 3 y
(
3
2 1 < 1,
)
(89)
A 3V < < 1. y y
3
In [5] we considered Fermat's equation for degree p 3 . This equation reduces to a transcendental equation, which has no solution in integers. We represent equation (87) in the form of a transcendental equation for the degree p = 3 , [5]:
u 2 k ( 3 ( 2 k + 1 ) R 2 k  2 k u R 2 k +1 ) =
(90)
= 2 k ( 3 ( 2 k + 1)  2 k ) , k 1 .
Equation (90) must hold for all k 1 . The coefficient R i , i 2 is equal to:
R i = (1  B i ) 3i1 , i 2 ,
B1 = 1 ,
Bi < 1 , i 2 ,
(91)
Bi =
2 5 ( 3i  4 ) , i2 . 3i 1 ( i  1)!
47
Transform the relation (90) to the form:
u
=
2k
Dk , k 1 .
(92)
The coefficient Dk , k 1 is:
3 ( 2 k + 1 ) R 2 k  2 k u R 2 k +1 , k 1. 3( 2 k + 1 )  2 k
Dk =
(93)
For integers
(a , y )
the coefficient Dk is a rational number.
The lefthand side of (92), we present in explicit form:
3V = y u
y , A
3
y > A > 3V , V 1 .
(94)
Under the condition A = a , ( a  a prime number), the relation (94) has the form:
3 y = , y a u
3
y > a 3.
(95)
The equality (92) must hold for all values of k 1 . For integers
( y ,V , A)
,
according to (94), the lefthand side (92), is a proper fraction, the righthand side is a radical degree 2 k from of a rational proper fraction Dk . For the equality (92), the radical
righthand side must be a rational number for all values of k 1 . This requirement is actually not feasible for an infinite sequence k = 1 , 2 , 3 , . . . . In other words, the radical in the righthand side of (91), at least for one of k 1 , is an irrational quantity.
48
In this case, according to (94), at least one of the variables irrational quantity. Since, by hypothesis,
( y ,V , A)
is an
(V , A)
are integers, then the variable y is an
irrational number. Consequently, equation (8) has no solution in integers
( y ,V , A)
.
Under the condition A = a , ( a  a prime number), according to (95), at least one of the variables
(a , y )
is an irrational quantity. Since, by hypothesis, a is integer, then
the variable y is an irrational number. Consequently, equation (21) has no solution in integers ( a , y ) .
2. Under condition k = 1 , the expression (92) is:
1
9  8u 2 = 9  2 . u
(96)
In view of (3) and (88), expression (96) can be written in the form:
2V 8 A9  9 y 3 A6 + 81 y 7 V 2 1  3y = 0.
(97)
Equation (97), we will present in the form of a cubic equation:
t3 + 3 p t + 2 q = 0 ,
(98)
U = t + ,
=
3 y3 , >1, 8V
(99)
p =  2 ,
q =  3 (1  ) ,
(100)
2V = 6 1  3y
4V y .
2
(101)
49
2. For the solution of (98), we apply the method of auxiliary variables ( , ) , [1]. The numbers of real roots of equation (98) depends on the sign of the discriminante D :
D = q 2 + p3 =  ( 2  ) 6 .
(102)
Auxiliary value is equal to:
= ±
p
= ± .
(103)
Sign must coincide with the sign of q . Auxiliary value is determined depending on the signs of
( p , D ) . According to (100), we have
p<0 .
Provided that D 0 , the value of and the solutions of equation (98) are:
cos = ,
=
q
3
,
(104)
t1 =  2 cos , 3
(105)
ti = + 2 cos 3
, i = 2, 3 . 3
Under the condition D > 0 , we have the expressions:
ch = ,
=
q
3
,
(106)
t1 =  2 ch . 3
(107)
The solutions
( t2 , t3 ) are imaginary.
50
A. Provided that > 2
, we have the following values:
q >0, D >0,
= + ,
= 1 .
According to (106), the value
is:
= ln  1 +
(
(  2 ) , > 2 , > 0.
)
According to (99) and (107), the real solution of equation (97) has the form:
U =  2 ch  1 , > 0 , > 0 . 3
(108)
So, provided > 2 , the equation (97) has no positive solutions.
B. Provided that = 2 , we have the following values:
q =3 , D = 0,
= + ,
= 1, = 0 .
In this case, the solutions (105) are:
t1 =  2 ,
t 2 = t3 = .
In view of (99), equation (97) has one positive solution:
U = 2 .
(109)
With (3) and (99), the solution (109), we will present in the form of:
3
A =
6 y. 2
(110)
51
The righthand side of (110) is an irrational quantity, by definition. Consequently, at least one of the variables
( y , A)
is an irrational number.
