IN GAUSSIAN INTEGERS X 3 + Y 3 = Z 3 HAS ONLY TRIVIAL SOLUTIONS A NEW APPROACH

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1 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008), #A2 IN GAUSSIAN INTEGERS X + Y = Z HAS ONLY TRIVIAL SOLUTIONS A NEW APPROACH Elis Lmpkis Lmpropoulou (Term), Kiprissi, T.K: 24500, GREECE lbkis@otenet.gr Received: 2/10/08, Revised: 7/7/08, Accepted: 7/1/08, Published: 7/18/08 Abstrct It will be shown vi new method tht in the Ring of Gussin Integers Zi] the solutions of the Diophntine eqution x + y = z re trivil, nmely, x y z = 0. The result is chieved by showing tht the existence of nontrivil Gussin Integer solutions implies the existence of rtionl points on the elliptic curve y 2 = x + 42, which is lredy known to hve none. 1. Introduction In this pper we present new method tht ddresses the question of the existence of nontrivil solutions to the Fermt type Diophntine eqution x + y = z (1) with x, y, z in the ring of Gussin integers Zi]. By nontrivil we men solutions x, y, z in Zi] for which x y z 0. In L.E. Dickson s History of the Theory of Numbers 1, p. 550], one cn find tht this question hs lredy been nswered by R. Feuter 2] within the frme of lgebric number theory. Nmely, he hs proven tht if ξ +η +ζ = 0 is solvble by numbers 0 of n imginry qudrtic domin k( m), where m < 0, m 2 (mod ), then the clss number of k is divisible by. Now k = Z, m = 1 nd Z is Principl Idel Domin hving clss number 1 not divisible by. Our method is new since it trnsfers, vi elementry polynomil theory, the question of the existence of nontrivil solution of (1) in Zi], to the question of the existence of rtionl point on the elliptic curve y 2 = x + 42, (2)

2 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008),#A2 2 which is lredy known 1] to hve none. Eqution (2) is curve 42A in Cremon s tbles with rnk 0 nd order of torsion subgroup 1; nmely, its torsion subgroup contins only the point t infinity. Thus (2) hs no rtionl points of infinite or finite order. So fr vrious extensions of Fermt s Lst Theorem (FLT) hve been treted in vst number of publictions. Mny references my be found in 5]. Some concern the nture of the exponents, others, the nture of the solution s underlying ring. We mention, for exmple, ] for rtionl exponents nd 4] for exponents in Zi]. On the other hnd, in 5, 6] one cn find the solution of the cubic exponent cse of FLT in Zω], ω = 1, ω 1; in 5, 7, 8] the solution of the fourth exponent cse in Zi]; nd in 9] the solution of the n th exponent cse in the ring of polynomils of n lgebriclly closed field with chrcteristic 0. There re lso mixed cses s in 10] where the exponents re not necessrily equl, the eqution my hve coefficients other thn 1 but the solutions re in Z, or in 11] with two exponents equl to 4, one exponent equl to 2, not ll coefficients equl to 1, nd underlying ring of solutions Zi]. The ltter is generliztion of the fourth exponent cse in 5, 7, 8]. 2. The Eqution x + y = z in Zi] In order for us to demonstrte the bove-mentioned connection between the nontrivil solutions of (1) in Zi] nd the rtionl points on (2), we introduce the following nottions nd ssumptions. Let (x 0, y 0, z 0 ) = ( 1 + b 1 i, 2 + b 2 i, + b i) be nontrivil solution of (1) in Zi]. Also let I = {1, 2, }. If m = 0 or b m = 0 for ll m in I, (1) implies either b 1 + b 2 = b or =, respectively. Since m, b m re in Z, the ltter holds only when t lest one of the b m s or the m s, respectively, is 0. Then t lest one of the m + b m i is 0, contrdiction. Thus nd b 1 + b 2 + b 0. Additionlly, let p m (x) = m + b m x, f(x) = p 1(x) + p 2(x) p (x) be polynomils in Zx]. Finlly, set { 1, m = 1, 2 k m = 1, m =. Our first prtil result bout (1), Theorem 1, is bsed upon the kind of roots tht f(x) possesses. For this piece of informtion we need the following lemm tht determines the degree of f(x) nd its behvior t 0. Notice the ctlytic ppernce of (2) in the proof. Lemm 1 If (1) hs nontrivil solution (x 0, y 0, z 0 ) in Zi], then f(x) hs degree nd f(0) 0. Proof. The coefficient of x in f(x) is t = b 1 + b 2 b, wheres the constnt term is t 0 = Assume t = 0. Then (b 1, b 2, b ) = (0, b, b) or (b, 0, b) or (b, b, 0) or

