Fermat Last Theorem And Riemann Hypothesis(3) Automorphic Functions And Fermat s Last Theorem(3) (Fermat s Proof of FLT)

Transcription

1 Fermat Last Theorem Ad Rema Hypothess(3) Automorphc Fuctos Ad Fermat s Last Theorem(3) (Fermat s Proof of FLT) Chu-Xua Jag P. O. Box 394, Beg 854, P. R. Cha agchuxua@sohu.com Abstract I 637 Fermat wrote: It s mpossble to separate a cube to two cubes, or a bquadrate to two bquadrates, or geeral ay power hgher tha the secod to powers of lke degree: I have dscovered a truly marvelous proof, whch ths marg s too small to cota. Ths meas: x + y = z ( > ) has o teger solutos, all dfferet from (.e., t has oly the trval soluto, where oe of the tegers s equal to ). It has bee called Fermat s last theorem (FLT). It suffces to prove FLT for expoet 4 ad every prme expoet P. Fermat proved FLT for expoet 4. Euler proved FLT for expoet 3. I ths paper usg automorphc fuctos we prove FLT for expoets 4P ad P, where P s a odd prme. We redscover the Fermat proof. The proof of FLT must be drect. But drect proof of FLT s dsbelevg. I 974 Jag foud out Euler formula of the cyclotomc real umbers the cyclotomc felds

2 exp = 4m 4m tj SJ, () = = 4m where J deotes a 4m th root of uty, J =, m=,,3,, t are the real umbers. S s called the automorphc fuctos(complex hyperbolc fuctos) of order 4m wth 4m varables [,5,7]. where =,...,4m; ( ) π ( ) π = + cos + + cos θ + 4 m m A B H S e e β e m = ( ) m ( ) A D ( ) π + e + e cos φ 4m = m () 4m 4m α α α, α= α= A = t, A = t ( ) m m α α α β α, α= α= H = t ( ), = t ( ) 4m 4m α π α π B = t cos, θ = tα s, m α α= m α= α π α α π D = t ( ) cos, = tα( ) s, m 4m 4m α α φ α= m α= m A+ A + H + ( B + D) =. (3) = From () we have ts verse trasformato[5,7] 4m 4m A A + =, = ( ) = = e S e S m m H + H cos β = ( ), s β = ( ) = =, e S e S B π π e cosθ S S cos, e s S s m 4m 4m B = + + θ = + = m =, D π π e cos φ = S + S ( ) cos, e s = S ( ) s m (3) ad (4) have the same form. 4m 4m D + φ + = m =. (4)

3 From (3) we have m exp A+ A + H + ( B + D) = (5) = From (4) we have S S L S 4m m S S L S3 exp A+ A + H + ( B + D) = = L L L L S S L S 4m 4m where S ( S) = [7] t = S ( S ) L ( S ) 4m S ( S ) L ( S ) 4m L L L L S ( S ) L ( S ) 4m 4m 4m 4m (6) From (5) ad (6) we have crculat determat S S L S 4m S S L S m 3 exp A+ A + H + ( B + D) = = = L L L L S S L S 4m 4m (7) Assume S, S, S =, where = 3,...,4 m. S = are (4m ) determate equatos wth (4m ) varables. From (4) we have B π e = S + S + SScos, m Example []. Let 4m =. From (3) we have e = S + S, e = S S, e = S + S A A H A = ( t + t ) + ( t + t ) + ( t + t ) + ( t + t ) + ( t + t ) + t, D π e = S + S SScos (8) m A = ( t + t ) + ( t + t ) ( t + t ) + ( t + t ) ( t + t ) + t, H = ( t + t ) + ( t + t ) t, π π 3π 4π 5π B = ( t+ t)cos + ( t + t)cos + ( t3+ t9)cos + ( t4 + t8)cos + ( t5 + t7)cos t6, π 4π 6π 8π π B = ( t+ t) cos + ( t + t) cos + ( t3+ t9) cos + ( t4 + t8) cos + ( t5 + t7) cos + t6,

4 π π 3π 4π 5π D = ( t+ t) cos + ( t + t) cos ( t3+ t9) cos + ( t4 + t8)cos ( t5 + t7) cos t6, π 4π 6π 8π π D = ( t+ t)cos + ( t + t)cos ( t3+ t9)cos + ( t4 + t8)cos ( t5 + t7)cos + t6, A+ A + ( H + B+ B + D+ D) =, A + B = 3( t3+ t6 t9). (9) From (8) ad (9) we have exp[ A + A + ( H + B + B + D + D )] = S S = ( S ) ( S ) =. () From (9) we have From (8) we have exp( A B ) [exp( t t t )] 3 + = () exp( A + B ) = ( S S )( S + S + S S ) = S S. () 3 3 From () ad () we have Fermat s equato exp( A + B ) = S S = [exp( t + t t )]. (3) Fermat proved that () has o ratoal solutos for expoet 4 [8]. Therefore we prove we prove that (3) has o ratoal solutos for expoet 3. [] Theorem. Let 4m= 4P, where P s a odd prme, ( P )/ s a eve umber. From (3) ad (8) we have From (3) we have From (8) we have P 4P 4P P 4 P 4 = exp[ A + A + H + ( B + D )] = S S = ( S ) ( S ) =. (4) P 4 P exp[ A + ( B + D )] = [exp( t + t t )]. (5) 4 4 P P 3P = From (5) ad (6) we have Fermat s equato P 4 P P = = exp[ A ( B D )] S S. (6) P 4 P P P = = P + P 3P = exp[ A ( B D )] S S [exp( t t t )]. (7) Fermat proved that (4) has o ratoal solutos for expoet 4 [8]. Therefor we prove that (7) has o ratoal solutos for prme expoet P. Remark. Mathematcas sad Fermat could ot possbly had a proof, because they do ot uderstad FLT.I complex hyperbolc fuctos let expoet be = Π P, = Π P ad = 4Π P. Every factor of expoet has Fermat s equato [-7]. Usg modular ellptc curves Wles ad Taylor prove FLT [9,]. Ths s ot the proof that Fermat thought to have had. The 4