So, provided = 2 , the equation (97) has no solution in integers ( y , A ) .
C. Provided that = 1 , we have the following values:
q =0, D<0,
= ,
= 0, =
2
.
In this case, the solutions (105) are:
t1 = 
3 ,
t2 = + 3 ,
t3 = 0 .
According to (99), equation (97) has only one rational solution:
U = .
(111)
Using (3) and (99), we will present the solution (111) in the form:
3
A =
3 y. 2
(112)
For integers of the variables
( y , A ) , the equality (112) is not feasible. Consequently, at least one
is an irrational number.
( y , A)
So, provided = 1 , the equation (97) has no solution in integers ( y , A ) .
D. Provided that 1 < < 2 , we have the following values:
q>0, D<0,
= ,
=  1,
= arc cos (  1 ) , 0 <  1 < 1 .
(113)
52
In this case, the solutions (105) are:
t1 =  2 cos , 3
(114)
ti = + cos ± 3
3 sin
, i = 2,3 . 3
In view of (113), for the parameter of , we have the inequality:
0 <
3
<
6
,
(115)
3 < cos < 1, 2 3 0 < 3 sin
3
<
3 . 2
Using (99) and (114), we will present the solutions of equation (97) in the form:
t A3 = 1 + i , i = 1, 2 , 3 .
(116a)
By (115), the functions cos , sin , under the condition 0 < < , 3 3 3 6
take only the irrational values. Since is a rational number, then the solutions
( t1 , t2 , t3 )
number.
are the irrational values. Hence, according to (116a), A3 is also an irrational
In view of (99), we represent the expression (116a) in the form:
1 3 A 3 t 3 = 1 + i , i = 1, 2 , 3 . y 2
(116b)
The righthand side of (116b) is an irrational quantity, by definition. Hense, at least one of the variables
( y , A)
is an irrational number. So, provided 1 < < 2 , the equation
(97) has no solution in integers
( y , A) .
53
E. Provided that 0 < < 1 , we have the following values:
q<0, D<0,
= ,
=1 ,
= arc cos ( 1  ) , 0 < 1  < 1 .
In this case, the solutions (105) are:
t1 = + 2 cos , 3
(117)
(118)
ti =  cos ± 3
3 sin
, i = 2, 3 . 3
In view of (117), for the parameter of , we have the inequalities (115). In view of (118), solutions of equation (97) have the form (116a) and (116b). So, provided 0 < < 1 , the equation (97) also has no solution in integers Since the equation (97) has no solution in integers integers
( y , A)
.
( y ,V ,U )
, then for
( p , y , A,V ) ( p , y , A,V )
equality (96) is not feasible, the lefthand side there is a proper
fraction, while the righthand side is an irrational quantity, by definition. Consequently, for integers and under the condition k 1 , equality (92) also is impossible: the
lefthand side there is a rational number, and the right side is an irrational quantity. Thus, for the degree three, Fermat's Last Theorem is true.
54
APPENDIX 1 Consider the functions ( 1 , 2 ) . Expressions (54a) and (54b), we present the general form:
=
3
c + p
N +
3
c  p
N ,
p 1,
(1.1)
( c , p )  Integers. Integer
N is not a perfect square.
We show that the right side of (1.1) is an irrational expression. Consider a linear dependence on irrational value w . Suppose we have the relations:
3
c + p
N
= (u + w ),
(1.2)
3
c  p
N
= (v  w ).
(u , v )
 Integers, w  the irrational part of expressions ( u + w ) and ( v  w ) .
In view of (1.2), solutions (38) and (40) are:
y = l + u + v,
l = ( 2n + 1, 2n ) , n 1 .
(1.3)
So, by (1.2), we have the expression:
2 c = u 3 + v 3 + 3 u 2  v 2 w + 3 ( u + v ) w2 ,
(
)
(1.4)
2 p
N = u 3  v 3 + 3 u 2 + v 2 w + 3 ( u  v ) w2 + 2 w3 . (1.5)
(
)
If u > v , then the equality (1.4) is impossible c  integer, the righthand side of equality is an irrational expression.
55
Provided that u = v , the equalities (1.4) (1.5) have form:
c = u 3 + 3 w2 u ,
(1.6)
p
N = w3 + 3 u 2 w .
(1.7)
According to (1.6), c  integer, provided that:
w = t a , t 1,
(1.8)
t  Integer, a  the prime number of (52).
If condition (1.8) is not feasible, then for integers ( u , c ) , equality (1.6) also is impossible, since the righthand side of (1.6) is an irrational expression, by definition. Therefore, the equalities (1.2) are not possible. In view of (1.8), the expressions (1.6) (1.7) are:
u3 + 3a t 2 u  c = 0 ,
(1.9)
p
N = a t 2 + 3u 2 t a .