3 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008),#A2 ( b, b, 0) for some b, nd (1) implies, respectively, b 2 ( 2 ) ] b ( )] i = 0 () b 2 ( 1 ) ] b ( )] i = 0 (4) b 2 ( ) ] b ( )] i = 0 (5) b 2 ( ) ] + b ( )] i = 0. (6) The imginry prt in ll four cses is 0. If b = 0, then b 1 = b 2 = b = 0 in (x 0, y 0, z 0 ), which contrdicts b 1 + b 2 + b 0. Thus b 0 nd m = ± j, m j, m, j in I. Now we tret cse () in detil. All other cses follow long the sme resoning. If 2 =, then 1 = 0 nd 1 + b 1 i = 0, contrdiction. If 2 =, then 6 b 2 = ( 1 ) + 2 (2 b) 2 (2 ) = 4 ( 1 ) + (2 ) (2 b) 2 (2 ) = 4 ( 1 ) (2 ) (72 b) 2 (2 ) = ( 12 1 ) + 42 (2 ). (7) If = 0, then 1 = 0 nd 1 + b 1 i = 0, contrdiction. Thus 0 nd dividing both sides of (7) by (2 ) we conclude tht (x, y) = ( 6 1 /, 6 b/ ) is rtionl point on (2), contrdiction. Finlly, t 0 nd f(x) hs degree. The cse when t 0 = 0 cn be discrded using the sme resoning s bove. Thus f(0) 0. The following theorem estblishes the reltion tht must be stisfied by the rel nd imginry prts of the nonzero Gussin integers tht solve (1). Theorem 1 If (1) hs nontrivil solution (x 0, y 0, z 0 ) in Zi] then λ in Q {0} nd exctly one vlue of m in I exist such tht for m l j, l, j in I, m = λ ( l + k m j ), b m = λ (b l + k m b j ) Proof. (1) implies tht f(i) = 0 nd f(i) = f( i) = 0. Thus f(x) = (c + d x) (x 2 + 1), c, d in Z {0} since the degree of f(x) is nd f(0) 0. We hve the following cses,. p m ( c/d) 0 for ll m in I. Then m d b m c 0 for ll m in I. f( c/d) = 0 implies ( 1 d b 1 c) + ( 2 d b 2 c) = ( d b c) contrdiction. b. p m ( c/d) = 0 for ll m in I. Then m d = b m c for ll m in I. Note tht b m = 0 implies tht m = 0 nd vice vers. In tht cse, m + b m i = 0, contrdiction. Thus m 0, b m 0 for ll m in I. Now m = (c/d) b m, c+d i 0 nd (1) implies b 1 +b 2 = b, contrdiction. c. p m ( c/d) = 0 for exctly two vlues of m in I. Then f( c/d) = 0 implies p m ( c/d) = 0 for the third vlue of m in I, contrdicting cse (b). d. p m ( c/d) = 0 for exctly one vlue of m in I. Let l, j be the other two elements of I. m d = b m c nd m 0, b m 0, since c, d in Z {0}. And f( c/d) = 0 implies ( l d b l c) + k m ( j d b j c) = 0 ( l d b l c) = k m ( j d b j c)