5 classcal theory of automorphc fuctos,created by Kle ad Pocare, was cocered wth the study of aalytc fuctos the ut crcle that are varat uder a dscrete group of trasformato. Automorphc fuctos are the geeralzato of trgoometrc, hyperbolc ellptc, ad certa other fuctos of elemetary aalyss. The complex trgoometrc fuctos ad complex hyperbolc fuctos have a wde applcato mathematcs ad physcs. Ackowledgmets. We thak Chey ad Moshe Kle for ther help ad suggesto. Refereces [] Jag, C-X, Fermat last theorem had bee proved, Potetal Scece ( Chese),.7- (99), Preprts ( Eglsh) December (99). [] Jag, C-X, Fermat last theorem had bee proved by Fermat more tha 3 years ago, Potetal Scece ( Chese), 6.8-(99). [3] Jag, C-X, O the factorzato theorem of crculat determat, Algebras, Groups ad Geometres, (994), MR. 96a: 3, [4] Jag, C-X, Fermat last theorem was proved 99, Preprts (993). I: Fudametal ope problems scece at the ed of the mlleum, T.Gll, K. Lu ad E. Trell (eds). Hadroc Press, 999, P [5] Jag, C-X, O the Fermat-Satll theorem, Algebras, Groups ad Geometres, (998) [6] Jag, C-X, Complex hyperbolc fuctos ad Fermat s last theorem, Hadroc Joural Supplemet, (). [7] Jag, C-X, Foudatos of Satll Isoumber Theory wth applcatos to ew cryptograms, Fermat s theorem ad Goldbach s Coecture. Iter. Acad. Press.. MR4c:, [8] Rbebom,P, Fermat last theorem for amateur, Sprger-Verlag, (999). [9] Wles A, Modular ellptc curves ad Femat s last theorem, A of Math, () 4 (995), [] Taylor, R. ad Wles, A., Rg-theoretc propertes of certa Hecke algebras, A. of Math., () 4(995),

6 I press at Algebras, Groups ad Geometres, Vol., 4 DISPROOFS OF RIEMANN S HYPOTHESIS Chu-Xua, Jag P.O.Box 394, Beg 854, Cha ad Isttute for Basc Research P.O.Box 577, Palm Harbor, FL 3468, U.S.A. lukx@publc3.bta.et.c Abstract As t s well kow, the Rema hypothess o the zeros of the ζ(s) fucto has bee assumed to be true varous basc developmets of the -th cetury mathematcs, although t has ever bee proved to be correct. The eed for a resoluto of ths ope hstorcal problem has bee voced by several dstgushed mathematcas. By usg precedg works, ths paper we preset comprehesve dsproofs of the Rema hypothess. Moreover, 994 the author dscovered the arthmetc fucto J (ω) that ca replace Rema s ζ(s) fucto vew of ts proved features: f J (ω), the the fucto has ftely may prme solutos; ad f J (ω) =, the the fucto has ftely may prme solutos. By usg the Jag J (ω) fucto we prove the tw prme theorem, Goldbach s theorem ad the prme theorem of the form x +. Due to the mportace of resolvg the hstorcal ope ature of the Rema hypothess, commets by terested colleagues are here solcted. AMS mathematcs subect classfcato: Prmary M6.

7 . Itroducto I 859 Rema defed the zeta fucto[] ζ(s) = p ( p s ) = = s, () where s = σ + t, =, σ ad t are real, p rages over all prmes. ζ(s) satsfes the fuctoal equato [] From () we have π s s ( s) Γ( )ζ(s) = π Γ( s )ζ( s). () ζ(t). (3) Rema coectured that ζ(s) has ftely may zeros σ, called the crtcal strp. Rema further made the remarkable coecture that the zeros of ζ(s) the crtcal strp all le o the cetral le σ = /, a coecture called the famous Rema hypothess (RH). It was stated by Hardy 94 that ftely may zeros le o the le; A. Selberg stated 94 that a postve proporto at least of all the zeros le o the le; Levso stated 974 that more tha oe thrd of the zeros le o the le; Corey stated 989 that at least two ffths of the zeros le o the le. The use of the RH the lead to may mathematcal problems, such as the geeralzed Rema coecture, Art s coecture, Wel s coecture, Laglads program, quatum chaos, the hypothetcal Rema flow [3, 4], the zeta fuctos ad L-fuctos of a algebrac varety ad other studes. Smlarly, t s possble to prove may theorems by usg the RH. However, the RH remas a bascally uproved coecture to ths day. I fact, Hlbert properly stated 99 that the problem of provg or dsprovg the RH s oe of the most mportat problems cofrotg th cetury mathematcas. I Grffths ad Graham poted out that the RH s the frst challegg problem for the st cetury. The proof of the RH the become the mlleum prze problem. I 997 we studed the tables of the Rema zeta fucto [5] ad reached prelmary results dcatg that the RH s false [6, 7, 8]. I

8 ths paper we preset a comprehesve dsproof of the RH ad show that the computato of all zeros of the ζ(/ + t) fucto doe durg the past years s error, as prelmarly dcated Refs. [9,, ]. Sce the RH s false, all theorems ad coectures based to the same are also false.. Dsproofs of Rema s Hypothess Theorem. ζ(s) has o zeros the crtcal strp, that s ζ(s), where σ. Proof. From () we have ζ(s) = p ( p s ) = Reθ, (4) where R = p cos(t log p) R p, R p = +, (5) p σ pσ θ = θ p, θ p = ta s(t log p) p p σ cos(t log p). (6) If σ =, from (5) we have R p = cos(t log p). If cos(t log p) =, we have R p = the R =. If σ > from (5) we have R p. ζ(s) = f ad oly f Re ζ(s) = ad Im ζ(s) =, that s R =. From (5) we have that f cos(t log p) the R p > ad f cos(t log p) > the R p <. cos(t log p) s depedet of the real part σ, but may well deped o prmes p ad magary part t. We wrte m + (t) for the umber of prmes p satsfyg cos(t log p) >, m (t) for the umber of prmes p satsfyg cos(t log p). For cos(t log p) >, we have > R p ( + t) > R p (.5 + t). (7) If m + (t ) s much greater tha m (t ) such that R(.5 + t ) = m. From (5), (6) ad (7) we have for gve t mr(σ + t ) > mr( + t ) > mr(.5 + t ) > mr(σ + t ), 3

