Fermat Last Theorem And Riemann Hypothesis(2)vixra.org › pdf › 1101.0079v1.pdfa conjecture...

1

Fermat Last Theorem And Riemann Hypothesis(2) Automorphic Functions

And Fermat’s Last Theorem（2）

Chun-Xuan Jiang

P. O. Box 3924, Beijing 100854, P. R. China [email protected]

Abstract

In 1637 Fermat wrote: “It is impossible to separate a cube into two cubes, or a biquadrate into two biquadrates, or in general any power higher than the second into powers of like degree: I have discovered a truly marvelous proof, which this margin is too small to contain.”

This means: ( 2)n n nx y z n+ = > has no integer solutions, all different from 0(i.e., it has only the trivial solution, where one of the integers is equal to 0). It has been called Fermat’s last

theorem (FLT). It suffices to prove FLT for exponent 4. and every prime exponent P . Fermat proved FLT for exponent 4. Euler proved FLT for exponent 3.

In this paper using automorphic functions we prove FLT for exponents 6P and P , where P is an odd prime. The proof of FLT must be direct .But indirect proof of FLT is disbelieving.

In 1974 Jiang found out Euler formula of the cyclotomic real numbers in the cyclotomic

fields

2 1 2

1

1 1

expn n

i ii i

i i

t J S J−

−

= =

⎛ ⎞ =⎜ ⎟⎝ ⎠∑ ∑ （1）

2

where J denotes a 2n th root of unity, 2 1nJ = , n is an odd number, it are the real numbers.

iS is called the automorphic functions(complex hyperbolic functions) of order 2n with 2 1n − variables [5,7].

1

12

( 1)

1

1 ( 1)2 ( 1) cos ( 1)2

j

n

i jBA ji j

j

i jS en n

πθ

−

−

=

⎡ ⎤−⎛ ⎞⎢ ⎥= + − + −⎜ ⎟⎢ ⎥⎝ ⎠

⎢ ⎥⎣ ⎦∑

2

1( 1) 2

( 1) 1

1

( 1) ( 1)2 ( 1) cos ( 1)2

j

ni

DA i j jj

j

i je en n

πφ

−−

− +

=

⎡ ⎤− −⎛ ⎞⎢ ⎥+ + − + −⎜ ⎟⎢ ⎥⎝ ⎠

⎢ ⎥⎣ ⎦∑ , （2）

where 1,..., 2i n= ;

2 1 2 1 2 1

( 1)1

1 1 1, ( 1) cos , ( 1) ( 1) sin

n n nj j j

j jj jA t B t tn n

α αα α α

α α α

α π α πθ− − −

+

= = =

= = − = − −∑ ∑ ∑ ,

2 1 2 1

( 1)2

1 1( 1) , ( 1) cos

n nj

jjA t D tn

α αα α

α α

α π− − −= =

= − = −∑ ∑ ,

12 1 2

( 1)1 2

1 1

( 1) ( 1) sin , 2 ( ) 0

nn

j jj j j

j

jt A A B Dn

αα

α

α πφ

−−

−

= =

= − − + + + =∑ ∑ （3）

From (2) we have its inverse transformation[5,7]

1 22 2

1

1 1

, ( 1)n n

A A ii i

i i

e S e S += =

= = −∑ ∑

2 1

1 11

cos ( 1) cosjn

B ijj i

i

ije S Snπθ

−

+=

= + −∑ ,

2 1

( 1)1

1

sin ( 1) ( 1) sinjn

B j ijj i

i

ije Snπθ

−+

+=

= − −∑ ,

2 1

( 1)1 1

1

cos ( 1) cosjn

D j ij i

i

ije S Snπφ

−−

+=

= + −∑

2 1

( 1)1

1

sin ( 1) ( 1) sinjn

D j j ij i

i

ije Snπφ

−−

+=

= − −∑ （4）

(3) and (4) have the same form. From (3) we have

12

1 21

exp 2 ( ) 1

n

j jj

A A B D

−

=

⎡ ⎤⎢ ⎥+ + + =⎢ ⎥⎢ ⎥⎣ ⎦

∑ （5）

3

From (4) we have

1 2 212

2 1 31 2

1

2 2 1 1

exp 2 ( )

nn

j jj

n n

S S SS S S

A A B D

S S S

−

=

−

⎡ ⎤⎢ ⎥+ + + =⎢ ⎥⎢ ⎥⎣ ⎦

∑

L

L

L L L L

L

1 1 1 1 2 1

2 2 1 2 2 1

2 2 1 2 2 1

( ) ( )( ) ( )

( ) ( )

n

n

n n n n

S S SS S S

S S S

−

−

−

=

L

L

L L L L

L

(6)

where ( ) ii jj

SSt∂

=∂

[7]..

From (5) and (6) we have circulant determinant

1 2 212

2 1 31 2

1

2 2 1 1

exp 2 ( ) 1

nn

j jj

n n

S S SS S S

A A B D

S S S

−

=

−

⎡ ⎤⎢ ⎥+ + + = =⎢ ⎥⎢ ⎥⎣ ⎦

∑

L

L

L L L L

L

（7）

If 0≠iS ，where ni 2,...,3,2,1= , then (7) have infinitely many rational solutions.

Let 1=n . From (3) we have 11 tA = and 12 tA −= . From (2) we have

11 ch tS = 12 sh tS = （8）

we have Pythagorean theorem

1shch 12

12 =− tt （9）

(9) has infinitely many rational solutions.

Assume 0,0,0 21 ≠≠≠ iSSS , where ni 2,...,3= . 0=iS are )22( −n indeterminate

equations with )12( −n variables. From (4) we have

njSSSSeSSeSSe jBAA j πcos)1(2,, 21

22

21

22121

21 −++=−=+= ,

njSSSSe jD j πcos)1(2 121

22

21

2 +−++= （10）

Example. Let 15=n . From (3) and (10) we have Fermat’s equation

1)()()](2exp[ 3102310

1302

301

7

121 =−=−=+++ ∑

=

SSSSDBAA jjj

（11）

From (3) we have

4

555

1631 )][exp()22exp( j

jtBBA ∑

=

=++ （12）

From (10) we have

5251631 )22exp( SSBBA +=++ (13)

From (12) and (13) we have Fermat’s equation

555

1

52

51631 )][exp()22exp( j

jtSSBBA ∑

=

=+=++ (14)

Euler prove that (19) has no rational solutions for exponent 3 [8]. Therefore we prove that (14) has no rational solutions for exponent 5. Theorem. Let Pn 3= where P is an odd prime. From (7) and (8) we have Fermat’s equation

1)()()](2exp( 32232

162

61

213

121 =−=−=+++ ∑

−

=

PPPPjj

P

j

SSSSDBAA (15)

From (3) we have

P

jPj

j

P

j

tBA⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+ ∑∑

=

−

=

5

13

21

11 exp2exp (16)

From (10) we have

PPj

P

j

SSBA 2132

1

11 2exp +=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+ ∑

−

=

(17)


P

jPj

PPj

P

j

tSSBA⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛=+=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛+ ∑∑

=

−

=

5

1213

21

11 exp2exp (18)

Euler prove that (15) has no rational solutions for exponent 3[8]. Therefore we prove that (18) has no rational solutions for prime exponent P [5,7]. Remark. It suffices to prove FLT for exponent 4. Let Pn 4= , where P is an odd prime. We have the Fermat’s equation for exponent P4 and the Fermat’s equation for exponent P [2,5,7]. This is the proof that Fermat thought to have had. In complex hyperbolic functions let exponent n be Pn Π= , Pn Π= 2 and Pn Π= 4 . Every factor of exponent n has the Fermat’s equation [1-7]. In complex trigonometric functions let exponent n be Pn Π= , Pn Π= 2 and Pn Π= 4 . Every factor of exponent n has Fermat’s equation [1-7]. Using modular elliptic curves Wiles and Taylor prove FLT [9, 10]. This is not the proof that Fermat thought to have had. The classical theory of automorphic functions, created by Klein and Poincare, was concerned with the study of analytic functions in the unit circle that are invariant under a discrete group of

5

transformation. Automorphic functions are the generalization of trigonometric, hyperbolic, elliptic, and certain other functions of elementary analysis. The complex trigonometric functions and complex hyperbolic functions have a wide application in mathematics and physics. Acknowledgments. We thank Chenny and Moshe Klein for their help and suggestion. References [1] Jiang, C-X, Fermat last theorem had been proved, Potential Science (in Chinese), 2.17-20

(1992), Preprints (in English) December (1991). http://www.wbabin.net/math/xuan47.pdf. [2] Jiang, C-X, Fermat last theorem had been proved by Fermat more than 300 years ago,

Potential Science (in Chinese), 6.18-20(1992). [3] Jiang, C-X, On the factorization theorem of circulant determinant, Algebras, Groups and

Geometries, 11. 371-377(1994), MR. 96a: 11023, http://www.wbabin.net/math/xuan45.pdf [4] Jiang, C-X, Fermat last theorem was proved in 1991, Preprints (1993). In: Fundamental open

problems in science at the end of the millennium, T.Gill, K. Liu and E. Trell (eds). Hadronic Press, 1999, P555-558. http://www.wbabin.net/math/xuan46.pdf.

