Chapter 2 Even Exponents

Transcription

1 Chater 2 Even Exonents In this chater we consider Catalan s equation x y q D 1 when one of the exonents and q is even. This reduces to the case when one of and q is equal to 2, and the other is an odd rime number. 2.1 The Equation x D y 2 C 1 Six years after Catalan s note extraite, the French mathematician Lebesgue [67] made the first ste in the long way towards the solion of Catalan s roblem. He roved that Catalan s equation x y q D 1 has no solions with q D 2. Theorem 2.1 (V.A. Lebesgue). Let 3 be an odd number. Then the equation x D y 2 C 1 has no solions in nonzero integers x and y. (One may further assume that is rime, b this is not needed for the roof.) Proof. If y is odd and x is even, then y 2 C 1 2 mod 4 and x 0 mod 8, a contradiction. Hence y is even and x is odd. Write x D.1 C iy/.1 iy/: (2.1) The greatest common divisor of 1 C iy and 1 iy (in the ring of Gaussian integers ZŒi ) divides the sum.1 C iy/ C.1 iy/ D 2. Since1 C y 2 is odd, the numbers 1 C iy and 1 iy are corime. Since ZŒi is a unique factorization ring, every factor in (2.1) is equal to a th ower times a unit of ZŒi ; thatis, 1 C iy D " ; 1 iy DN" N ; Sringer International Publishing Switzerland 2014 Y.F. Bilu et al., The Problem of Catalan, DOI /

2 12 2 Even Exonents where 2 ZŒi and " 2 ZŒi Df 1; ig. Since is odd, every unit of ZŒi is a th ower of another unit. Writing " D " 1 and ˇ D " 1, we obtain 1 C iy D ˇ; 1 iy D Nˇ: Write ˇ D a C ib, wherea; b 2 Z. Since is odd, the number 2a D ˇ C Nˇ divides ˇ C Nˇ D 2. Hence a D 1. Further,1 C y 2 D.a 2 C b 2 / D.1 C b 2 / is an odd number, which imlies that b is even. It follows that 1 C iy D.a C ib/ a C ia 1 b mod 4; and, in articular, 1 a mod 4, which rules o the ossibility a D 1. Thus, ˇ D 1 C ib. Comaring the real arts in the equality 1 C iy D.1 C ib/ ; we obtain 1 D. 1/=2 X kd0. 1/ k b 2k ; 2k which can be rewritten as b 2 C 2. 1/=2 X kd2. 1/ k 2k b 2k D 0: (2.2) We shall use Lemma A.1 to show that (2.2) is imossible. For 1 k. 1/=2 we have b 2k D b 2 1 2k 2 k.2k 1/ 2 b 2k 2 : 2k 2 Hence Ord 2 b 2k Ord 2 2k b 2.2k 2/Ord 2 b Ord 2 k 2 2k 2 log 2 k: Since 2k 2>log 2 k for k 2, we have Ord 2 2k b 2k > Ord 2 2 b 2 for k D 2;:::;. 1/=2. Now Lemma A.1 imlies that the left-hand side of (2.2) cannot vanish. The theorem is roved.