(
)
(1.10)
Equation (1.10) holds if:
N = m a , m 1,
(1.11)
m  Integer.
In view of (1.11), equation (1.10) is:
a t 3 + 3u 2 t  m p = 0 .
(1.12)
56
The equation (1.9) has solution:
3
2 u =
3
c2 + 4 a3 t 6 + c 
3
c 2 + 4 a3 t 6  c .
(1.13)
Subject to a = 2 , the expression (1.13) takes the form:
3
2 u =
3
g + c 
3
g c ,
(1.14)
c2 + 2 4t3
(
)
2
= g2 .
(1.15)
If
( c , 4t
3
,g
) (
are the reciprocal prime numbers, and also
(c , g )
are the odd
integers, then the equation (1.15) has no solution in positive integers (77):
c = ± k 2  2l 2 , 2t3 = k l , g = k 2 + 2l 2 ,
)
(1.16)
( k , l )  The reciprocal prime integers,
k  an odd integer, c  a positive integer.
In view of (1.16), the expression (1.14) is:
u = ±
(
3
k2 
3
2l 2
).
(1.17)
3 3 Provided that k = ko , l = 2lo , we have a system of equations:
u = ko2  2 lo2 ,
6 6 c = ko  8 lo .
( ko , lo )  The reciprocal prime integers,
The equality (1.12) has solution:
ko  an odd integer,
(u , c )
 the odd
integers. Coefficient values c are given by expressions (39) and (41).
3
2 a t =
3
p2 N + 4u6 + p N 
3
p2 N + 4u6  p N
. (1.18)
57
With (1.11), expression (1.18) can be written in the form:
3
2 a t =
3
a ( p m) + 4 u 6 + p m a 
2
3
a ( p m ) + 4 u 6  p m a . (1.19)
2
If a = 2 , then the expression (1.19) takes the form:
3
4 t =
3
r + pm 
3
r  pm ,
(1.20)
( p m )2
If
+ 2 u3
( )
2
= r2 .
(1.21)
( p m , u
3
,r
) (
are the reciprocal prime numbers, and also
( pm, r )
are the
odd integers, then the equation (1.21) has no solution in positive integers (1.16):
p m = ± k 2  2l 2 , u3 = 2 k l , r = k 2 + 2l 2 ,
)
(1.22)
In view of (1.22), the expression (1.20) is:
3
2 t = ±
(
3
k2 
3
2l 2
).
(1.23)
Since k is an odd integer, then for integers ( t , k , l ) , the equality (1.23) is not feasible. Thus, according to (1.20), the variable t is an irrational quantity. So, the condition (1.8) is not feasible. Consequently, equality (1.6) also is not feasible. And so, the equalities (1.2) are impossible. Provided a = 2 A or a = A , A  an odd number (52), we have the expression (1.18). Since the values of
(
2A , A
)
are the irrational numbers, then according to
(1.18), t is an irrational quantity. Thus, the function is an irrational expression. Note that Fermat's last theorem for degree three was proved by Euler [3]. Hence, the solution (1.3) is impossible.
58
APPENDIX 2
Consider the solution (86):
3y = d +
3
4 3 R b + s 
3
R b  s .
(2.1)
Suppose we have the relations:
3
4R b ± 4 s =
(w
± u ),
(2.2)
u  Integer, w  the irrational part of expressions
( w ± u ).
In view of (2.2), the solution (2.1) is an integer:
3y = d + 2u .
So, by (2.2), we have the equalities:
4 s = u 3 + 3 w 2 u ,
(2.3)
4 R b = w3 + 3 u 2 w .
(2.4)
According to (2.3), provided (1.8), s is integer. In view of (1.8), the equalities (1.26) (1.27) take the form:
u3 + 3a t 2 u  4 s = 0 ,
(2.5)
4 R b = ( a t 2 + 3u 2 ) t a .
(2.6)
Equality (2.6) holds under the condition that
b =
a .
59
In this case, equation (2.6) is:
a t 3 + 3u 2 t  4 R = 0 .
(2.7)
The equation (2.5) has solution:
u =
3
4 s 2 + a 3t 6 + 2 s 
3
4 s 2 + a 3t 6  2 s ,
(2.8)
If a = 2 , then the expression (2.8) takes the form:
u =
3
2 3
s2 + 2 t 6 + 2 s 
3
s2 + 2t 6  s .
(2.9)
Consider equation:
s2 + 2t 6 = f 2 .