[5] Jiang, C-X, On the Fermat-Santilli theorem, Algebras, Groups and Geometries, 15. 319-349(1998)

[6] Jiang, C-X, Complex hyperbolic functions and Fermat’s last theorem, Hadronic Journal Supplement, 15. 341-348(2000).

[7] Jiang, C-X, Foundations of Santilli’s Isonumber Theory with applications to new cryptograms, Fermat’s theorem and Goldbach’s Conjecture. Inter. Acad. Press. 2002. MR2004c:11001, http://www.wbabin.net/math/xuan13.pdf. http://www.i-b-r.org/docs/jiang.pdf

[8] Ribenboim,P, Fermat last theorem for amateur, Springer-Verlag, (1999). [9] Wiles A, Modular elliptic curves and Fenmat’s last theorem, Ann of Math, (2) 141 (1995),

443-551. [10] Taylor, R. and Wiles, A., Ring-theoretic properties of certain Hecke algebras, Ann. of Math.,

(2) 141(1995), 553-572.

In press atAlgebras, Groups and Geometreis, Vol. 21, 2004

DISPROOFS OF RIEMANN’S HYPOTHESIS

Chun-Xuan, JiangP.O.Box 3924, Beijing 100854, China

and Institute for Basic ResearchP.O.Box 1577, Palm Harbor, FL 34682, U.S.A.

[email protected]

AbstractAs it is well known, the Riemann hypothesis on the zeros of the ζ(s)function has been assumed to be true in various basic developments ofthe 20-th century mathematics, although it has never been proved tobe correct. The need for a resolution of this open historical problemhas been voiced by several distinguished mathematicians. By using pre-ceding works, in this paper we present comprehensive disproofs of theRiemann hypothesis. Moreover, in 1994 the author discovered the arith-metic function Jn(ω) that can replace Riemann’s ζ(s) function in view ofits proved features: if Jn(ω) 6= 0, then the function has infinitely manyprime solutions; and if Jn(ω) = 0, then the function has finitely manyprime solutions. By using the Jiang J2(ω) function we prove the twinprime theorem, Goldbach’s theorem and the prime theorem of the formx2 + 1. Due to the importance of resolving the historical open natureof the Riemann hypothesis, comments by interested colleagues are heresolicited.

AMS mathematics subject classification: Primary 11M26.

1. Introduction

In 1859 Riemann defined the zeta function[1]

ζ(s) =∏p

(1− p−s)−1 =∞∑

n=1

1

ns, (1)

where s = σ + ti, i =√−1, σ and t are real, p ranges over all primes.

ζ(s) satisfies the functional equation [2]

π−s2 Γ(

s

2)ζ(s) = π−

(1−s)2 Γ(

1− s2

)ζ(1− s). (2)

From (2) we haveζ(ti) 6= 0. (3)

Riemann conjectured that ζ(s) has infinitely many zeros in 0 ≤ σ ≤ 1,called the critical strip. Riemann further made the remarkable conjecturethat the zeros of ζ(s) in the critical strip all lie on the central line σ = 1/2,a conjecture called the famous Riemann hypothesis (RH).

It was stated by Hardy in 1914 that infinitely many zeros lie on theline; A. Selberg stated in 1942 that a positive proportion at least of allthe zeros lie on the line; Levinson stated in 1974 that more than onethird of the zeros lie on the line; Conrey stated in 1989 that at least twofifths of the zeros lie on the line.

The use of the RH then lead to many mathematical problems, suchas the generalized Riemann conjecture, Artin’s conjecture, Weil’s con-jecture, Langlands’ program, quantum chaos, the hypothetical Riemannflow [3, 4], the zeta functions and L-functions of an algebraic variety andother studies. Similarly, it is possible to prove many theorems by usingthe RH.

However, the RH remains a basically unproved conjecture to thisday. In fact, Hilbert properly stated in 199 that the problem of provingor disproving the RH is one of the most important problems confronting20th century mathematicians. In 2000 Griffiths and Graham pointed outthat the RH is the first challenging problem for the 21st century. Theproof of the RH then become the millennium prize problem.

In 1997 we studied the tables of the Riemann zeta function [5] andreached preliminary results indicating that the RH is false [6, 7, 8]. In

2

this paper we present a comprehensive disproof of the RH and show thatthe computation of all zeros of the ζ(1/2 + ti) function done during thepast 100 years is in error, as preliminarily indicated in Refs. [9, 10, 11].Since the RH is false, all theorems and conjectures based to the same arealso false.

2. Disproofs of Riemann’s Hypothesis

Theorem 1. ζ(s) has no zeros in the critical strip, that is ζ(s) 6= 0,where 0 ≤ σ ≤ 1.

Proof 1. From (1) we have

1

ζ(s)=

∏p

(1− 1ps

) = Reθi, (4)

where

R =∏p

Rp, Rp =

√1− 2 cos(t log p)

pσ+

1

p2σ, (5)

θ =∑p

θp, θp = tan−1 sin(t log p)

pσ − cos(t log p) . (6)

If σ = 0, from (5) we have Rp =√

2√

1− cos(t log p). If cos(t log p) =1, we have Rp = 0 then R = 0. If σ > 0 from (5) we have Rp 6= 0.ζ(s) = 0 if and only if Re ζ(s) = 0 and Im ζ(s) = 0, that is R = ∞. From(5) we have that if cos(t log p) ≤ 0 then Rp > 1 and if cos(t log p) > 0then Rp < 1. cos(t log p) is independent of the real part σ, but maywell depend on primes p and imaginary part t. We write m+(t) for thenumber of primes p satisfying cos(t log p) > 0, m−(t) for the number ofprimes p satisfying cos(t log p) ≤ 0.

For cos(t log p) > 0, we have

1 > Rp(1 + ti) > Rp(0.5 + ti). (7)

If m+(t1) is much greater than m−(t1) such that R(0.5 + t1i) = min.From (5), (6) and (7) we have for given t1

minR(σ1 + t1i) > minR(1 + t1i) > minR(0.5 + t1i)

> minR(σ2 + t1i) → 0,

3

(8)

θ(σ1 + t1i) = θ(1 + t1i) = θ(0.5 + t1i) = θ(σ2 + t1i) = const (9)

where σ1 > 1 and 0 ≤ σ2 < 0.5.Since | ζ(s) |= 1

Rfrom (8) we have

max | ζ(σ1 + t1i) |< max | ζ(1 + t1i) |

< max | ζ(0.5 + t1i) |< max | ζ(σ2 + t1i) |→ ∞. (10)For cos(t log p) < 0 we have

1 < Rp(0.5 + ti) < Rp(0.4 + ti) < Rp(0.3 + ti). (11)

If m−(t1) is much greater than m+(t1) such that R(0.5 + t1i) = max.From (5), (6) and (11) we have for given t1

maxR(σ1 + t1i) < maxR(0.5 + t1i) < maxR(0.4 + t1i),

< maxR(0.3 + t1i) < maxR(σ2 + t1i) 6= ∞, (12)

θ(σ1 + t1i) = θ(0.5 + t1i) = θ(0.4 + t1i)

= θ(0.3 + t1i) = θ(σ2 + t1i) = const,

(13)

where σ1 > 0.5 and 0 ≤ σ2 < 0.3.Since | ζ(s) |= 1

Rfrom (12) we have

min | ζ(σ1 + t1i) |> min | ζ(0.5 + t1i) |> min | ζ(0.4 + t1i) |>

min | ζ(0.3 + t1i) |> min | ζ(σ2 + t1i) |6= 0. (14)Proof 2. We define the beta function

β(s) =∏p

(1 + p−s)−1 =∞∑

n=1

λ(n)

ns, (15)

where λ(1) = 1, λ(n) = (−1)a1+···+ak if n = pa11 · · · pakk , t 6= 0.