3 2.2 Units of Real Quadratic Rings Units of Real Quadratic Rings This section is auxiliary. In it we recall the structure of the unit grou of the ring ZŒ D, whered is a ositive integer, which is not a square (we do not assume that D is square-free). Theorem 2.2. The multilicative grou ZŒ D C ZŒ D is infinite cyclic. of ositive units of the ring It follows that the grou ZŒ D of all units of ZŒ D is 1 times an infinite cyclic grou. The ring ZŒ D is not, in general, the ring of integers of the quadratic field Q. D/. Hence Theorem 2.2 is not a formal consequence of the Dirichlet unit theorem, as stated in Aendix A.2. Of course, it may be deduced from the slightly more general Dirichlet unit theorem for maximal orders;see[12,sect.2.4]. However, we sketch here a short indeendent roof, for the reader s convenience. The roof uses the famous Dirichlet aroximation theorem. Theorem 2.3 (Dirichlet aroximation theorem). Let be a real number and Y>0. Then there exist integers x;y such that 0<y Y and jy xj <Y 1. Informally, this means that the rational number x=y is a good aroximation for. Proof. It is well known and simle: on the quotient grou R=Z consider the images of the real numbers m, wherem runs through the integers satisfying 0 m Y. We obtain by cc1oints on R=Z, which slit R=Z into by cc1disjoint intervals. Since by cc1>y, at least one of these intervals is of length <Y 1. This means that there exist integers m 1, m 2,andx such that 0 m 1 <m 2 Y and jm 2 m 1 xj <Y 1. Pting y D m 2 m 1, we comlete the roof. The following consequence is immediate. Corollary 2.4. Let be a real number. Then there exist infinitely many coules.x; y/ 2 Z 2 with y>0and jy xj <y 1. We return to the roof of Theorem 2.2. We denote the grou ZŒ D C of ositive units by U. We denote by 7 N the nontrivial aomorhism of ZŒ D (that is, x C y D D x y D) and by N W ZŒ D Z the norm ma: N D N. Proof of Theorem 2.2. First of all, we rove that U f1g. By the Dirichlet aroximation theorem, there exist infinitely many coules of ositive integers x and y ˇ such that ˇx y Dˇ y 1. For any such x and y we have x y D C 1. Hence D x C y D satisfies 0< 2y D C 1 and jn j y 1, which imlies jn j Dj N j 2 D C 1. We have roved that ZŒ D contains infinitely many ositive elements of norm bounded by 2 D C 1. Therefore there exists a nonzero a 2 Z such that ZŒ D contains infinitely many ositive elements of norm equal to a. Since there are only

4 14 2 Even Exonents finitely many residue classes mod a, there exist distinct ositive ; ˇ 2 ZŒ D such that N D N ˇ D a and ˇ mod a.then Nˇ ˇ Nˇ a 0 mod a: P D =ˇ. Then 1, because and ˇ are distinct. On the other hand, D Nˇ=a 2 ZŒ D, and, similarly, 1 2 ZŒ D. We have found an element 2 U distinct from 1. Hence U f1g. Next, we rove that U is an infinite cyclic grou. The logarithmic ma log W U R defines an injective homomorhism of U into the additive grou of real numbers. Since U f1g, the image log U is a nonzero subgrou of R. Further, if 2 U satisfies >1,thenDxCy D with x;y > 0, whence > D. It follows that any ositive element of log U is greater than log D, which imlies that log U is a discrete subgrou of R. Since any nonzero discrete subgrou of R is infinite cyclic, the theorem follows. The unit >1, generating the grou ZŒ D C, is called the basic (or fundamental) unit of the ring ZŒ D. Usually, it is not easy to find the basic unit or to decide whether a given unit is basic. In some cases this can be done using the following simle observation. Proosition 2.5. Let D a C b D be the basic unit of ZŒ D, and let D x C y D be any other unit. Then b j y. Proof. We may assume that x;y > 0. Then D n,wheren is a ositive integer. It follows that 2b D D N divides 2y D D n N n, whence the result. The following consequence is immediate. Corollary 2.6. Assume that D D a 2 1, wherea is a ositive integer (satisfying a>1if D D a 2 1). Then a C D is a basic unit of ZŒ D. It is worth mentioning that Corollary 2.6 is the simlest articular case of the famous theorem of Størmer (1897). Theorem 2.7 (Størmer). Let a C b D be a unit of ZŒ D such that a; b > 0 and every rime divisor of b divides D. Then it is a basic unit. We do not rove this theorem since we do not need it. An interested reader can find the roof in Ribenboim s book [117, Sect. A.4]. To conclude, let us mention that the results of this section are often interreted in terms of the Pell Diohantine equation x 2 Dy 2 D 1.