(2.10)
If
(s , t
3
, f
)
are the reciprocal prime numbers, and also
(s , f )
are the odd
integers, then the equation (2.10) has no solution in positive integers (1.16):
s = ± k 2  2l 2 , t3 = 2 k l ,
(
)
f = k 2 + 2l 2 .
(2.11)
In view of (2.11), the expression (2.9) is:
u = ±
3
4
(
3
k2 
3
2l2
).
(2.12)
Since k is an odd integer, then for integers ( u , k , l ) , the equality (2.12) is not feasible. Thus, according to (2.9), the variable u is an irrational quantity. Consequently, equality (2.3) is not feasible. And so, the equalities (2.2) are impossible.
60
The equation (2.7) has solution:
a t =
3
h + 2 a R
+
3
h  2 a R ,
(2.13)
h2 = 4 a R 2 + u 6 .
(2.14)
If a = 2 , then the expressions (2.13) (2.14) take the form:
2t =
3
h + 2 2 R
+
3
h  2 2 R ,
(2.15)
h 2 = 2( 2 R ) + u 6 .
2
(2.16)
If
(u
3
, 2R , h
)
are the reciprocal prime numbers, and also
(u , h )
are the odd
integers, then the equation (2.16) has no solution in positive integers (1.16):
u3 = ± k 2  2l 2 , R = k l , h = k 2 + 2l 2 .
(
)
(2.17)
In view of (2.17), the solution (2.15) is:
2t = k +
(
2l
)
2 3
 k 
(
2l
)
2 3
.
(2.18)
For integers ( t , k , l ) the equality (2.18) is impossible. Thus, according to (2.15), the variable t is an irrational quantity. So, the condition (1.8) is not feasible. Consequently, equality (2.3) also is not feasible. And so, the equalities (2.2) are impossible. Provided a = 2 A , A  an odd number (52), solution (2.13) takes the form:
2A t =
3
h + 2 2A R
+
3
h  2 2A R ,
(2.19)
h2 = 8 A R 2 + u 6 .
61
Since the values of
2 A are an irrational number, then according to (2.19), t is
an irrational quantity. Thus, the right part of the solution (2.1) is an irrational expression.
APPENDIX 3
Consider the expressions (73) and (75). With (77) and (78), we have a system of equations:
3y =
3
k m ± 2
(
3
k 2  2 3 m2 ,
)
(3.1)
3 A3 = 7 k m ±
( 8m
2
 k2 .
)
(3.2)
Under condition k 2 > 8 m 2 , we have a sign (+ ) . Under condition 8 m 2 > k 2 , we have a sign ( ) .
3 3 Provided that k = ko , m = mo ,
( ko , mo )
 integers, we obtain the equation:
2 3 y = ko mo ± 2 ( ko2  2 mo ) ,
(3.3)
3 A3 = 7 ( ko mo )
3
±
( 8m
6 o
6  ko .
)
(3.4)
2 Under condition ko2 > 2 mo , we have a sign (+ ) . 2 Under condition 2 mo > ko2 , we have a sign ( ) .
According to (3.3), 3 y there is a positive integer. However, the system of equations (3.3)  (3.4) is not solvable in integers ( ko , mo , y , A ) . Consequently, for integers ( k , m , A ) , equality (79) is impossible.
62
REFERENCES 1. . ., . . . . 1986. 2. . . . . 1983. 3. . . . .  . 1934. 4. Ovchinnikov G. I. The solution the equation of Pythagoras in integers 5. Ovchinnikov G. I. Proof of Fermat's theorem www.fermattheorem.net November 2011, Tartu
63
THE PROOF OF FERMAT'S THEOREM FOR A PRIME NUMBER BY METHOD OF DECOMPOSITION
1. We consider the equation:
xp + yp = zp ,
p3
(1)
Fermat's theorem: the equation (1) has no solutions in integers ( p , x , y , z ) . We consider the equation (1), provided that the variables integers. Assuming that x = y , the equation (1) becomes:
( x , y , z ) are positive
2 yp = zp ,
p 3.
(2)
From (2), we have the expression:
1
y 1p = , z 2
p 3.
(3)
For integers
( y,z )
, the equality (3) is not feasible, since the righthand side is
an irrational number by definition. Consequently, at least one of the numbers irrational number. Provided that 0 < x < y < z , we write equation (1) as follows:
( y,z )
is an
zp  yp = xp , z > y > x.
p 3,
(4)
If x is an integer, then by Ferman's theorem, at least one of the numbers ( y , z ) is an irrational number.