4

From (15) we have

1

β(s)=

∏p

(1 +1

ps) = R̄eθ̄i, (16)

where

R̄ =∏p

R̄p, R̄p =

√1 +

2 cos(t log p)

pσ+

1

p2σ, (17)

θ̄ =∑p

θ̄p, θ̄p = tan−1 − sin(t log p)

pσ + cos(t log p). (18)

For cos(t log p) < 0, we have

1 > R̄p(1 + ti) > R̄p(0.5 + ti). (19)

If m−(t1) is much greater than m+(t1) such that R̄(0.5 + t1i) = min.From (17), (18) and (19) we have for given t1

minR̄(σ1 + t1i) > minR̄(1 + t1i) > minR̄(0.5 + t1i)

> minR̄(σ2 + t1i) → 0, (20)θ̄(σ1 + t1i) = θ̄(1 + t1i) = θ̄(0.5 + t1i) = θ̄(σ2 + t1i) = const, (21)

where σ1 > and 0 ≤ σ2 < 0.5.Since | β(s) |= 1

R̄from (20) we have

max | β(σ1 + t1i) |< max | β(1 + t1i) |

< max | β(0.5 + t1i) |< max | β(σ2 + t1i) |→ ∞. (22)For cos(t log p) > 0 we have

1 < R̄p(0.5 + ti) < R̄p(0.4 + ti) < R̄p(0.3 + ti). (23)

If m+(t1) is much greater than m−(t1) such that that R̄(0.5+t1i) = max.From (17), (18) and (23) we have for given t1

maxR̄(σ1 + t1i) < maxR̄(0.5 + t1i) < maxR̄(0.4 + t1i)

< maxR̄(0.3 + t1i) < maxR̄(σ2 + t1i) 6= ∞, (24)

5

θ̄(σ1 + t1i) = θ̄(0.5 + t1i) = θ̄(0.4 + t1i)

= θ̄(0.3 + t1i) = θ̄(σ2 + t1i) = const,

(25)

where σ1 > 0.5 and 0 ≤ σ2 < 0.3.Since | β(s) |= 1

R̄from (24) we have

min | β(σ1 + t1i) |> min | β(0.5 + t1i) |

> min | β(0.4 + t1i) |>

min | β(0.3 + t1i) |> min | β(σ2 + t1i) |6= 0. (26)From (1) and (15) we have

ζ(2s) = ζ(s)β(s). (27)

In 1896 J. Hadamard and de la Vallee Poussin proved independently| ζ(1 + ti) |6= 0. From (27) we have

| ζ(1 + 2ti) |=| ζ(12

+ ti) || β(12

+ ti) |6= 0. (28)

From (28) we have

| ζ(12

+ ti) |6= 0 and | β(12

+ ti) |6= 0. (29)

ζ(s) and β(s) are the dual functions. From (22) we have

| β(12

+ ti) |6= ∞. (30)

Therefore we have

| ζ(12

+ ti) |6= 0. (31)In the same way we have

| ζ(12

+ 2ti) |=| ζ(14

+ ti) || β(14

+ ti) |6= 0. (32)

6

From (32) we have

| ζ(14

+ ti) |6= 0 and | β(14

+ ti) |6= 0. (33)

In the same way we have

| ζ( 12n

+ ti) |6= 0. (34)

As n →∞ we have| ζ(ti) |6= 0. (35)

Proof 3. For σ > 1 we have

log ζ(s) =∑p

∞∑

m=1

m−1p−mσexp(−itm log p). (36)

If ζ(s) had a zero at 12+ ti, then log | ζ(σ + ti) | would tend to −∞ as σ

tends to 12

from the right. From (36) we have

log | ζ(s) |= ∑p

∞∑

m=1

m−1p−mσ cos(tm log p), (37)

with t replaced by 0, t, 2t, · · · , Ht, it givesH−1∑

j=0

(2Hj

)log | ζ(σ + (H − j)ti) | +1

2

(2HH

)

log ζ(σ) =∑p

∞∑

m=1

m−1p−mσA ≥ 0, (38)

where

A =H−1∑j=0

(2Hj

)cos ((H − j)tm log p) + 1

2

(2HH

)

= 2H−1 [1 + cos(tm log p)]H ≥ 0,(39)

H is an even number.

7

From (38) we have

(ζ(σ))

12

(2HH

)H−1∏

j=0

|ζ(σ + (H − j)ti)|

(2Hj

)

≥ 1. (40)

Since | ζ(12

+ eti) |6= ∞[5], where e = 1, 2, · · · , H, from (40) we have| ζ(1

2+ eti) |6= 0 for sufficiently large even number H.

Min | ζ(12+ ti) ≈ 0 but 6= 0. The computation of all zeros of ζ(1

2+ ti)

is error, which satisfies the the error RH.From (39) we have

cos 2θ + 4 cos θ + 3 = 2(1 + cos θ)2,

cos 4θ + 8 cos 3θ + 28 cos 2θ + 56 cos θ + 35 = 8(1 + cos θ)4,

cos 5θ + 12 cos 5θ + 66 cos 4θ + 220 cos 3θ

+495 cos 2θ + 792 cos θ + 462 = 32(1 + cos θ)6.

3. The Arithmetic Function Jn(ω) Replacing Rie-mann’s Hypothesis

In view of the preceding results, the RH has no value for the study ofprime distributions. In 1994 the author discover the arithmetic functionJn(ω) [12, 13] that is able to take the place of Riemann’s zeta-functionsand L-functions because of the following properties: Jn(ω) 6= 0, then thefunction has infinitely many prime solutions; and if Jn(ω) = 0, then thefunction has finitely many prime solutions.

By using Jiang’s Jn(ω) function we have proved numerous theoremsincluding the twin prime theorem, Goldbach’s theorem, the prime k-tuples theorem, Santilli’s theory for a prime table, the theorem of finiteFermat primes, the theorem of finite Mersenne primes, the theorem offinite repunit primes, there are infinitely many triples of n, n+1 and n+2that each is the product of k distinct primes, there are infinitely manyCarmichael numbers which are product of exactly five primes, there thereare finitely many Carmichael numbers which are product of exactly sixprimes · · · in the prime distributions [14]. We give some theorems below

Theorem 2. Twin prime theorem: p1 = p + 2.

8

We have the arithmetic function

J2(ω) =∏

3≤p≤pi(p− 2) 6= 0.

Since J2(ω) 6= 0, there are infinitely many primes p such that p1 is aprime.

Theorem 3. Goldbach theorem: N = p1 + p2.We have the arithmetic function

J2(ω) =∏

3≤p≤pi(p− 2) ∏

p|N

p− 1p− 2 6= 0.

Since J2(ω) 6= 0, every even number N greater than 4 is the sum of twoodd primes.

Theorem 4. p1 = (p + 1)2 + 1.


J2(ω) =∏

3≤p≤pi(p− 2− (−1) p−12 ) 6= 0.


Theorem 5. p1 = p2 − 2.

We have

J2(ω) =∏

3≤p≤pi(p− 2− (2

p)) 6= 0.


Theorem 6. p1 = p + 4 and p2 = 4p + 1.We have

J2(ω) = 3∏

7≤p≤pi(p− 3) 6= 0.

Since J2(ω) 6= 0, there are infinitely many primes p such that p1 and p2are primes.

9

Theorem 7. p1 = (p + 1)2 + 1 and p2 = (p + 1)

2 + 3.We have

J2(ω) =∏

5≤p≤pi(p− 3− (−1) p−12 − (−3

p)) 6= 0.

Since J2(ω) 6= 0, there are infinitely many primes p such that p1 and p2are primes.

Theorem 8. The prime 13-tuples theorem: p+b: b = 0, 4, 6, 10, 16, 18,24, 28, 34, 40, 46, 48, 90.Since J2(13) = 0, there are no prime 13-tuples if p 6= 13.

Theorem 9. The prime 14-tuples theorem: p+b: b = 0, 2, 6, 8, 12, 18,20, 26, 30, 32, 36, 42, 48, 50.We have

J2(ω) = 300∏

29≤p≤pi(p− 14) 6= 0.

Since J2(ω) 6= 0, there are infinitely many prime 14-tuples.Theorem 10. p1 = 6m + 1, p2 = 12m + 1, p3 = 18m + 1, p4 =

36m + 1, p5 = 72m + 1.We have

J2(ω) = 12∏

13≤p≤pi(p− 6) 6= 0.

Since J2(ω) 6= 0, there are infinitely many integers m such that p1, p2, p3, p4and p5 are primes. n = p1p2p3p4p5 is the Carmichael numbers.

Theorem 11. p3 = p1 + p2 + p1p2.We have

J3(ω) =∏

3≤p≤pi(p2 − 3p + 3) 6= 0.

Since J3(ω) 6= 0, there are infinitely many primes p1 and p2 such that p3is a prime.

Theorem 12. p3 = (p1 + 1)5 + p2.

We haveJ3(ω) =

∏

3≤p≤pi(p2 − 3p + 3) 6= 0.

10

Since J3(ω) 6= 0, there are infinitely many primes p1 and p2 such that p3is a prime.

Theorem 13. p4 = p1(p2 + p3) + p2p3.We have

J4(ω) =∏

3≤p≤pi

((p− 1)4 − 1

p+ 1

)6= 0.

Since J4(ω) 6= 0, there are infinitely many primes p1, p2 and p3 such thatp4 is a prime.

Theorem 14. Each of n, n + 1 and n + 2 is the product of k distinctprimes.

Suppose that each of m1,m2 = m1+1 and m3 = m1+2 is the productof k − 1 distinct primes. We define

p1 = 2m2m3x + 1, p2 = 2m1m3x + 1, p3 = 2m1m2x + 1. (41)


J2(ω) =∏

3≤p≤pi(p− 4− χ(p)) 6= 0, (42)

where χ(p) = −2 if p | m1m2m3; χ(p) = 0 otherwise.Since J2(ω) 6= 0, there exist infinitely many integers x such that p1, p2and p3 are primes.