9 (8) θ(σ + t ) = θ( + t ) = θ(.5 + t ) = θ(σ + t ) = cost (9) where σ > ad σ <.5. Sce ζ(s) = from (8) we have R max ζ(σ + t ) < max ζ( + t ) < max ζ(.5 + t ) < max ζ(σ + t ). () For cos(t log p) < we have < R p (.5 + t) < R p (.4 + t) < R p (.3 + t). () If m (t ) s much greater tha m + (t ) such that R(.5 + t ) = max. From (5), (6) ad () we have for gve t maxr(σ + t ) < maxr(.5 + t ) < maxr(.4 + t ), < maxr(.3 + t ) < maxr(σ + t ), () θ(σ + t ) = θ(.5 + t ) = θ(.4 + t ) where σ >.5 ad σ <.3. Sce ζ(s) = from () we have R = θ(.3 + t ) = θ(σ + t ) = cost, m ζ(σ + t ) > m ζ(.5 + t ) > m ζ(.4 + t ) > (3) m ζ(.3 + t ) > m ζ(σ + t ). (4) Proof. We defe the beta fucto β(s) = p ( + p s ) = = λ() s, (5) where λ() =, λ() = ( ) a + +a k f = p a p a k k, t. 4

10 From (5) we have β(s) = p ( + p s ) = Re θ, (6) where R = p θ = p For cos(t log p) <, we have R p, R cos(t log p) p = + +, (7) p σ pσ θ p, θ p = ta s(t log p) p σ + cos(t log p). (8) > R p ( + t) > R p (.5 + t). (9) If m (t ) s much greater tha m + (t ) such that R(.5 + t ) = m. From (7), (8) ad (9) we have for gve t m R(σ + t ) > m R( + t ) > m R(.5 + t ) > m R(σ + t ), () θ(σ + t ) = θ( + t ) = θ(.5 + t ) = θ(σ + t ) = cost, () where σ > ad σ <.5. Sce β(s) = R from () we have max β(σ + t ) < max β( + t ) < max β(.5 + t ) < max β(σ + t ). () For cos(t log p) > we have < R p (.5 + t) < R p (.4 + t) < R p (.3 + t). (3) If m + (t ) s much greater tha m (t ) such that that R(.5+t ) = max. From (7), (8) ad (3) we have for gve t max R(σ + t ) < max R(.5 + t ) < max R(.4 + t ) < max R(.3 + t ) < max R(σ + t ), (4) 5

11 θ(σ + t ) = θ(.5 + t ) = θ(.4 + t ) = θ(.3 + t ) = θ(σ + t ) = cost, (5) where σ >.5 ad σ <.3. Sce β(s) = R from (4) we have From () ad (5) we have m β(σ + t ) > m β(.5 + t ) > m β(.4 + t ) > m β(.3 + t ) > m β(σ + t ). (6) ζ(s) = ζ(s)β(s). (7) I 896 J. Hadamard ad de la Vallee Pouss proved depedetly ζ( + t). From (7) we have From (8) we have ζ( + t) = ζ( + t) β( + t). (8) ζ( + t) ad β( + t). (9) ζ(s) ad β(s) are the dual fuctos. From () we have β( + t). (3) Therefore we have ζ( + t). (3) I the same way we have ζ( + t) = ζ( 4 + t) β( + t). (3) 4 6

12 From (3) we have I the same way we have As we have ζ( 4 + t) ad β( + t). (33) 4 Proof 3. For σ > we have ζ( + t). (34) ζ(t). (35) log ζ(s) = p m p mσ exp( tm log p). (36) m= If ζ(s) had a zero at + t, the log ζ(σ + t) would ted to as σ teds to from the rght. From (36) we have log ζ(s) = p m p mσ cos(tm log p), (37) m= wth t replaced by, t, t,, Ht, t gves H = ( H ) log ζ(σ + (H )t) + ( H H ) where A = H = log ζ(σ) = p ( H ) m p mσ A, (38) m= cos ((H )tm log p) + = H [ + cos(tm log p)] H, ( H H ) (39) H s a eve umber. 7

13 From (38) we have (ζ(σ)) ( H H ) H = ( H ζ(σ + (H )t) ). (4) Sce ζ( + et) [5], where e =,,, H, from (4) we have ζ( + et) for suffcetly large eve umber H. M ζ( + t) but. The computato of all zeros of ζ( + t) s error, whch satsfes the the error RH. From (39) we have cos θ + 4 cos θ + 3 = ( + cos θ), cos 4θ + 8 cos 3θ + 8 cos θ + 56 cos θ + 35 = 8( + cos θ) 4, cos 5θ + cos 5θ + 66 cos 4θ + cos 3θ +495 cos θ + 79 cos θ + 46 = 3( + cos θ) The Arthmetc Fucto J (ω) Replacg Rema s Hypothess I vew of the precedg results, the RH has o value for the study of prme dstrbutos. I 994 the author dscover the arthmetc fucto J (ω) [, 3] that s able to take the place of Rema s zeta-fuctos ad L-fuctos because of the followg propertes: J (ω), the the fucto has ftely may prme solutos; ad f J (ω) =, the the fucto has ftely may prme solutos. By usg Jag s J (ω) fucto we have proved umerous theorems cludg the tw prme theorem, Goldbach s theorem, the prme k- tuples theorem, Satll s theory for a prme table, the theorem of fte Fermat prmes, the theorem of fte Mersee prmes, the theorem of fte reput prmes, there are ftely may trples of, + ad + that each s the product of k dstct prmes, there are ftely may Carmchael umbers whch are product of exactly fve prmes, there there are ftely may Carmchael umbers whch are product of exactly sx prmes the prme dstrbutos [4]. We gve some theorems below Theorem. Tw prme theorem: p = p +. 8