64
2. Let x = A . A  The integer. In this case, the equation (4) we write as:
z p  y p = Ap , z > y > A.
p 3,
(5)
Any composite integer can be uniquely represented as a product of prime factors. We represent the integer A p in the form of decomposition into the factors ( V , U
) provided that U > V 1 :
A p = V U , U >V 1, p3 ,
(6)
V U =
{ Vk U k },
k = (1 , N ) ,
(7)
V1 U1 = ... = Vk U k = ... = VN U N , U k > Vk , V1 = 1 , U1 = U .
For an odd number A , integer multipliers ( V , U ) are odd number. For an even number B , integer multipliers ( V , U ) are either even integers or integers of different parity. 3. We consider the equation (5) for even and odd values of p 3 . A. Under the condition p = 2n , n 2 , equation (5) is:
z 2 n  y 2 n = A2 n , n 2 .
(8)
In equation (8), we introduce the new variables ( yo , zo ) :
yo = y n ,
zo = z n , n 2 .
(9)
65
In view of (9), the equation (8) becomes:
2 2 z o  yo = A 2 n , n 2 .
(10)
According to (9), if ( y , z ) are integers, then ( yo , zo ) are also integers. However, if ( yo , zo ) are the integers, then ( y , z ) are or integers or irrational numbers. According to (9), the ratio ( y / z ) is equal to:
y y = o z z o
1/ n
, n2 .
(11)
From (11), we have: if the radical righthand side is an irrational value, then the ration ( y / z ) is an irrational number. Consequently, in equation (8) at least one of the numbres ( y , z ) is an irrational number. Taking into account the expansion (6), we represent equation (10) as follows:
( zo
 yo ) ( z o + yo ) = V U ,
(12)
A2 n = V U , V < U , n 2 .
If ( yo , zo , V , U )  integers, then subject to zo > yo > 0 , U > V > 0 we can represent equation (12) as a system of equations:
z o  yo = V ,
(13)
zo + yo = U .
For the equations (13), we have the solutions:
yo =
U V , 2
zo =
U +V . 2
(14)
66
From solutions (14), we have: If (V , U ) are integers of one parity, then ( yo , zo ) are integers. If (V , U ) are integers of different parity, then ( 2 yo , 2 zo ) are odd integers. In any case, the ratio ( yo / zo ) is a rational fraction. In view of (14), expression (11) is:
y U V = z U +V
1/ n
, n2,
(15)
A2 n = V U , V < U .
B. Under the condition p = 2n + 1 , n 1 , equation (5) is:
z 2 n+1  y 2 n+1 = A2 n+1 , n 1 .
(16)
Let ( y , z ) are perfect squares of integers (Y , Z ) :
y = Y 2, z = Z 2 .
(17)
We introduce the variables:
yo = Y 2 n+1 , zo = Z 2 n+1 , n 1 .
(18)
In view of (17) and (18), the equation (16) is:
2 2 zo  yo = A2 n+1 , n 1 .
(19)
From (18), we have: if (Y , Z ) are integers, then ( yo , zo ) are also integers. However, if ( yo , zo ) are the integers, then ( y , z ) are or integers or irrational numbers.
67
According to (17)  (18), the ratio ( y / z ) is equal to:
y y = o z z o
2 / 2 n +1
, n 1 .
(20)
If ( yo , zo )  integers, then using (12) (13), for equation (19) are valid solutions (14). In view of (14), expression (20) becomes:
y U V = z U +V
2 / 2 n +1
, n 1,
(21)
A2 n +1 = V U , V < U .
4. Consider the equation (5), provided A = a , a  a prime number. In this case A p = a p , ( p 3) and the equation (5) has the form:
z p  y p = a p,
p3 .
(22)
According to (9) ( p = 2n, n 2 ) and (18) ( p = 2n + 1, n 1 ), we introduce in equation (22) the variables ( yo , zo ) :
2 2 zo  yo = a p ,
p3 .
(23)
We represent a p ,
p 3 in the form of decomposition into the factors ( V , U
).
Under the condition p = 2n , n 2 , we have the factors:
a 2 n = a ni a n+i , i = ( 0 , n ) ,
(24)
V = a ni , U = a n +i , i = (1 , n ) .
68
Under the condition p = 2n + 1 , n 1 , we have the factors:
a 2 n+1 = a ni a n+1+i , i = ( 0 , n ) ,
(25)
V = a ni , U = a n +1+ i , i = (0 , n) .
Given the decompositions (24) and (25), we obtain the solutions (14) for the variables ( yo , zo ) . Under the condition p = 2n , n 2 , we have the solutions:
yo =
an a i  a i , 2
(
)
zo =
an a i + a i , 2
(
)
i = (1 , n ) .
(26)
Under the condition p = 2n + 1 , n 1 , we have the solutions:
yo =
an a i+1  a i , 2
(
)
zo =
an a i+1 + a i 2
(
)
,
i = (0 , n) .