¿From (41) we have n = m1p1 = 2m1m2m3x+m1, n+1 = m1p1 +1 =2m1m2m3x + m1 + 1 = m2(2m1m3x + 1) = m2p2, n + 2 = m1p1 + 2 =2m1m2m3x + m1 + 2 = m3(2m1m2x + 1) = m3p3. If p1, p2 and p3 areprimes, then each of n, n + 1 and n + 2 is the product of k distinctprimes. For example, n = 1727913 = 3× 11× 52361, n + 1 = 1727914 =2× 17× 50821, n + 2 = 1727915 = 5× 7× 49369.

Jn(ω) is a generalization of Euler’s proof for the existence of infinitelymany primes. It has a wide application in various fields.

Acknowledgements

The Author would like to express his deepest appreciation to ProfessorsR. M. Santilli, G. Weiss, D. Zuckerman, Ke-xi Liu, Mao-xian Zuo, Zhong-dian Wang, Chang-pei Wang, and Xin-ping Tong for their helps andsupports.

11

References

[1] B. Riemann, “Uber die Anzahl der Primzahlen under einergegebener Grösse,” Monatsber. Akad. Berlin. 671-680 (1859).

[2] H. Davenport, “Multiplicative Number Theory, ” Springer Verlag(1980).

[3] N. Katz and P. Sarnak, “Zeroes of zeta functions and symmetry,”Bull. AMS, 36, 1-26 (1999).

[4] A. Connes,“Trace formula in noncommutative geometry and thezeros of the Riemann zeta function,” Sel. Math., New Ser. 5, 29-106(1999).

[5] C. B. Haslgrove,“Tables of the Riemann zeta function,” Roy. Soc.Math. Tables, Vol.6, Cambridge Univ. Press, London and New York(1960).

[6] Chun-xuan, Jiang, “The study of the Riemann Zeta function,” Un-published Oct. (1997).

[7] Chun-xuan,Jiang,“Foundations of Santilli’s isonumber theory II,”Algebras Groups and Geometries, 15, 509-544 (1998).

[8] Chun-xuan, Jiang,“Foundations of Santilli’s isonumber theory.” In:Foundamental open problems in sciences at the end of the millen-nium, T. Gill, K. Liu and E. Trell (Eds) Hadronic Press, USA, 105-139 (1999).

[9] Richard P. Brent,“On the zeros of the Riemann zeta function in thecritical strip,” Math. Comp., 33, 1361-1372(1979).

[10] J. van de Lune, H. J. J. te Riele and D. T. Winter, “On the zeros ofthe Riemann zeta function in the critical strip,” Math. Comp. 46,667-681 (1986).

[11] A. M. Odlyzko, “1020-th zero of the Riemann zeta function and itsneighbors,” Preprint (1989).

12

[12] Chun-xuan, Jiang, “On the Yu-Goldbach prime theorem,” GuangxiSciences (in Chinese) 3. 9-12 (1996).

[13] Chun-xuan, Jiang,“Foundations of Santilli’s isonumber theory I,”

[14] C.X.Jiang. Foundation of Santilli’s isonumber theory with applica-tions to new cryptograms, Fermat’s theorem and Goldbach’s con-jecture. International Academic Press, 2002 (also available in dpfformat at http://www.i-b-r.org/docs/magneh.pdf

13

1

Fermat Last Theorem And

Riemann Hypothesis(3)

Automorphic Functions And

Fermat’s Last Theorem（3） (Fermat’s Proof of FLT)

Chun-Xuan Jiang P. O. Box 3924, Beijing 100854, P. R. China

[email protected]

Abstract In 1637 Fermat wrote: “It is impossible to separate a cube into two cubes, or a biquadrate into

two biquadrates, or in general any power higher than the second into powers of like degree: I have discovered a truly marvelous proof, which this margin is too small to contain.”


theorem (FLT). It suffices to prove FLT for exponent 4 and every prime exponent P . Fermat proved FLT for exponent 4. Euler proved FLT for exponent 3.

In this paper using automorphic functions we prove FLT for exponents 4P and P , where P is an odd prime. We rediscover the Fermat proof. The proof of FLT must be direct. But indirect proof of FLT is disbelieving.

In 1974 Jiang found out Euler formula of the cyclotomic real numbers in the cyclotomic fields

2

4 1 4

1

1 1

expm m

i ii i

i i

t J S J−

−

= =

⎛ ⎞ =⎜ ⎟⎝ ⎠∑ ∑ , （1）

where J denotes a 4m th root of unity, 4 1mJ = , m=1,2,3,…, it are the real numbers.

iS is called the automorphic functions(complex hyperbolic functions) of order 4m with 4 1m − variables [2,5,7].

11

1

1 ( 1) ( 1)2 cos 2 cos4 2 2

jm

BA Hi j

j

i i jS e e em m

π πβ θ−

=

⎡ ⎤− −⎛ ⎞ ⎛ ⎞= + + + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎣ ⎦

∑

2

( 1) 1

1

( 1) ( 1)2 cos4 2

ji m

DAj

j

i je em m

πφ− −

=

⎡ ⎤− −⎛ ⎞+ + −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

∑ （2）

where 1,..., 4i m= ;

4 1 4 1

1 21 1

, ( 1)m m

A t A t αα αα α

− −

= =

= = −∑ ∑ , 2 1 2

2 2 11 1

( 1) , ( 1)m m

H t tα αα αα α

β−

−= =

= − = −∑ ∑ ,

4 1 4 1

1 1cos , sin ,

2 2

m m

j jj jB t tm mα αα α

α π α πθ− −

= =

= = −∑ ∑

4 1 4 1

1 1( 1) cos , ( 1) sin

2 2

m m

j jj jD t tm m

α αα α

α α

α π α πφ− −

= =

= − = −∑ ∑ ,

1

1 21

2 2 ( ) 0m

j jj

A A H B D−

=

+ + + + =∑ . （3）


1 24 4

1

1 1

, ( 1)m m

A A ii i

i i

e S e S += =

= = −∑ ∑

2 2

12 1 2

1 1

cos ( 1) , sin ( 1)m m

H i H ii i

i i

e S e Sβ β+−= =

= − = −∑ ∑ ,

4 1 4 1

1 1 11 1

cos cos , sin sin2 2

j jm m

B Bj i j i

i i

ij ije S S e Sm mπ πθ θ

− −

+ += =

= + = −∑ ∑ ,

4 1 4 1

1 1 11 1

cos ( 1) cos , sin ( 1) sin2 2

j jm m

D Di ij i j i

i i

ij ije S S e Sm mπ πφ φ

− −

+ += =

= + − = −∑ ∑ . （4）

(3) and (4) have the same form.

3

From (3) we have

1

1 21

exp 2 2 ( ) 1m

j jj

A A H B D−

=

⎡ ⎤+ + + + =⎢ ⎥

⎣ ⎦∑ （5）

From (4) we have

1 4 2

12 1 3

1 21

4 4 1 1

exp 2 2 ( )

m

m

j jj

m m

S S SS S S

A A H B D

S S S

−

=

−

⎡ ⎤+ + + + =⎢ ⎥

⎣ ⎦∑

L

L

L L L L

L

1 1 1 1 4 1

2 2 1 2 4 1

4 4 1 4 4 1

( ) ( )( ) ( )

( ) ( )

m

m

m m m m

S S SS S S

S S S

−

−

−

=

L

L

L L L L

L

(6)

where

( ) ii jj

SSt∂

=∂

[7]


1 4 2

12 1 3

1 21

4 4 1 1

exp 2 2 ( ) 1

m

m

j jj

m m

S S SS S S

A A H B D

S S S

−

=

−

⎡ ⎤+ + + + = =⎢ ⎥

⎣ ⎦∑

L

L

L L L L

L

（7）

Assume 1 20, 0, 0iS S S≠ ≠ = , where 3,..., 4 .i m= 0iS = are (4 2)m− indeterminate

equations with (4 1)m− variables. From (4) we have

1 2 2 2 21 2 1 2 1 2, ,A A He S S e S S e S S= + = − = +

2 2 21 2 1 22 cos 2jB je S S S S

mπ

= + + , 2 2 21 2 1 22 cos 2jD je S S S S

mπ

= + − （8）

Example [2]. Let 4 12m = . From (3) we have

1 1 11 2 10 3 9 4 8 5 7 6( ) ( ) ( ) ( ) ( )A t t t t t t t t t t t= + + + + + + + + + + ,

2 1 11 2 10 3 9 4 8 5 7 6( ) ( ) ( ) ( ) ( )A t t t t t t t t t t t= − + + + − + + + − + + ,

2 10 4 8 6( ) ( )H t t t t t= − + + + − ,

1 1 11 2 10 3 9 4 8 5 7 62 3 4 5( )cos ( ) cos ( ) cos ( ) cos ( ) cos ,

6 6 6 6 6B t t t t t t t t t t tπ π π π π= + + + + + + + + + −

2 1 11 2 10 3 9 4 8 5 7 62 4 6 8 10( )cos ( ) cos ( ) cos ( ) cos ( ) cos ,6 6 6 6 6