14 We have the arthmetc fucto J (ω) = (p ). 3 p p Sce J (ω), there are ftely may prmes p such that p s a prme. Theorem 3. Goldbach theorem: N = p + p. We have the arthmetc fucto J (ω) = (p ) 3 p p p N p p. Sce J (ω), every eve umber N greater tha 4 s the sum of two odd prmes. Theorem 4. p = (p + ) +. We have the arthmetc fucto J (ω) = (p ( ) p ). 3 p p Sce J (ω), there are ftely may prmes p such that p s a prme. Theorem 5. p = p. We have J (ω) = (p ( )). 3 p p p Sce J (ω), there are ftely may prmes p such that p s a prme. Theorem 6. p = p + 4 ad p = 4p +. We have J (ω) = 3 (p 3). 7 p p Sce J (ω), there are ftely may prmes p such that p ad p are prmes. 9

15 Theorem 7. p = (p + ) + ad p = (p + ) + 3. We have J (ω) = (p 3 ( ) p ( 3 )). 5 p p p Sce J (ω), there are ftely may prmes p such that p ad p are prmes. Theorem 8. The prme 3-tuples theorem: p+b: b =, 4, 6,, 6, 8, 4, 8, 34, 4, 46, 48, 9. Sce J (3) =, there are o prme 3-tuples f p 3. Theorem 9. The prme 4-tuples theorem: p+b: b =,, 6, 8,, 8,, 6, 3, 3, 36, 4, 48, 5. We have J (ω) = 3 (p 4). 9 p p Sce J (ω), there are ftely may prme 4-tuples. Theorem. p = 6m +, p = m +, p 3 = 8m +, p 4 = 36m +, p 5 = 7m +. We have J (ω) = (p 6). 3 p p Sce J (ω), there are ftely may tegers m such that p, p, p 3, p 4 ad p 5 are prmes. = p p p 3 p 4 p 5 s the Carmchael umbers. Theorem. p 3 = p + p + p p. We have J 3 (ω) = (p 3p + 3). 3 p p Sce J 3 (ω), there are ftely may prmes p ad p such that p 3 s a prme. Theorem. p 3 = (p + ) 5 + p. We have J 3 (ω) = (p 3p + 3). 3 p p

16 Sce J 3 (ω), there are ftely may prmes p ad p such that p 3 s a prme. Theorem 3. p 4 = p (p + p 3 ) + p p 3. We have J 4 (ω) = ( (p ) 4 ) +. p 3 p p Sce J 4 (ω), there are ftely may prmes p, p ad p 3 such that p 4 s a prme. Theorem 4. Each of, + ad + s the product of k dstct prmes. Suppose that each of m, m = m + ad m 3 = m + s the product of k dstct prmes. We defe p = m m 3 x +, p = m m 3 x +, p 3 = m m x +. (4) We have the arthmetc fucto J (ω) = (p 4 χ(p)), (4) 3 p p where χ(p) = f p m m m 3 ; χ(p) = otherwse. Sce J (ω), there exst ftely may tegers x such that p, p ad p 3 are prmes. From (4) we have = m p = m m m 3 x+m, + = m p + = m m m 3 x + m + = m (m m 3 x + ) = m p, + = m p + = m m m 3 x + m + = m 3 (m m x + ) = m 3 p 3. If p, p ad p 3 are prmes, the each of, + ad + s the product of k dstct prmes. For example, = 7793 = 3 536, + = 7794 = 7 58, + = 7795 = J (ω) s a geeralzato of Euler s proof for the exstece of ftely may prmes. It has a wde applcato varous felds. Ackowledgemets The Author would lke to express hs deepest apprecato to Professors R. M. Satll, G. Wess, D. Zuckerma, Ke-x Lu, Mao-xa Zuo, Zhogda Wag, Chag-pe Wag, ad X-pg Tog for ther helps ad supports.

17 Refereces [] B. Rema, Uber de Azahl der Prmzahle uder eer gegebeer Grösse, Moatsber. Akad. Berl (859). [] H. Daveport, Multplcatve Number Theory, Sprger Verlag (98). [3] N. Katz ad P. Sarak, Zeroes of zeta fuctos ad symmetry, Bull. AMS, 36, -6 (999). [4] A. Coes, Trace formula ocommutatve geometry ad the zeros of the Rema zeta fucto, Sel. Math., New Ser. 5, 9-6(999). [5] C. B. Haslgrove, Tables of the Rema zeta fucto, Roy. Soc. Math. Tables, Vol.6, Cambrdge Uv. Press, Lodo ad New York (96). [6] Chu-xua, Jag, The study of the Rema Zeta fucto, Upublshed Oct. (997). [7] Chu-xua,Jag, Foudatos of Satll s soumber theory II, Algebras Groups ad Geometres, 5, (998). [8] Chu-xua, Jag, Foudatos of Satll s soumber theory. I: Foudametal ope problems sceces at the ed of the mlleum, T. Gll, K. Lu ad E. Trell (Eds) Hadroc Press, USA, 5-39 (999). [9] Rchard P. Bret, O the zeros of the Rema zeta fucto the crtcal strp, Math. Comp., 33, 36-37(979). [] J. va de Lue, H. J. J. te Rele ad D. T. Wter, O the zeros of the Rema zeta fucto the crtcal strp, Math. Comp. 46, (986). [] A. M. Odlyzko, -th zero of the Rema zeta fucto ad ts eghbors, Preprt (989).

18 [] Chu-xua, Jag, O the Yu-Goldbach prme theorem, Guagx Sceces ( Chese) (996). [3] Chu-xua, Jag, Foudatos of Satll s soumber theory I, [4] C.X.Jag. Foudato of Satll s soumber theory wth applcatos to ew cryptograms, Fermat s theorem ad Goldbach s coecture. Iteratoal Academc Press, (also avalable dpf format at 3

19 Fermat Last Theorem Ad Rema Hypothess(4) Automorphc Fuctos Ad Fermat s Last Theorem(4) Chu-Xua Jag P. O. Box 394, Beg 854, P. R. Cha agchuxua@sohu.com Absract 637 Fermat wrote: It s mpossble to separate a cube to two cubes, or a bquadrate to two bquadrates, or geeral ay power hgher tha the secod to powers of lke degree: I have dscovered a truly marvelous proof, whch ths marg s too small to cota. Ths meas: x + y = z ( > ) has o teger solutos, all dfferet from (.e., t has oly the trval soluto, where oe of the tegers s equal to ). It has bee called Fermat s last theorem (FLT). It suffces to prove FLT for expoet 4. ad every prme expoet P. Fermat proved FLT for expoet 4. Euler proved FLT for expoet 3. I ths paper usg automorphc fuctos we prove FLT for expoets 3P ad P, where P s a odd prme. We fd the Fermat proof. The proof of FLT must be drect. But drect proof of FLT s dsbelevg..