(27)
Substituting (26) and (27), respectively, in (15) and (16), we obtain expressions for the ratio ( y / z ) :
2
a  1 p y = a +1 , z
p 3,
(28)
i = (1, n) , 2i, = 2i + 1, i = (0, n) ,
p = 2 n,
n2
.
p = 2n + 1, n 1
(29)
69
5. Under the condition a = 2 , the integer 2 R = 2 , ( 1) is multiple of 2. In this case the expression (28) is equal to:
2
y 2R 1 p = , z 2R + 1
p 3.
(30)
Since (2 R  1) , (2 R + 1) are mutually prime odd numbers, then the radical of the right side of (30) is an irrational quantity. Consequently, at least one of the numbers ( y , z ) is an irrational number. 6. If a 3 , then an odd number of a can be written as:
a = 2S + 1 .
(31)
In this case the expression (28) has the form:
2
y S p = , z S +1
p 3.
(32)
Since S and ( S + 1) are mutually prime integers of different parity, then the radical of the right side of (32) is an irrational value. Consequently, at least one of the numbers
( y, z ) is an irrational number. So, if in the equation (1) subject to 0 < x < y < z , x is a
prime number, then at least one of the numbers ( y , z ) is an irrational number. This assertion is true if p = 2n , n 2 , and also if p = 2n + 1 , n 1 provided that the variables ( y , z ) are perfect squares of integers.
December 2005. Tartu
70
THE SOLUTION THE EQUATION OF PYTHAGORAS IN INTEGERS
1. We consider the equation of Pythagoras:
x2 + y2 = z 2 .
(1)
The equation (1) has the solutions in the form [1, 2]:
x = nm,
y=
m2  n2 , 2
z=
m2 + n2 . 2
(2)
(m > n ) , the expressions (2) are the solutions of the equation (1). If ( m , n ) are integers of one parity, then ( x , y , z ) are the integers. If ( m , n ) are integers of different parity, then ( 2 y , 2 z ) are the odd integers. 2. Consider the equation (1) provided that the numbers ( x , y , z ) are positive
For all values of n and m integers. Under the condition x = y , the equation (1) is:
z = y 2
(3)
For integers
( y,z )
, equality (3) is impossible by definition.
3. We consider the equation (1), subject to:
z > y > x > 0.
Let x = A , where A  integer. In this case, the equation (1) can be written as:
z 2  y 2 = A2 .
(4)
71
The integer A we can be to represent in the form of decomposition into prime factors:
A = 1 a1 a2 ai am ,
(5)
ai+1 ai , i = ( 1 , m  1 ) , a1 2 , ai  A prime number.
Given (5), we represent the integer A2 as an expansion in integer factors ( V , U provided that U > V 1 :
)
A2 = V U , U > V 1 ,
(6)
V U = { Vk U k
} , U k > Vk ,
k = (1 , N
),
V1 U1 = ... = Vk U k = ... = VN U N , V1 = 1 , U1 = U .
In view of (6), we present equation (4) as follows:
( z  y ) ( z + y )
Since
= V U .
(7)
( y , z , V , U )  integers,
z> y>0
U > V > 0 then the equation (7) can be
represented as a system of equations:
z  y = V,
(8)
z + y = U.
From the equations (8), we have the solutions:
y =
U V , 2
z =
U +V , 2
U > V 1.
(9)
72
If
( V ,U )
 the integers of the one parity then
( y,z )
are the integers. If
( V ,U )
 the integers of different parity then
( 2 y , 2z )
are the odd integers.
4. In view of (8), we represent equation (4) as follows:
(V (U
+ y
)2 )2
 y 2 = A2 ,
(10)
 y  y 2 = A2 .
If A  odd integer, then ( V , U
)
the odd integers. If A  even integer, then
( V ,U )
 either the even integers, or the integers of different parity. If
(V , A )
and
(U , A )
 the integers of the one parity, then for a whole number y (any parity), the
equalities (10) are taking place: both sides are the integers of one parity. So, instead of the equation (4) we have the equations (10), where
( V , U , A ) the integers of one parity. ( V , U ) the integers of one parity.
From (10), we have the solutions (9), where
In this case ( y , z ) are whole numbers only. If A  an odd integer, then the multipliers
(V ,U ) are odd integers. If A  an even integer, then the multipliers (V ,U ) are even
integers only. 5. We will compare the solutions (2) and (9), with each other. As a result, we obtain the relations:
m2 + n2 = U + V ,
(11)
m2  n2 = U  V .
From the expressions (11), we get the condition under which the integer solutions (2) and (9) are equal:
V = n2 , U = m2 .