B t t t t t t t t t t tπ π π π π= + + + + + + + + + +

4

1 1 11 2 10 3 9 4 8 5 7 62 3 4 5( ) cos ( ) cos ( ) cos ( )cos ( ) cos ,

6 6 6 6 6D t t t t t t t t t t tπ π π π π= − + + + − + + + − + −

2 1 11 2 10 3 9 4 8 5 7 62 4 6 8 10( )cos ( )cos ( )cos ( )cos ( )cos ,6 6 6 6 6

D t t t t t t t t t t tπ π π π π= − + + + − + + + − + +

1 2 1 2 1 22( ) 0A A H B B D D+ + + + + + = , 2 2 3 6 92 3( )A B t t t+ = − + − . （9）

From (8) and (9) we have

12 12 3 4 3 41 2 1 2 1 2 1 2 1 2exp[ 2( )] ( ) ( ) 1A A H B B D D S S S S+ + + + + + = − = − = . (10)

From (9) we have

32 2 3 6 9exp( 2 ) [exp( )]A B t t t+ = − + − . (11)

From (8) we have

2 2 3 32 2 1 2 1 2 1 2 1 2exp( 2 ) ( )( )A B S S S S S S S S+ = − + + = − . (12)


3 3 32 2 1 2 3 6 9exp( 2 ) [exp( )]A B S S t t t+ = − = − + − . (13)

Fermat proved that (10) has no rational solutions for exponent 4 [8]. Therefore we prove we prove that (13) has no rational solutions for exponent 3. [2] Theorem . Let 4 4m P= , where P is an odd prime, ( 1) / 2P − is an even number. From (3) and (8) we have

1

4 4 4 41 2 1 2 1 2

1exp[ 2 2 ( )] ( ) ( ) 1

PP P P P

j jj

A A H B D S S S S−

=

+ + + + = − = − =∑ . (14)

From (3) we have

14

2 4 2 4 2 31

exp[ 2 ( )] [exp( )]

P

Pj j P P P

j

A B D t t t

−

−=

+ + = − + −∑ . (15)

From (8) we have

14

2 4 2 4 1 21

exp[ 2 ( )]

P

P Pj j

j

A B D S S

−

−=

+ + = −∑ . (16)


14

2 4 2 4 1 2 2 31

exp[ 2 ( )] [exp( )]

P

P P Pj j P P P

j

A B D S S t t t

−

−=

+ + = − = − + −∑ . (17)

Fermat proved that (14) has no rational solutions for exponent 4 [8]. Therefor we prove that (17) has no rational solutions for prime exponent P . Remark. Mathematicians said Fermat could not possibly had a proof, because they do not understand FLT.In complex hyperbolic functions let exponent n be n P= Π ， 2n P= Π and

4n P= Π . Every factor of exponent n has Fermat’s equation [1-7]. Using modular elliptic curves Wiles and Taylor prove FLT [9,10]. This is not the proof that Fermat thought to have had. The

5

classical theory of automorphic functions,created by Klein and Poincare, was concerned with the study of analytic functions in the unit circle that are invariant under a discrete group of transformation. Automorphic functions are the generalization of trigonometric, hyperbolic elliptic, and certain other functions of elementary analysis. The complex trigonometric functions and complex hyperbolic functions have a wide application in mathematics and physics. Acknowledgments. We thank Chenny and Moshe Klein for their help and suggestion. References [1] Jiang, C-X, Fermat last theorem had been proved, Potential Science (in Chinese), 2.17-20 (1992),

Preprints (in English) December (1991). http://www.wbabin.net/math/xuan47.pdf. [2] Jiang, C-X, Fermat last theorem had been proved by Fermat more than 300 years ago, Potential

Science (in Chinese), 6.18-20(1992). [3] Jiang, C-X, On the factorization theorem of circulant determinant, Algebras, Groups and





[7] Jiang, C-X, Foundations of Santilli Isonumber Theory with applications to new cryptograms, Fermat’s theorem and Goldbach’s Conjecture. Inter. Acad. Press. 2002. MR2004c:11001, http://www.wbabin.net/math/xuan13.pdf. http://www.i-b-r.org/docs/jiang.pdf

[8] Ribenboim,P, Fermat last theorem for amateur, Springer-Verlag, (1999). [9] Wiles A, Modular elliptic curves and Fenmat’s last theorem, Ann of Math, (2) 141 (1995),

443-551. [10] Taylor, R. and Wiles, A., Ring-theoretic properties of certain Hecke algebras, Ann. of Math., (2)

141(1995), 553-572.

1

Fermat Last Theorem

And Riemann Hypothesis(4)


Fermat’s Last Theorem(4) Chun-Xuan Jiang

P. O. Box 3924, Beijing 100854, P. R. China [email protected]

Absract

1637 Fermat wrote: “It is impossible to separate a cube into two cubes, or a biquadrate into two biquadrates, or in general any power higher than the second into powers of like degree: I have discovered a truly marvelous proof, which this margin is too small to contain.”


theorem (FLT). It suffices to prove FLT for exponent 4. and every prime exponent P . Fermat proved FLT for exponent 4. Euler proved FLT for exponent 3.

In this paper using automorphic functions we prove FLT for exponents 3P and P , where P is an odd prime. We find the Fermat proof. The proof of FLT must be direct. But indirect proof of FLT is disbelieving..

2

In 1974 Jiang found out Euler formula of the cyclotomic real numbers in the cyclotomic fields

1

1

1 1

expn n

i ii i

i i

t J S J−

−

= =

⎛ ⎞ =⎜ ⎟⎝ ⎠∑ ∑ （1）

where J denotes a n th root of negative unity, 1nJ = − , n is an odd number, it are the real numbers.

iS is called the automorphic functions(complex trigonometric functions) of order n with 1n − variables [1-7].

11 2

( 1)

1

( 1) ( 1)[ 2 ( 1) cos( ( 1) )]jn

iBA i j j

i jj

i jS e en n

πθ

−−

−

=

− −= + − + −∑ （2）

where i=1,2,3,…,n;

1

1( 1)

n

A t ααα

−

=

= −∑ , 1

( 1)

1( 1) cos

nj

jjB tn

αα

α

α π− −=

= −∑ , （3）

11 ( 1)

1( 1) ( 1) sin

nj j

jjtn

αα

α

α πθ−

+ −

=

= − −∑ , 1

2

1

2 0

n

jj

A B

−

=

+ =∑

(2) may be written in the matrix form

1

2

3

2

1 1 0 0( 1)1 cos sin sin

22 2 ( 1)1 1 cos sin sin

( 1) ( 1) ( 1)1 cos sin sin2

n

nSn n nS

nSn n nn

S n n nn n n

π π π

π π π

π π π

⎡ ⎤⎢ ⎥−⎡ ⎤ ⎢ ⎥−

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ −= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ − − −⎢ ⎥−⎢ ⎥⎣ ⎦

L

L

L

LL L L L L

L

1

1

1

1

1 12 2

2 cos2 sin

2exp sin

A

B

B

n n

eee

B

θθ

θ− −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

L

(4)

where ( 1) / 2n − is an even number.

From (4) we have its inverse transformation

1

1

1

1

1 1 22 2

1 1 1 12 ( 1)1 cos cos cos

cos2 ( 1)sin 0 sin sin sin

exp( )sin( )( 1) ( 1) ( 1)0 sin sin sin

2 2

A

B

B

n n

e nn n ne

nen n n

Bn n n

n n n

π π πθ

π π πθ

θπ π π

− −

−⎡ ⎤⎢ ⎥⎡ ⎤ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ = ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ − − −⎣ ⎦ ⎢ ⎥− −⎢ ⎥⎣ ⎦

L

L

LL

L L L L L

L

1

2

3

n

SSS

S

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

L

(5)

3

From (5) we have

11

( 1)n

A ii

ie S +

=

= −∑ , 1

( 1)1 1

1cos ( 1) cosj

nB j i

j ii

ije S Snπθ

−−

+=

= + −∑

1

1 ( 1)1

1sin ( 1) ( 1) sinj

nB j j i

j ii

ije Snπθ

−+ −

+=

= − −∑ , （6）

In (3) and (6) it and iS have the same formulas. (4) and (5) are the most critical formulas of

proofs for FLT. Using (4) and (5) in 1991 Jiang invented that every factor of exponent n has the Fermat equation and proved FLT [1-7] .Substituting (4) into (5) we prove (5).