20 I 974 Jag foud out Euler formula of the cyclotomc real umbers the cyclotomc felds exp tj = SJ () = = where J deotes a th root of egatve uty, J =, s a odd umber, t are the real umbers. S s called the automorphc fuctos(complex trgoometrc fuctos) of order wth varables [-7]. A ( ) B θ = ( ) ( ) π S = [ e + ( ) e cos( + ( ) )] () where =,,3,,; α A= tα ( ), α = B α π ( ) α tα ( ) cos, (3) α = = α π θ = ( ) ( ) s, + ( ) α tα α = () may be wrtte the matrx form A+ B = = L S π π ( ) π cos s L s S π π ( ) π S 3 = cos s L s L L L L L L S ( ) π ( ) π ( ) π cos s L s A e B e cosθ B e sθ L expb sθ (4) where ( )/ s a eve umber. From (4) we have ts verse trasformato L A e π π ( ) π cos cos L cos e cosθ B B e sθ π π ( ) π = s s L s L exp( B )s( θ ) L L L L L ( ) ( ) ( ) s s L s π π π S S S 3 L S (5)

21 From (5) we have e A + S ( ), = = π e cos S S ( ) cos B θ = + + ( ) = e π s ( ) ( ) s, (6) B + θ = S + ( ) = I (3) ad (6) t ad S have the same formulas. (4) ad (5) are the most crtcal formulas of proofs for FLT. Usg (4) ad (5) 99 Jag veted that every factor of expoet has the Fermat equato ad proved FLT [-7].Substtutg (4) to (5) we prove (5). L A e π π ( ) π cos cos L cos e cosθ B B e sθ π π ( ) π = s s L s L exp( B )s( θ ) L L L L L s s L s ( ) π ( ) π ( ) π L A π π ( ) π e cos s L s B e cosθ π π ( ) π B cos s s e sθ L L L L L L L exp( B )s( θ ) ( ) π ( ) π ( ) π cos s L s L A e L B e cosθ B e sθ = L L K L L L L exp( B )s( θ ) L 3

22 e e cosθ B B e sθ = A, (7) L exp( B )s( θ ) where π + (cos ) =, = π (s ) =. = From (3) we have From (6) we have exp( A+ B ) =. (8) = = S S L S S ( S ) L ( S ) S S L S S ( S ) L ( S ) exp( A+ B ) = = L L L L L L L L 3 S S L S S ( S ) L ( S ), (9) S where ( S) = [7]. t From (8) ad (9) we have the crculat determat S S L S S S L S3 exp( A+ B ) = = = L L L M S S L S () If S, where =,, L,, the () has ftely may ratoal solutos. Assume S, S, S = where = 3, 4, L,. S = are determate equatos wth varables. From (6) we have A B π e = S S, e = S + S + SS( ) cos. () From (3) ad () we have the Fermat equato π = = exp( A+ B ) = ( S S ) Π ( S + S + S S ( ) cos ) = S S = () Example[]. Let = 5. From (3) we have A= ( t t ) + ( t t ) ( t t ) + ( t t ) ( t t ) + ( t t ) ( t t )

23 3 4 B = ( t t4)cos π + ( t t3 )cos π + ( t3 t)cos π + ( t4 t)cos π π 6π 7π + ( t5 t)cos + ( t6 t9)cos + ( t7 t8)cos, π 4π 6π 8π B = ( t t4)cos + ( t t3)cos ( t3 t)cos + ( t4 t)cos π π 4π ( t5 t)cos + ( t6 t9)cos ( t7 t8)cos, B3 = ( t t4)cos π + ( t t3)cos π + ( t3 t)cos π + ( t4 t)cos π π 8π π + ( t5 t)cos + ( t6 t9)cos + ( t7 t8)cos, π 8π π 6π B4 = ( t t4)cos + ( t t3)cos ( t3 t)cos + ( t4 t)cos π 4π 8π ( t5 t) cos + ( t6 t9) cos ( t7 t8) cos, B5 = ( t t4)cos π + ( t t3)cos π + ( t3 t)cos π + ( t4 t)cos π π 3π 35π ( t5 t)cos + ( t6 t9)cos + ( t7 t8)cos, B6 = ( t t4)cos π + ( t t3)cos π ( t3 t)cos π + ( t4 t)cos π π 36π 4π ( t5 t)cos + ( t6 t9)cos ( t7 t8)cos, B7 = ( t t4)cos π + ( t t3)cos π + ( t3 t)cos π + ( t4 t)cos π π 4π 49π + ( t5 t)cos + ( t6 t9)cos + ( t7 t8)cos, (3) = A+ B =, A+ B + B = 5( t + t ) Form () we have the Fermat equato From (3) we have From () we have (4) = exp( A+ B ) = S S = ( S ) ( S ) = exp( A B B ) [exp( t t )] From (5) ad (6) we have the Fermat equato = 5 +. (5) exp( A + B + B ) = S S. (6) exp( A + B + B ) = S S = [exp( t + t )]. (7)