4 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008),#A2 4 or ( l + k m j ) d (b l + k m b j ) c = 0. Since d = (b m / m ) c, we tke ( l + k m j ) b m (b l + k m b j ) m = 0 l + k m j b l + k m b j m b m = 0. The vectors ( l + k m j, b l + k m b j ) nd ( m, b m ) re linerly dependent in Q. For the unique vlue of m I, there exist λ Q {0} such tht m = λ ( l + k m j ), b m = λ (b l + k m b j ). Remrk 1 If (1) hs nontrivil solutions in Zi] so do the equtions y + x = z (8) ( z) + y = ( x) (9) nd vice vers. Thus, without loss of generlity, we my ssume in Theorem 1 tht the exct vlue of m I is 1. If m is 2 or we substitute (8) or (9) in (1), denoting by (y 0, x 0, z 0 ) or ( z 0, y 0, x 0 ) the nontrivil solution ( 1 + b 1 i, 2 + b 2 i, + b i) of ech one in Zi], respectively. Our next result estblishes the connection between the existence of nontrivil solutions of type (1) eqution in Zi] nd the existence of rtionl points on (2). Theorem 2 If (1) hs nontrivil solution (x 0, y 0, z 0 ) in Zi] then (2) hs rtionl point. Proof. Theorem 1 nd Remrk 1 imply 1 = λ ( l j ), b 1 = λ (b l b j ), l j in {2, }, λ in Q {0}. Since we cn write 1 = λ ( j l ), b 1 = λ (b j b l ), it is obvious tht, without loss of generlity, we my ssume l = 2, j =. Set w = ( 2 + b 2 i)/( + b i) in Q(i) {0}. We hve w 1, else (1) implies 1 + b 1 i = 0, contrdiction. On the other hnd, dividing both sides of (1) by ( + b i), we tke ] 1 + b 1 i + + b i ] 2 + b 2 i = 1 λ (w 1) + w = 1 + b i (λ + 1) w 2 + ( 2 λ + 1) w + (λ + 1) ] = 0. We hve λ 1, else the ltter would imply w = 0, 2 + b 2 i = 0, contrdiction. The discriminnt of the ltter is 12 λ, nd w = 2 λ (λ + 1) ± λ + 2 λ = 2 (λ + 1) 2 (λ + 1) ± λ + 2 (λ + 1) i.

5 INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 8 (2008),#A2 5 Since w is in Q(i) {0}, there exists µ Q such tht µ = ( µ ) 2 ( µ ) 2 12 λ + 9 = 12 λ + = 4 λ + 1 ( µ ) = 4 λ (12 µ) 2 = (12 λ) + 42, implying tht (2) hs the rtionl solution (x, y) = (12 λ, 12 µ). Now we stte our min result bout the solutions of (1) in Zi]. Theorem If (x, y, z) is solution of (1) in Zi], then x y z = 0. Proof. Let (x, y, z) be solution of (1) in Zi] with x y z 0. Theorem 2 implies tht (2) hs rtionl point. We hve lredy mentioned tht (2) hs no rtionl points 1]. Thus x y z = 0. Acknowledgments We re thnkful to professor J. Silvermn of Brown University for communicting to us the conductor of y 2 = x + 42, something tht simplified the investigtion in Cremon s tbles. We would lso like to thnk the nonymous referee for vluble suggestions concerning the historicl sttus of the problem. References 1. Cremon s Tbles, 2. R. Feuter, Sitzungsber. Akd. Wiss. Heidelberg (Mth.), 4, A, 191 No C. Bennett, A. M. W. Glss nd G. Székely, Fermt s Lst Theorem for Rtionl Exponents, Americn Mthemticl Monthly 111 (2004) J. Zuehlke, Fermt s Lst Theorem for Gussin Integer Exponents, Americn Mthemticl Monthly 106 (1999) P. Ribenboim, Fermt s Lst Theorem For Amteurs, Springer Verlg, New York, K. Irelnd nd M. Rosen, A Clssicl Introduction to Modern Number Theory, Springer Verlg, New York, D. Hilbert, Jhresbericht d. Deutschen Mth. Vereinigung, 4 ( ) J. T. Cross, In the Gussin Integers α 4 + β 4 γ 4, Mth. Mgzine 66 (199) S. Lng, Old nd new conjectured Diophntine Inequlities, Bulletin (New Series) of the AMS 2 (1990) M. Bennett nd C. M. Skinner, Ternry Diophntine Equtions vi Glois Representtions nd Modulr Forms, CndinMJ 56 (2004) S. Szbó, Some Fourth Degree Diophntine Equtions in Gussin Integers, Integers: Electronic Journl of Combintoril Number Theory 4 (2004) A16, 17 pges.