1

1

1

1

1 1 22 2

1 1 1 12 ( 1)1 cos cos cos

cos2 ( 1)1sin 0 sin sin sin

exp( )sin( )( 1) ( 1) ( 1)0 sin sin sin

2 2

A

B

B

n n

e nn n ne

nen n nn

Bn n n

n n n

π π πθ

π π πθ

θπ π π

− −

−⎡ ⎤⎢ ⎥⎡ ⎤ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎢ ⎥ = ×⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ − − −⎣ ⎦ ⎢ ⎥− −⎢ ⎥⎣ ⎦

L

L

LL

L L L L L

L

1

1

1

1

1 122 2

1 1 0 0( 1)1 cos sin sin

2 2 cos2 2 ( 1) 2 sin1 cos sin sin

2exp( )sin( )( 1) ( 1) ( 1)1 cos sin sin

2

A

B

B

n n

enn n n e

n en n n

Bn n n

n n n

π π πθ

π π π θ

θπ π π

− −

⎡ ⎤⎢ ⎥ ⎡ ⎤−⎢ ⎥− ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥− − − ⎣ ⎦⎢ ⎥−⎢ ⎥⎣ ⎦

L

L

LL

L L L L L

L

1

1

1

1

1 12 2

0 0 0

0 0 02 2 cos

1 2 sin0 0 02

2exp( )sin( )0 0 0

2

A

B

B

n n

nen

en e

n

Bn

θθ

θ− −

⎡ ⎤⎢ ⎥ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎣ ⎦

L

L

LL

K L L L L

L

4

1

1

1

1

1 12 2

cossin

exp( )sin( )

A

B

B

n n

eee

B

θθ

θ− −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

L, （7）

where 1

2

11 (cos )

2

n

j

j nnπ−

=

+ =∑ , 1

2

1(sin )

2

n

j

j nnπ−

=

=∑ .

From (3) we have

12

1

exp( 2 ) 1

n

jj

A B

−

=

+ =∑ . （8）

From (6) we have

1 2 1 1 1 1 112

2 1 3 2 2 1 2 1

1

1 1 1 1

( ) ( )( ) ( )

exp( 2 )

( ) ( )

n nn

nj

j

n n n n n n

S S S S S SS S S S S S

A B

S S S S S S

−−

−

=

− −

− −−

+ = =∑

L L

L L

L L L L L L L L

L L

, （9）

where ( ) ii jj

SSt∂

=∂

[7].

From (8) and (9) we have the circulant determinant

1 212

2 1 3

1

1 1

exp( 2 ) 1

nn

jj

n n

S S SS S S

A B

S S S

−

=

−

− −−

+ = =∑

L

L

L L L M

L

（10）

If 0iS ≠ , where 1,2, ,i n= L , then (10) has infinitely many rational solutions.

Assume 1 0S ≠ , 2 0S ≠ , 0iS = where 3, 4, , . 0ii n S= =L are 2n − indeterminate

equations with 1n − variables. From (6) we have

1 2Ae S S= − , 2 2 2 11 2 1 22 ( 1) cosj

B j je S S S Snπ−= + + − . （11）

From (3) and (11) we have the Fermat equation 1 1

2 2 2 2 11 2 1 2 1 2 1 211

exp( 2 ) ( ) ( 2 ( 1) cos ) 1

n n

j n nj jj

jA B S S S S S S S Snπ

− −

−

==

+ = − Π + + − = − =∑ （12）

Example[1]. Let 15n = . From (3) we have

1 14 2 13 3 12 4 11 5 10 6 9 7 8( ) ( ) ( ) ( ) ( ) ( ) ( )A t t t t t t t t t t t t t t= − − + − − − + − − − + − − −

5

1 1 14 2 13 3 12 4 112 3 4( )cos ( ) cos ( ) cos ( )cos

15 15 15 15B t t t t t t t tπ π π π= − + − + − + −

5 10 6 9 7 85 6 7( )cos ( ) cos ( ) cos15 15 15

t t t t t tπ π π+ − + − + − ,

2 1 14 2 13 3 12 4 112 4 6 8( )cos ( )cos ( ) cos ( )cos15 15 15 15

B t t t t t t t tπ π π π= − − + − − − + −

5 10 6 9 7 810 12 14( )cos ( ) cos ( ) cos15 15 15

t t t t t tπ π π− − + − − − ,

3 1 14 2 13 3 12 4 113 6 9 12( )cos ( )cos ( )cos ( ) cos15 15 15 15

B t t t t t t t tπ π π π= − + − + − + −

5 10 6 9 7 815 18 21( )cos ( ) cos ( ) cos15 15 15

t t t t t tπ π π+ − + − + − ,

4 1 14 2 13 3 12 4 114 8 12 16( )cos ( )cos ( ) cos ( )cos15 15 15 15

B t t t t t t t tπ π π π= − − + − − − + −

5 10 6 9 7 820 24 28( )cos ( ) cos ( ) cos15 15 15

t t t t t tπ π π− − + − − − ,

5 1 14 2 13 3 12 4 115 10 15 20( )cos ( ) cos ( )cos ( )cos15 15 15 15

B t t t t t t t tπ π π π= − + − + − + −

5 10 6 9 7 825 30 35( )cos ( )cos ( )cos15 15 15

t t t t t tπ π π− + − + − ,

6 1 14 2 13 3 12 4 116 12 18 24( )cos ( )cos ( ) cos ( )cos15 15 15 15

B t t t t t t t tπ π π π= − − + − − − + −

5 10 6 9 7 830 36 42( )cos ( ) cos ( )cos15 15 15

t t t t t tπ π π− − + − − − ,

7 1 14 2 13 3 12 4 117 14 21 28( )cos ( ) cos ( )cos ( )cos15 15 15 15

B t t t t t t t tπ π π π= − + − + − + −

5 10 6 9 7 835 42 49( )cos ( ) cos ( ) cos15 15 15

t t t t t tπ π π+ − + − + − ,

7

3 6 5 101

2 0, 2 2 5( )jj

A B A B B t t=

+ = + + = − +∑ . (13)

Form (12) we have the Fermat equation

7

15 15 5 3 5 31 2 1 2

1exp( 2 ) ( ) ( ) 1j

jA B S S S S

=

+ = − = − =∑ . (14)

From (13) we have

53 6 5 10exp( 2 2 ) [exp( )]A B B t t+ + = − + . （15）

From (11) we have

5 53 6 1 2exp( 2 2 )A B B S S+ + = − . (16)

From (15) and (16) we have the Fermat equation

5 5 53 6 1 2 5 10exp( 2 2 ) [exp( )]A B B S S t t+ + = − = − + . （17）

6

Euler proved that (14) has no rational solutions for exponent 3[8]. Therefore we prove that (17) has no rational solutions for exponent 5[1]. Theorem 1. Let 3n P= ,where 3P > is odd prime. From (12) we have the Fermat’s equation

3 1

3 3 3 31 2 1 2

1exp( 2 ) ( ) ( ) 1

PP P P P

jj

A B S S S S−

=

+ = − = − =∑ . (18)

From (3) we have

12

3 21

exp( 2 ) [exp( )]

P

Pj P P

jA B t t

−

=

+ = − +∑ . (19)

From (11) we have

12

3 1 21

exp( 2 )

P

P Pj

jA B S S

−

=

+ = −∑ . (20)

From (19) and (20) we have the Fermat equation

12

3 1 2 21

exp( 2 ) [exp( )]

P

P P Pj P P

jA B S S t t

−

=

+ = − = − +∑ . （21）

Euler proved that (18) has no rational solutions for exponent 3[8]. Therefore we prove that (21) has no rational solutions for 3P > [1, 3-7]. Theorem 2. We consider the Fermat’s equation

3 3 3P P Px y z− = （22）

we rewrite (22)

3 3 3( ) ( ) ( )P P Px y z− = (23)

From (24) we have

2 2 3( )( )P P P P P P Px y x x y y z− + + = （24）

Let 1xSz

= , 2ySz

= . From (20) and (24) we have the Fermat’s equation

2 2 2 2( [exp( )]P P P P P P

P Px x y y z t t+ + = − (25)

2[ exp( )]P P P

P Px y z t t− = × − + (26)

Euler proved that (23) has no integer solutions for exponent 3[8]. Therefore we prove that (26) has no integer solutions for prime exponent P . Fermat Theorem. It suffices to prove FLT for exponent 4. We rewrite (22)

3 3 3( ) ( ) ( )P P Px y z− = (27)

Euler proved that（23）has no integer solutions for exponent 3 [8]. Therefore we prove that (27) has no integer solutions for all prime exponent P [1-7]. We consider Fermat equation

4 4 4P P Px y z− = (28)

7

We rewrite (28)

4 4 4( ) (( ) ( )P P Px y z− = （29）

4 4 4( ) ( ) ( )P P Px y z− = （30）

Fermat proved that (29) has no integer solutions for exponent 4 [8]. Therefore we prove that (30) has no integer solutions for all prime exponent P [2,5,7].This is the proof that Fermat thought to have had. Remark. It suffices to prove FLT for exponent 4. Let 4n P= , where P is an odd prime. We have the Fermat’s equation for exponent 4P and the Fermat’s equation for exponent P [2,5,7]. This is the proof that Fermat thought to have had. In complex hyperbolic functions let exponent n be n P= Π , 2n P= Π and 4n P= Π . Every factor of exponent n has the Fermat’s equation [1-7]. In complex trigonometric functions let exponent n be n P= Π , 2n P= Π and 4n P= Π . Every factor of exponent n has Fermat’s equation [1-7]. Using modular elliptic Curves Wiles and Taylor prove FLT[9,10]. This is not the proof that Fermat thought to have had. The classical theory of automorphic functions, created by Klein and Poincare, was concerned with the study of analytic functions in the unit circle that are invariant under a discrete group of transformation. Automorphic functions are the generalization of trigonometric, hyperbolic elliptic and certain other functions of elementary analysis. The complex trigonometric functions and complex hyperbolic functions have a wide application in mathematics and physics. Acknowledgments. We thank Chenny and Moshe Klein for their help and suggestion. References [1] Jiang, C-X, Fermat last theorem had been proved, Potential Science (in Chinese), 2.17-20

(1992), Preprints (in English) December (1991). http://www.wbabin.net/math/xuan47.pdf. [2] Jiang, C-X, Fermat last theorem had been proved by Fermat more than 300 years ago,

Potential Science (in Chinese), 6.18-20(1992). [3] Jiang, C-X, On the factorization theorem of circulant determinant, Algebras, Groups and





[7] Jiang, C-X, Foundations of Santilli Isonumber Theory with applications to new cryptograms, Fermat’s theorem and Goldbach’s Conjecture. Inter, Acad. Press. 2002. MR2004c:11001, http://www.wbabin.net/math/xuan13.pdf. http://www.i-b-r.org/docs/jiang.pdf

[8] Ribenboim,P, Fermat last theorem for amateur, Springer-Verlag, (1999). [9] Wiles,A,Modular elliptic curves and Fermat last theorem ,Ann. of Math.,(2)141(1995),

443-551. [10] Taylor,R,and Wiles,A, Ring-theoretic properties of certain Hecke algebras, Ann. of Math., (2)141(1995),553-572.