24 Euler proved that (4) has o ratoal solutos for expoet 3[8]. Therefore we prove that (7) has o ratoal solutos for expoet 5[]. Theorem. Let = 3P,where P > 3 s odd prme. From () we have the Fermat s equato From (3) we have From () we have 3P 3P 3P P 3 P 3. (8) = exp( A+ B ) = S S = ( S ) ( S ) = P 3 P P P. (9) = exp( A+ B ) = [exp( t + t )] P P P 3. () = exp( A+ B ) = S S From (9) ad () we have the Fermat equato P P P 3 P P P. () = exp( A+ B ) = S S = [exp( t + t )] Euler proved that (8) has o ratoal solutos for expoet 3[8]. Therefore we prove that () has o ratoal solutos for P > 3 [, 3-7]. Theorem. We cosder the Fermat s equato we rewrte () From (4) we have 3P 3P 3P x y = z () P 3 P 3 P 3 ( x ) ( y ) = ( z ) (3) P P P P P P 3P ( x y )( x + x y + y ) = z (4) x y Let S =, S =. From () ad (4) we have the Fermat s equato z z P P P P P P ( x x y y z [exp( tp tp)] + + = (5) x y = [ z exp( t + t )] (6) P P P P P Euler proved that (3) has o teger solutos for expoet 3[8]. Therefore we prove that (6) has o teger solutos for prme expoet P. Fermat Theorem. It suffces to prove FLT for expoet 4. We rewrte () 3 P 3 P 3 P ( x ) ( y ) ( z ) = (7) Euler proved that(3)has o teger solutos for expoet 3 [8]. Therefore we prove that (7) has o teger solutos for all prme expoet P [-7]. We cosder Fermat equato 4P 4P 4P x y = z (8) 6

25 We rewrte (8) P 4 P 4 P 4 ( x ) (( y ) = ( z ) (9) 4 P 4 P 4 P ( x ) ( y ) ( z ) = (3) Fermat proved that (9) has o teger solutos for expoet 4 [8]. Therefore we prove that (3) has o teger solutos for all prme expoet P [,5,7].Ths s the proof that Fermat thought to have had. Remark. It suffces to prove FLT for expoet 4. Let = 4P, where P s a odd prme. We have the Fermat s equato for expoet 4P ad the Fermat s equato for expoet P [,5,7]. Ths s the proof that Fermat thought to have had. I complex hyperbolc fuctos let expoet be =Π P, = Π P ad = 4Π P. Every factor of expoet has the Fermat s equato [-7]. I complex trgoometrc fuctos let expoet be =Π P, = Π P ad = 4Π P. Every factor of expoet has Fermat s equato [-7]. Usg modular ellptc Curves Wles ad Taylor prove FLT[9,]. Ths s ot the proof that Fermat thought to have had. The classcal theory of automorphc fuctos, created by Kle ad Pocare, was cocered wth the study of aalytc fuctos the ut crcle that are varat uder a dscrete group of trasformato. Automorphc fuctos are the geeralzato of trgoometrc, hyperbolc ellptc ad certa other fuctos of elemetary aalyss. The complex trgoometrc fuctos ad complex hyperbolc fuctos have a wde applcato mathematcs ad physcs. Ackowledgmets. We thak Chey ad Moshe Kle for ther help ad suggesto. Refereces [] Jag, C-X, Fermat last theorem had bee proved, Potetal Scece ( Chese),.7- (99), Preprts ( Eglsh) December (99). [] Jag, C-X, Fermat last theorem had bee proved by Fermat more tha 3 years ago, Potetal Scece ( Chese), 6.8-(99). [3] Jag, C-X, O the factorzato theorem of crculat determat, Algebras, Groups ad Geometres, (994), MR. 96a: 3, [4] Jag, C-X, Fermat last theorem was proved 99, Preprts (993). I: Fudametal ope problems scece at the ed of the mlleum, T.Gll, K. Lu ad E. Trell (eds). Hadroc Press, 999, P [5] Jag, C-X, O the Fermat-Satll theorem, Algebras, Groups ad Geometres, (998) [6] Jag, C-X, Complex hyperbolc fuctos ad Fermat s last theorem, Hadroc Joural Supplemet, (). [7] Jag, C-X, Foudatos of Satll Isoumber Theory wth applcatos to ew cryptograms, Fermat s theorem ad Goldbach s Coecture. Iter, Acad. Press.. MR4c:, [8] Rbebom,P, Fermat last theorem for amateur, Sprger-Verlag, (999). [9] Wles,A,Modular ellptc curves ad Fermat last theorem,a. of Math.,()4(995), [] Taylor,R,ad Wles,A, Rg-theoretc propertes of certa Hecke algebras, A. of Math., ()4(995),

26 Fermat Last Theorem Ad Rema Hypothess(5) Automorphc Fuctos Ad Fermat s Last Theorem(5) Chu-Xua Jag P. O. Box 394, Beg 854, P. R. Cha agchuxua@sohu.com Abstract I 637 Fermat wrote: It s mpossble to separate a cube to two cubes, or a bquadrate to two bquadrates, or geeral ay power hgher tha the secod to powers of lke degree: I have dscovered a truly marvelous proof, whch ths marg s too small to cota. Ths meas: x + y = z ( > ) has o teger solutos, all dfferet from (.e., t has oly the trval soluto, where oe of the tegers s equal to ). It has bee called Fermat s last theorem (FLT). It suffces to prove FLT for expoet 4 ad every prme expoet P. Fermat proved FLT for expoet 4. Euler proved FLT for expoet 3[8]. I ths paper usg automorphc fuctos we prove FLT for expoets 6P ad P, where P s a odd prme. The proof of FLT must be drect. But drect proof of FLT s dsbelevg. I 974 Jag foud out Euler formula of the cyclotomc real umbers the cyclotomc felds exp = tj SJ, () = =

27 where J deotes a th root of egatve uty, J =, s a odd umber, t are the real umbers. S s called the automorphc fuctos (complex trgoometrc fuctos) of order wth ( ) varables [5,7]. ( ) ( ) ( )( + ) S = e + + e + where =,...,; 3 H π B π cos β cos θ = 3 D ( )( ) + π + e cos φ =, () H B α = t α ), β = α = α = ( t α ( ) + α α ( + ) απ + α ( + ) απ = tα ( ) cos, θ = tα ( ) s, α = α = D ( + ) απ = tα cos, φ = α = α = t α ( + ) απ s, 3 H + ( B + D ) =. (3) = From () we have ts verse trasformato[5,7] e H cos β + H = S ( ), e s β = = = S ( ) + e e e e ( + ) π cosθ = S + S + ( ) cos, B = + ( + ) π sθ = S + ( ) s, B = ( + ) π cosφ = S + S + cos, D = ( + ) π sφ = S + s. (4) D = (3) ad (4) have the same form. Let =. We have H = ad β = t. From () we have