(12)
According to (12), the factors ( V , U ) are perfect squares of integers.
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However, according to (5) and (6), the factors ( Vk , U k ) are not necessarily perfect squares of integers. Thus, the solutions (2) (in integers) are special case of more general solutions (9). 6. The condition (12) is completely feasible, if A = a , where a  a prime number. Consider the solutions (2) and (9), subject to (12). For solution (2), the factors
( n , m ) are equal to:
(13)
x = nm = a , n = 1, m = a .
For solution (8), the factors ( V , U
) are equal to:
(14)
A2 = V U = a 2 , V = 1 , U = a 2 .
In view of (15) and (16), the solutions (2) and (9) are equal to each other and have the form:
y =
a2  1 , 2
z =
a2 +1 . 2
(15)
7. Consider the solutions (15). Assuming that a = 1 , the solutions (15) are y = 0 , z = 1 . Assuming that a = 2 , the solutions (17) are 2 y = 3 , 2 z = 5 . So, under the condition A = 2 , equation (4) has no solution in integers. In other words, if in the equation (4), A = 2 and y is an integer, then a member of
( y 2 + 4 ) is not a perfect square integer, ie z  an irrational number, (Section 8).
Subject to a 3 , an odd number of a 2 , we will present in the form:
a2 = 2 S + 1 .
(16)
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In view of (16), the solutions (15) are:
y = S, z = S + 1.
(17)
According to (17),
( y,z )
 mutually prime numbers with different parity.
In this case, the first three integers for the solutions (17) there are x = a = 3 ,
y =4, z =5 .
8. Consider the equation (4) subject to A = 2 :
z2  y2 = 4 .
(18)
Equality (18) holds provided that A.
( y , z )  integers of the one parity.
( y,z )
 the even integers:
y = 2k , k 1,
(19)
z = 2l , l 1 , l > k .
In view of (19), the equation (18) is:
( l  k ) ( l + k )
B.
= 1, l > k 1 .
(20)
( y,z )
 the odd integers:
y = 2k + 1 , k 0 ,
(21)
z = 2l + 1 , l 1, l > k .
In view of (21), the equation (18) is:
( l  k ) ( l + k + 1 )
= 1, l > k 0 .
(22)
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So, for all values of the integers
( k , l ) , the equalities (20) and (22) are not
feasible: the left side is greater than 1. Consequently, in the equation (18) the numbers
( y,z )
are not integers of the one parity.
If y  an even integer, then in view of (19), the equality (18) has the form:
z2 = 4 k 2 + 1 , k 1
(
)
(23)
From the expression (23) implies that z is an irrational number, by definition:
z = 2 k2 + 1 , k 1.
(24)
If y  an odd integer, then in view of (21), the equality (18) has the form:
z2 = 4 k 2 + k +1 +1 , k 0 .
(
)
(25)
For the integers ( z , k ) , the equality (25) holds, if z  an odd number. However, according to (22), z is not an odd number. Consequently, for integers
( z , k ) , the
equality (25) is impossible: the left  an even number, and the right  an odd number. We represent (25) in the form of solution:
z =
4(k2 + k + 1 ) + 1 , k 0 .
(26)
If y  an integer, then according to (20) and (22) z can not be an integer. Consequently, the radical in the right of (26) is an irrational number.
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9. A simple example will show that the solutions (2) in integers are a special case of solutions (9). Consider equation (1) provided:
x = aN , a 3 ,
N 2.
(27)
In view of (27), equation (1) is:
z 2  y2 = a2N , a 2 , N 2 .
(28)
A) For equation (28) we define the factors
( n,m ) .
We distinguish the odd and even number of values N :
N = 2p , p 1;
N = 2 p +1, p 1.
Provided that N = 2 p , p 1 the factors
( n , m ) are:
(29)
n = a p i , m = a p + i , i = ( 1 , p ) , p 1 .
Provided that N = 2 p + 1 , p 1 , the facors
( n , m ) are:
(30)
n = a p i , m = a p +i +1 , i = ( 0 , p ) , p 1 .
In view of (29)  (30), we obtain the solutions (2) for variables Provided that N = 2 p , p 1 , we have the solutions:
( y,z )
.
y = a 2 ( p  i )  1 a 4i  1 ,
(
)
i = (1 , p ) , p 1 .
(31)
z = a 2( p  i ) 1 a 4i + 1 ,
(
)
i = (1 , p ) , p 1 .
(32)
Provided i = ( 1 , p  1 ) , p 2 solutions (31)  (32) are even numbers for all a 2 .
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Provided i = p , p 1 solutions (28)  (29) are as follows:
y =
a4 p  1 , 2
z =
a4 p + 1 , 2
p 1.