1




Fermat’s Last Theorem（5）


[email protected]

Abstract In 1637 Fermat wrote: “It is impossible to separate a cube into two cubes, or a biquadrate into two biquadrates, or in general any power higher than the second into powers of like degree: I have discovered a truly marvelous proof, which this margin is too small to contain.”

This means: ( 2)n n nx y z n+ = > has no integer solutions, all different from 0(i.e., it has only the trivial solution, where one of the integers is equal to 0). It has been called

Fermat’s last theorem (FLT). It suffices to prove FLT for exponent 4 and every prime

exponent P . Fermat proved FLT for exponent 4. Euler proved FLT for exponent 3[8]. In this paper using automorphic functions we prove FLT for exponents 6P and 2P , where P is an odd prime. The proof of FLT must be direct. But indirect proof of FLT is disbelieving. In 1974 Jiang found out Euler formula of the cyclotomic real numbers in the cyclotomic fields

2 1 2

1

1 1

expn n

i ii i

i i

t J S J−

−

= =

⎛ ⎞ =⎜ ⎟⎝ ⎠∑ ∑ , （1）

2

where J denotes a 2n th root of negative unity, 2 1nJ = − , n is an odd number, it are the real numbers.

iS is called the automorphic functions (complex trigonometric functions) of order 2n with (2 1)n − variables [5,7].

31 2

0

( 1) ( 1) ( 1)(2 1)cos cos2 2

j

ni

BHi j

j

i i jS e en n

π πβ θ

−−

=

⎡ ⎤− − − +⎛ ⎞ ⎛ ⎞⎢ ⎥= + + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠

⎢ ⎥⎣ ⎦∑

32

0

1 ( 1)(2 1)cos2

j

n

Dj

j

i jen n

πφ

−

=

− +⎛ ⎞+ −⎜ ⎟⎝ ⎠

∑ ，（2）

where 1,..., 2i n= ;

ααα

αα

α

β +−=

−

=

−=−= ∑∑ 1121

2

1

1)1(,)1( ttH

nn

n

jtn

jtBn

j

n

j 2)12(sin)1(,

2)12(cos)1( 1

12

1

12

1

απθαπ ααα

αα

α

+−=

+−= +

−

=

−

=∑∑ ，

n

jtn

jtDn

j

n

j 2)12(sin,

2)12(cos

12

1

12

1

απφαπ αα

αα

+=

+= ∑∑

−

=

−

=

，

0)(222

3

0=++ ∑

−

=jj

n

jDBH . （3）


iin

i

Hii

n

i

H SeSe +=

+−

=

−=−= ∑∑ 121

112

1)1(sin,)1(cos ββ

n

ijSSe iin

ij

B j

2)12(cos)1(cos 1

12

11

πθ +−+= +−

=∑ ，

n

ijSe iin

ij

B j

2)12(sin)1(sin 11

12

1

πθ +−= ++−

=∑ ，

n

ijSSe in

ij

D j

2)12(coscos 1

12

11

πφ ++= +−

=∑ ，

n

ijSe in

ij

D j

2)12(sinsin 1

12

1

πφ += +−

=∑ . （4）

(3) and (4) have the same form.

Let 1=n . We have 0=H and 1t=β . From (2) we have

3

1211 sin,cos tStS == （5）

From (5) we have

1sincos 12

12 =+ tt （6）

(6) is Pythagorean theorem. It has infinitely many rational solutions. From (3) we have

1)](22exp[2

3

0

=++ ∑−

=jj

n

j

DBH . （7）

From (4) we have

1 2 232

2 1 3

0

2 2 1 1

exp 2 2 ( )

nn

j jj

n n

S S SS S S

H B D

S S S

−

=

−

− −⎡ ⎤ −⎢ ⎥+ + =⎢ ⎥⎢ ⎥⎣ ⎦

∑

L

L

L L L L

L

1 1 1 1 2 1

2 2 1 2 2 1

2 2 1 2 2 1

( ) ( )( ) ( )

( ) ( )

n

n

n n n n

S S SS S S

S S S

−

−

−

=

L

L

L L L L

L

(8)

where

( ) ii jj

SSt∂

=∂

[7]


1 2 232

2 1 3

0

2 2 1 1

exp 2 2 ( ) 1

nn

j jj

n n

S S SS S S

H B D

S S S

−

=

−

− −⎡ ⎤ −⎢ ⎥+ + = =⎢ ⎥⎢ ⎥⎣ ⎦

∑

L

L

L L L L

L

（9）

If 0≠iS , where ni 2,...,2,1= , then (9) has infinitely many rational solutions.

Assume 1 20, 0, 0iS S S≠ ≠ = , where 3,..., 2i n= . 0iS = are (2 2)n − indeterminate

equations with (2 1)n − variables. From (4) we have

22 2 2 2 21 2 1 2 1 2(2 1), 2 cos

2jBH je S S e S S S S

nπ+

= + = + − ,

2 2 21 2 1 2

(2 1)2 cos2

jD je S S S Snπ+

= + + . （10）

Example. Let 15=n . From (9) and (10) we have Fermat’s equation

1)()()](22exp[ 3102310

1302

301

6

0=+=+=++ ∑

=

SSSSDBH jjj

. (11)

From (3) we have

10201013131

0)][exp()](22exp[ ttDBH jj

j+−=++ ++

=∑ . (12)

4

From (10) we have

1021011313

1

0)](22exp[ SSDBH jj

j+=++ ++

=∑ . (13)


102010102

1011313

1

0)][exp()](22exp[ ttSSDBH jj

j+−=+=++ ++

=∑ （14）

Euler prove that (11) has no rational solutions for exponent 3[8]. Therefore we prove that (14) has no rational solutions for exponent 10. Theorem [5,7]. Let Pn 3= , where P is an odd prime. From (9) and (10) we have Fermat’s equation.

1)()()](22exp[ 32232

162

61

233

0

=+=+=++ ∑−

=

PPPPjj

P

j

SSSSDBH . (15)

From (3) we have

PPPjj

P

j

ttDBH 24213132

3

0

)][exp()](22exp[ +−=++ ++

−

=∑ (16)

From (10) we have

PPjj

P

j

SSDBH 222

11313

23

0

)](22exp[ +=++ ++

−

=∑ . （17）


PPPPP

jj

P

j

ttSSDBH 24222

211313

23

0

)][exp()](22exp[ +−=+=++ ++

−

=∑ (18)

Euler prove that (15) has no rational solutions for exponent 3 [8]. Therefore we prove that (18) has no rational solutions for exponent P2 [5,7]. Remark. It suffices to prove FLT for exponent 4. Let 4n P= , where P is an odd prime. We have the Fermat’s equation for exponent 4P and the Fermat’s equation for exponent P [2,5,7]. This is the proof that Fermat thought to have had. In complex hyperbolic functions let exponent n be n P= Π ， 2n P= Π and 4n P= Π . Every factor of exponent n has Fermat’s equation [1-7]. In complex trigonometric functions let exponent n be n P= Π ， 2n P= Π and 4n P= Π . Every factor of exponent n has Fermat’s equation [1-7]. Using modular elliptic curves Wiles and Taylor prove FLT [9,10]. This is not the proof that Fermat thought to have had. The classical theory of automorphic functions, created by Klein and Poincarè, was concerned with the study of analytic functions in the unit circle that are invariant under a discrete group of transformation. Automorphic functions are the generalization of trigonometric, hyperbolic, elliptic, and certain other functions of elementary analysis. The automorphic functions (complex trigonometric functions and complex hyperbolic functions) have a wide application in mathematics and physics. Acknowledgments We thank Chenny and Moshe Klein for their help and suggestion.