28 S = = t (5) cost, S s From (5) we have cos t + s t = (6) (6) s Pythagorea theorem. It has ftely may ratoal solutos. From (3) we have 3 exp[h + ( B + D )] =. (7) = From (4) we have S S 3 L S S ( S ) L ( S ) S S S3 exp H ( B D) L S ( S ) L ( S ) + + = = = L L L L L L L L S S L S S ( S ) L ( S ) where S ( S) = [7] t From (7) ad (8) we have crculat determat S S L S 3 3 exp S S S H ( B D) L + + = = = L L L L S S L S (8) (9) If S, where =,,...,, the (9) has ftely may ratoal solutos. Assume S, S, S =, where = 3,...,. S = are ( ) determate equatos wth ( ) varables. From (4) we have H B ( ), + π e = S + S e = S + S SScos, D ( + ) π e = S + S + SScos. () Example. Let = 5. From (9) ad () we have Fermat s equato exp[h + 6 = ( B + D )] = S 3 + S 3 = ( S ) 3 + ( S ) 3 =. () From (3) we have exp[ H + ( B3 + + D3 + )] = [exp( t + t )]. () = 3

29 From () we have exp[ From () ad (3) we have Fermat s equato exp[ H H + ( B3 + + D3 + )] = S + S. (3) = + ( B3 + + D3 + )] = S + S = [exp( t + t )] (4) = Euler prove that () has o ratoal solutos for expoet 3[8]. Therefore we prove that (4) has o ratoal solutos for expoet. Theorem [5,7]. Let = 3P, where P s a odd prme. From (9) ad () we have Fermat s equato. 3P 3 6P 6P P 3 P 3 exp[h + ( B + D )] = S + S = ( S ) + ( S ) =. (5) From (3) we have From () we have = P 3 P + ( B3 + + D3 + )] = [exp( tp t4p )] (6) = exp[ H + P 3 P P + ( B3 + + D3 + )] = S S. (7) = exp[ H + From (6) ad (7) we have Fermat s equato P 3 P P P + ( B3 + + D3 + )] = S + S = [exp( tp t4p )] (8) = exp[ H + Euler prove that (5) has o ratoal solutos for expoet 3 [8]. Therefore we prove that (8) has o ratoal solutos for expoet P [5,7]. Remark. It suffces to prove FLT for expoet 4. Let = 4P, where P s a odd prme. We have the Fermat s equato for expoet 4P ad the Fermat s equato for expoet P [,5,7]. Ths s the proof that Fermat thought to have had. I complex hyperbolc fuctos let expoet be =Π P, = Π P ad = 4Π P. Every factor of expoet has Fermat s equato [-7]. I complex trgoometrc fuctos let expoet be = Π P, = Π P ad = 4Π P. Every factor of expoet has Fermat s equato [-7]. Usg modular ellptc curves Wles ad Taylor prove FLT [9,]. Ths s ot the proof that Fermat thought to have had. The classcal theory of automorphc fuctos, created by Kle ad Pocarè, was cocered wth the study of aalytc fuctos the ut crcle that are varat uder a dscrete group of trasformato. Automorphc fuctos are the geeralzato of trgoometrc, hyperbolc, ellptc, ad certa other fuctos of elemetary aalyss. The automorphc fuctos (complex trgoometrc fuctos ad complex hyperbolc fuctos) have a wde applcato mathematcs ad physcs. Ackowledgmets We thak Chey ad Moshe Kle for ther help ad suggesto. 4

30 Refereces [] Jag, C-X, Fermat last theorem had bee proved, Potetal Scece ( Chese),.7- (99), Preprts ( Eglsh) December (99). [] Jag, C-X, Fermat last theorem had bee proved by Fermat more tha 3 years ago, Potetal Scece ( Chese), 6.8-(99). [3] Jag, C-X, O the factorzato theorem of crculat determat, Algebras, Groups ad Geometres, (994), MR. 96a: 3, [4] Jag, C-X, Fermat last theorem was proved 99, Preprts (993). I: Fudametal ope problems scece at the ed of the mlleum, T.Gll, K. Lu ad E. Trell (eds). Hadroc Press, 999, P [5] Jag, C-X, O the Fermat-Satll theorem, Algebras, Groups ad Geometres, (998) [6] Jag, C-X, Complex hyperbolc fuctos ad Fermat s last theorem, Hadroc Joural Supplemet, (). [7] Jag, C-X, Foudatos of Satll Isoumber Theory wth applcatos to ew cryptograms, Fermat s theorem ad Goldbach s Coecture. Iter. Acad. Press.. MR4c:, [8] Rbebom, P, Femat s last theorem for amateur. Sprger, New York, 999. [9] Wles A, Modular ellptc curves ad Femat s last theorem, A of Math, () 4 (995), [] Taylor, R. ad Wles, A., Rg-theoretc propertes of certa Hecke algebras, A. of Math., () 4(995),

31 Fermat Last Theorem Ad Rema Hypothess(6) Automorphc Fuctos Ad Fermat s Last Theorem(6) Chu-Xua Jag P. O. Box 394, Beg 854, P. R. Cha agchuxua@sohu.com Abstract I 637 Fermat wrote: It s mpossble to separate a cube to two cubes, or a bquadrate to two bquadrates, or geeral ay power hgher tha the secod to powers of lke degree: I have dscovered a truly marvelous proof, whch ths marg s too small to cota. Ths meas: x + y = z ( > ) has o teger solutos, all dfferet from (.e., t has oly the trval soluto, where oe of the tegers s equal to ). It has bee called Fermat s last theorem (FLT). It suffces to prove FLT for expoet 4 ad every prme expoet P. Fermat proved FLT for expoet 4. Euler proved FLT for expoet 3[8]. I ths paper usg automorphc fuctos we prove FLT for expoets P ad 4P, where P s a odd prme. The proof of FLT must be drect. But drect proof of FLT s dsbelevg. I 974 Jag foud out Euler formula of the cyclotomc real umbers the cyclotomc felds exp = 4m 4m tj SJ, () = =