(33)
Under the condition a = 2 solutions (33) are rational fractions. If a prime number a 3 , then the solutions (33) are integers. Provided that N = 2 p + 1 , p 1 , we have the solutions:
y = a 2 ( p  i ) 1 a 4i + 2  1 ,
(
)
i = (0 , p) , p 1
(34)
z = a 2 ( p  i )  1 a 4i + 2 + 1 ,
(
)
i = (0 , p) , p 1
(35)
Provided i = ( 0 , p  1 ) , p 2 solutions (34)  (35) are even numbers for all a 2 . Provided i = p , p 1 solutions (34)  (35) are as follows:
y =
a4 p + 2  1 , 2
z =
a4 p + 2 + 1 , 2
p 1.
(36)
Under the condition a = 2 solutions (36) are rational fractions. If a prime number a 3 , then the solutions (36) are integers. B) For equation (28) we define the factors ( V , U
):
A = aN ,
A2 = V U = a 2 N , a 2 , N 2 ,
(37)
V = aN  j , U = aN + j ,
j = (1 , N ) .
In view of (37), the solutions (9) are:
y = a N  j 1 a 2 j  1 ,
(
)
j = (1 , N ) , N 2 ,
(38)
78
z = a N  j 1 a 2 j + 1 ,
(
)
j = (1 , N ) , N 2 .
(39)
We compare the solutions (38)  (39), respectively, with solutions (31)  (32). Under the condition N = 2 p , p 1 , the solutions (38)  (39) are:
y = a 2 p  j 1 a 2 j  1 ,
(
)
j = (1 , 2 p ) , p 1 ,
(40)
z = a 2 p  j 1 a 2 j + 1 ,
(
)
j = (1 , 2 p ) , p 1 .
(41)
Provided j = 2 i , i = ( 1 , p ) , p 1 solutions (40)  (41) coincide, respectively, with solutions (31)  (32). In addition, there are additional solutions in integers for odd values j = 1 , 3 , ... , 2 p  3 , 2 p  1 . We introduce the notation:
j = 2i + 1 , i = ( 0 , p 1 ) , p 1 .
Provide additional solutions
( yo , zo ) in the form:
i = ( 0 , p  1) , p 1 ,
yo = a 2 ( p  i  1 ) ( a 2 ( 2 i + 1 )  1 ) , zo = a 2 ( p  i  1 ) ( a 2 ( 2 i + 1 ) + 1 ) ,
(42)
i = ( 0 , p  1) , p 1 .
(43)
Provided i = ( 0 , p  2 ) , p 2 the solutions (42)  (43) are even numbers for all a 2 . Provided i = p  1 , p 1 the solutions (42)  (43) are as follows:
yo = a 2 ( 2 p1 )  1 ,
zo = a 2 ( 2 p1 ) + 1 ,
p 1.
(44)
Under the condition a = 2 the solutions (44) are odd numbers. If a prime number a 3 , then the solutions (44) are even numbers. We compare the solutions (38) (39), respectively, with solutions (34)  (35).
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Provided N = 2 p + 1 , p 1 , solutions (38)  (39) are:
y = a2 p  j a2 j  1 ,
(
)
j = (1 , 2 p + 1) , p 1 ,
(45)
z = a2 p  j a2 j + 1 ,
(
)
j = (1 , 2 p + 1) , p 1 .
(46)
Provided j = 2 i + 1 , i = ( 1 , p ) , p 1 solutions (45)  (46) coincide, respectively, with solutions (34)  (35). In addition, there are additional solutions in integers for even values j = 2 , 4 , ... , 2 ( p  1) , 2 p . We introduce the notation j = 2 i , i = ( 0 , p ) , p 1 and provide additional solutions
( yo , zo ) in the form:
yo = a 2 ( p  i ) ( a 4 j  1 ) , zo = a 2 ( p  i ) ( a 4 j + 1 ) , j = (1 , p ) , p 1 ,
(47)
j = (1 , p ) , p 1 .
(48)
Provided i = ( 1 , p  1 ) , p 1 the solutions (47)  (48) are even numbers for all a 2 . Provided i = p  1 , p 1 the solutions (47)  (48) are as follows:
yo = a 4 p  1 ,
zo = a 4 p + 1 ,
p 1.
(49)
Under the condition a = 2 the solutions (49) are odd numbers. If a prime number a 3 , then the solutions (49) are even numbers. So, with the additional solutions (42)  (43) and (47)  (48), the integral solutions (2) are special cases of more general solutions (9) in integers. 1. . . . .. 1934. 2. . . . .1983. August 2007. Tartu
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