5

References [1] Jiang, C-X, Fermat last theorem had been proved, Potential Science (in Chinese), 2.17-20 (1992),







[7] Jiang, C-X, Foundations of Santilli Isonumber Theory with applications to new cryptograms, Fermat’s theorem and Goldbach’s Conjecture. Inter. Acad. Press. 2002. MR2004c:11001, http://www.wbabin.net/math/xuan13.pdf. http://www.i-b-r.org/docs/jiang.pdf

[8] Ribenboim, P, Femat’s last theorem for amateur. Springer, New York, 1999. [9] Wiles A, Modular elliptic curves and Fenmat’s last theorem, Ann of Math, (2) 141 (1995),


141(1995), 553-572.

1




Fermat’s Last Theorem（6）


[email protected]

Abstract

In 1637 Fermat wrote: “It is impossible to separate a cube into two cubes, or a biquadrate into two biquadrates, or in general any power higher than the second into powers of like degree: I have discovered a truly marvelous proof, which this margin is too small to contain.”

This means: ( 2)n n nx y z n+ = > has no integer solutions, all different from 0(i.e., it has only the trivial solution, where one of the integers is equal to 0). It has been called

Fermat’s last theorem (FLT). It suffices to prove FLT for exponent 4 and every prime

exponent P . Fermat proved FLT for exponent 4. Euler proved FLT for exponent 3[8]. In this paper using automorphic functions we prove FLT for exponents 12P and 4P , where P is an odd prime. The proof of FLT must be direct. But indirect proof of FLT is disbelieving. In 1974 Jiang found out Euler formula of the cyclotomic real numbers in the cyclotomic fields

4 1 4

1

1 1

expm m

i ii i

i i

t J S J−

−

= =

⎛ ⎞ =⎜ ⎟⎝ ⎠∑ ∑ , （1）

2

where J denotes a 4m th root of negative unity, 4 1mJ = − , m=1, 2, 3,…, it are the real numbers.

iS is called the automorphic functions (complex trigonometric functions) of order 4m with (4 1)m− variables [5,7].

1

1

0

1 ( 1)(2 1)( 1) cos2 4

jm

Bii j

j

i jS em m

πθ−

−

=

⎡ − +⎛ ⎞= − +⎢ ⎜ ⎟⎝ ⎠⎣

∑

1

0

( 1)(2 1)cos4

jm

Dj

j

i jem

πφ−

=

− + ⎤⎛ ⎞+ −⎜ ⎟⎥⎝ ⎠⎦∑ ，（2）

where 1,..., 4i m= ;

4 1 4 1

1

1 1

(2 1) (2 1)( 1) cos , ( 1) sin4 4

m m

j jj jB t t

m mα α

α αα α

απ απθ− −

+

= =

+ += − = −∑ ∑ ，

4 1 4 1

1 1

(2 1) (2 1)cos , sin4 4

m m

j jj jD t t

m mα αα ααπ απφ

− −

= =

+ += =∑ ∑ ，

1

0

2 ( ) 0m

j jj

B D−

=

+ =∑ . （3）


4 1

1 11

(2 1)cos ( 1) cos4

jm

B ij i

i

j ie S Sm

πθ−

+=

+= + −∑ ，

4 1

11

1

(2 1)sin ( 1) sin4

jm

B ij i

i

j ie Sm

πθ−

++

=

+= −∑ ，

4 1

1 11

(2 1)cos cos4

jm

Dj i

i

j ie S Sm

πφ−

+=

+= + ∑ ，

4 1

11

(2 1)sin sin4

jm

Dj i

i

j ie Sm

πφ−

+=

+= ∑ . （4）

(3) and (4) have the same form. From (3) we have

1

0exp 2 ( ) 1

m

j jj

B D−

=

⎡ ⎤+ =⎢ ⎥

⎣ ⎦∑ . （5）

From (4) we have

3

1 4 2

12 1 3

0

4 4 1 1

exp 2 ( )

m

m

j jj

m m

S S SS S S

B D

S S S

−

=

−

− −−⎡ ⎤

+ =⎢ ⎥⎣ ⎦∑

L

L

L L L L

L

1 1 1 1 4 1

2 2 1 2 4 1

4 4 1 4 4 1

( ) ( )( ) ( )

( ) ( )

m

m

m m m m

S S SS S S

S S S

−

−

−

=

L

L

L L L L

L

(6)

where

( ) ii jj

SSt∂

=∂

[7]


1 4 2

12 1 3

0

4 4 1 1

exp 2 ( ) 1

m

m

j jj

m m

S S SS S S

B D

S S S

−

=

−

− −−⎡ ⎤

+ = =⎢ ⎥⎣ ⎦∑

L

L

L L L L

L

（7）

If 0≠iS , where 1,2,..., 4i m= , then (7) has infinitely many rational solutions.

Assume 1 20, 0,S S≠ ≠ and 0iS = , where 3,..., 4i m= . 0iS = are (4 2)m− indeterminate

equations with (4 1)m− variables. From (4) we have

2 22 2 2 21 2 1 2 1 2 1 2(2 1) (2 1)2 cos , 2 cos

4 4j jB Dj je S S S S e S S S S

m mπ π+ +

= + − = + + , （8）

Example. Let 15m = . From (3) and (8) we have Fermat’s equations

14

60 60 20 3 20 31 2 1 2

0exp[2 ( )] ( ) ( ) 1j j

jB D S S S S

=

+ = + = + =∑ . (9)

From (3) we have

4

203 1 3 1 20 40

0exp[2 ( )] [exp( )]j j

jB D t t+ +

=

+ = − +∑ . (10)

From (8) we have

4

20 203 1 3 1 1 2

0exp[2 ( )]j j

jB D S S+ +

=

+ = +∑ （11）


4

20 20 203 1 3 1 1 2 20 40

0exp[2 ( )] [exp( )]j j

jB D S S t t+ +

=

+ = + = − +∑ （12）

Euler prove that (9) has no rational solutions for exponent 3[8]. Therefore we prove that (12) has no rational solutions for exponent 20. Theorem. Let 3m P= , where P is an odd prime. From (3) and (8) we have Fermat’s equation.

4

3 112 12 4 3 4 31 2 1 2

0exp[2 ( )] ( ) ( ) 1

PP P P P

j jj

B D S S S S−

=

+ = + = + =∑ (13)

From (3) we have

3 1

43 1 3 1 4 8

0exp[2 ( )] [exp( )]

PP

j j P Pj

B D t t−

+ +=

+ = − +∑ . (14)

From (8) we have

1

4 43 1 3 1 1 2

0exp[2 ( )]

PP P

j jj

B D S S−

+ +=

+ = +∑ . (15)


1

4 4 43 1 3 1 1 2 4 8

0exp[2 ( )] [exp( )]

PP P P

j j P Pj

B D S S t t−

+ +=

+ = + = − +∑ (16)

Euler prove that (13) has no rational solutions for exponent 3 [8]. Therefore we prove that (16) has no rational solutions for exponent 4P [5,7]. Remark. It suffices to prove FLT for exponent 4. Let 4n P= , where P is an odd prime. We have the Fermat’s equation for exponent 4P and the Fermat’s equation for exponent P [5,7]. This is the proof that Fermat thought to have had. In complex hyperbolic functions let exponent n be n P= Π ， 2n P= Π and 4n P= Π . Every factor of exponent n has Fermat’s equation [1-7]. In complex trigonometric functions let exponent n be n P= Π ， 2n P= Π and 4n P= Π . Every factor of exponent n has Fermat’s equation [1-7]. Using modular elliptic curves Wiles and Taylor prove FLT [9,10]. This is not the proof that Fermat thought to have had. The classical theory of automorphic functions, created by Klein and Poincarè, was concerned with the study of analytic functions in the unit circle that are invariant under a discrete group of transformation. Automorphic funciton are the generalization of trigonometric, hyperbolic, elliptic, and certain other functions of elementary analysis. The automorphic functions (complex trigonometric functions and complex hyperbolic functions) have a wide application in mathematics and physics. Acknowledgments We thank Chenny and Moshe Klein for their help and suggestion. References [1] Jiang, C-X, Fermat last theorem had been proved, Potential Science (in Chinese), 2.17-20 (1992),





5



[7] Jiang, C-X, Foundations of Santilli’s Isonumber Theory with applications to new cryptograms, Fermat’s theorem and Goldbach’s Conjecture. Inter. Acad. Press. 2002. MR2004c:11001, http://www.wbabin.net/math/xuan13.pdf. http://www.i-b-r.org/docs/jiang.pdf

[8] Ribenboim, P, Femat’s last theorem for amateur. Springer, New York, 1999. [9] Wiles A, Modular elliptic curves and Fenmat’s last theorem, Ann of Math, (2) 141 (1995),


141(1995), 553-572.

auto_2_.pdfJiangRiemann.pdfauto_3_.pdfauto_4_.pdfAuto_5_.pdfauto_6_.pdf

Fermat Last Theorem And Riemann Hypothesis(2)vixra.org › pdf › 1101.0079v1.pdfa conjecture...

Documents

Transcript of Fermat Last Theorem And Riemann Hypothesis(2)vixra.org › pdf › 1101.0079v1.pdfa conjecture...