32 where J deotes a 4m th root of egatve uty, 4 m J =, m=,, 3,, t are the real umbers. S s called the automorphc fuctos (complex trgoometrc fuctos) of order 4m wth (4m ) varables [5,7]. m B ( )( ) + π S = ( ) e cos θ + m = 4m where =,...,4m; m D ( )( ) + π + e cos φ = 4m, () ( + ) απ ( + ) απ B = t = t 4 4m 4m 4m α + α α( ) cos, θ α( ) s α= m α=, 4m 4m ( + ) απ ( + ) απ D = t cos, φ = tα s 4 4m, α α= m α= m ( B + D) =. (3) = From () we have ts verse trasformato[5,7] 4m B ( ) + π e cos θ = S + S ( ) cos, 4m e + = ( + ) π, 4m 4m B + s θ = S + ( ) s = D 4m e cosφ S S e D + = ( + ) π cos 4m = +, 4m ( + ) π sφ = S + s. (4) 4m = (3) ad (4) have the same form. From (3) we have m exp ( B + D) =. (5) = From (4) we have

33 m exp ( B + D) = = S S L S 4m S S L S 3 L L L L S S L S 4m 4m = S ( S ) L ( S ) 4m S ( S ) L ( S ) 4m L L L L S ( S ) L ( S ) 4m 4m 4m 4m (6) where S ( S) = [7] t From (5) ad (6) we have crculat determat S S L S 4m S S L S m 3 exp ( B + D) = = = L L L L S S L S 4m 4m (7) If S, where =,,...,4m, the (7) has ftely may ratoal solutos. Assume S, S, ad S =, where = 3,...,4m. S = are (4m ) determate equatos wth (4m ) varables. From (4) we have B ( + ) π ( ) D + π e = S + S SScos, e = S + S + SScos, (8) 4m 4m Example. Let m = 5. From (3) ad (8) we have Fermat s equatos From (3) we have From (8) we have B + D = S + S = S + S =. (9) = exp[ ( )] ( ) ( ) 4 B3+ + D3+ = t + t4. () = exp[ ( )] [exp( )] 4 B3+ + D3+ = S + S () = exp[ ( )] From () ad () we have Fermat s equato 4 B3+ + D3+ = S + S = t + t4 () = exp[ ( )] [exp( )] Euler prove that (9) has o ratoal solutos for expoet 3[8]. Therefore we prove that () has o ratoal solutos for expoet. Theorem. Let m= 3P, where P s a odd prme. From (3) ad (8) we have Fermat s equato. 3

34 From (3) we have From (8) we have 3P P P 4P 3 4P 3 B + D = S + S = S + S = (3) = exp[ ( )] ( ) ( ) 3P B3+ + D3+ = t4p + t8p 4P. (4) = exp[ ( )] [exp( )] P 4P 4P B3+ + D3+ = S + S. (5) = exp[ ( )] From (4) ad (5) we have Fermat s equato P 4P 4P B3+ + D3+ = S + S = t4p + t8p 4P (6) = exp[ ( )] [exp( )] Euler prove that (3) has o ratoal solutos for expoet 3 [8]. Therefore we prove that (6) has o ratoal solutos for expoet 4P [5,7]. Remark. It suffces to prove FLT for expoet 4. Let = 4P, where P s a odd prme. We have the Fermat s equato for expoet 4P ad the Fermat s equato for expoet P [5,7]. Ths s the proof that Fermat thought to have had. I complex hyperbolc fuctos let expoet be =Π P, = Π P ad = 4Π P. Every factor of expoet has Fermat s equato [-7]. I complex trgoometrc fuctos let expoet be = Π P, = Π P ad = 4Π P. Every factor of expoet has Fermat s equato [-7]. Usg modular ellptc curves Wles ad Taylor prove FLT [9,]. Ths s ot the proof that Fermat thought to have had. The classcal theory of automorphc fuctos, created by Kle ad Pocarè, was cocered wth the study of aalytc fuctos the ut crcle that are varat uder a dscrete group of trasformato. Automorphc fucto are the geeralzato of trgoometrc, hyperbolc, ellptc, ad certa other fuctos of elemetary aalyss. The automorphc fuctos (complex trgoometrc fuctos ad complex hyperbolc fuctos) have a wde applcato mathematcs ad physcs. Ackowledgmets We thak Chey ad Moshe Kle for ther help ad suggesto. Refereces [] Jag, C-X, Fermat last theorem had bee proved, Potetal Scece ( Chese),.7- (99), Preprts ( Eglsh) December (99). [] Jag, C-X, Fermat last theorem had bee proved by Fermat more tha 3 years ago, Potetal Scece ( Chese), 6.8-(99). [3] Jag, C-X, O the factorzato theorem of crculat determat, Algebras, Groups ad Geometres, (994), MR. 96a: 3, [4] Jag, C-X, Fermat last theorem was proved 99, Preprts (993). I: Fudametal ope problems scece at the ed of the mlleum, T.Gll, K. Lu ad E. Trell (eds). Hadroc Press, 999, P

35 [5] Jag, C-X, O the Fermat-Satll theorem, Algebras, Groups ad Geometres, (998) [6] Jag, C-X, Complex hyperbolc fuctos ad Fermat s last theorem, Hadroc Joural Supplemet, (). [7] Jag, C-X, Foudatos of Satll s Isoumber Theory wth applcatos to ew cryptograms, Fermat s theorem ad Goldbach s Coecture. Iter. Acad. Press.. MR4c:, [8] Rbebom, P, Femat s last theorem for amateur. Sprger, New York, 999. [9] Wles A, Modular ellptc curves ad Femat s last theorem, A of Math, () 4 (995), [] Taylor, R. ad Wles, A., Rg-theoretc propertes of certa Hecke algebras, A. of Math., () 4(995),