5 2.3 The Equation x 2 y q D 1 with q The Equation x 2 y q D 1 with q 5 The equation x 2 y q D 1 with an odd (rime) exonent q is much more difficult than x y 2 D 1. The case q D 3 was settled already by Euler, b, in site of some artial results, the general case remained oen until 1965, when the Chinese mathematician Ko Chao 1 roved [55, 56] that, for a rime q 5, the equation x 2 y q D 1 has no solions in ositive integers x and y. In 1976 Chein [23] discovered another roof, simler than Ko Chao s and based on a totally different idea. Chein s roof is reroduced below (with some changes). Ko Chao s roof can be found in Mordell s book [96, Sect. 30]. Both the arguments of Ko Chao and Chein work for a rime q 5 anddonot extend to q D 3. This case is solved in Sect. 2.5 by a totally different argument. Most of the known roofs of the theorem of Ko Chao rely on a result of Nagell abo the arithmetical structure of the solions of the equation x 2 y q D 1. We rove this theorem in Sect The theorem of Ko Chao will be roved (following Chein) in Sect In Sect we briefly describe the history of the equation Nagell s Theorem We start with an elementary lemma, which will be used in the next chater as well. Lemma 2.8. Let A and B be distinct corime rational integers and arime number. 1. If divides one of the numbers.a B /=.A B/ and A B, then it divides the other as well. 2. P d D gcd..a B /=.A B/;A B/.Thend 2f1; g. 3. If >2and d D,thenOrd..A B /=.A B// D 1. Proof. All the three statements easily follow from the identity A B A B D..A B/ C B/ B A B D X kd1 k.a B/ k 1 B k : (2.3) Rewriting it as A B A B D 1 X kd1.a B/ k 1 B k C.A B/ 1 ; k we obtain.a B /=.A B/.A B/ 1 mod, which roves art (1). 1 Sometimes selled as Ko Zhao.

6 16 2 Even Exonents Rewriting (2.3)as A B A B D B 1 C.A B/ X kd2.a B/ k 2 B k ; (2.4) k we observe that d divides B 1.SinceA and B are corime, d and B are corime as well, and we conclude that d j, which roves art (2). Finally, if d D then divides A B and does not divide B. Rewriting (2.3)as A B A B D B 1 C.A B/ 1 X kd2.a B/ k 2 B k C.A B/ 2 ; k we obtain.a B /=.A B/ B 1 mod 2, which roves art (3). Next, we establish the following reliminary result, due to Nagell [100]. Theorem 2.9 (Nagell). Let x;y be ositive integers and q an odd rime number satisfying x 2 y q D 1. Then2 j y and q j x. Proof. Write.x 1/.x C 1/ D y q. The greatest common divisor of x 1 and x C 1 divides 2. Ify is odd then they are corime; hence both are qth owers: x 1 D a q and x C 1 D b q. We obtain b q a q D 2, which is imossible. This roves that 2 j y. Further, write y q C 1 y C 1.y C 1/ D x2 : By Lemma 2.8, the greatest common divisor of the factors in the left-hand side is either 1 or q. Ifx is not divisible by q then the factors are corime, which means that each of them is a comlete square. Thus, there exist ositive integers a and b such that y C 1 D a 2 ; y q C 1 y C 1 D b2 ; x D ab : On the other hand, equality x 2 y q D 1 means that x C y.q 1/=2 y is a unit of the ring ZΠy, and Corollary 2.6 imlies that a C y is the basic unit of this ring. Hence there exists a ositive integer n such that We want to show that this is imossible. x C y.q 1/=2 y D.a C y/ n : (2.5)

7 2.3 The Equation x 2 y q D 1 with q 5 17 First of all, let us rove that n is even. Exanding.a C y/ n by the binomial formula, and reducing mod y, wefind.a C y/ n a n C na n 1 y mod y in the ring ZŒ y. Combining this with (2.5), we obtain na n 1 y.q 1/=2 0 mod y: Since y is even and a is odd, this imlies that n is even. Next, reducing (2.5) moda and using the congruences we obtain x D ab 0 mod a; y D a mod a;. 1/.q 1/=2 y. 1/ n=2 mod a; which means that a divides one of the numbers 1 y in the ring ZŒ y. Since a>1, this is imossible. The theorem is roved. We learned this argument from Hendrik Lenstra (rivate communication). It was also indeendently discovered by Nesterenko and Zudilin [106]. Nagell himself used the Theorem of Størmer (see Theorem 2.7) to show that both units a C y and x C y.q 1/=2 y should be basic, which is a contradiction Chein s Proof of the Theorem of Ko Chao Now we are ready to rove the theorem of Ko Chao. Theorem 2.10 (Ko Chao). The equation x 2 y q D 1 has no solions in ositive integers x;y and rime q 5. Proof (Chein). Theorem 2.9 imlies that x is odd. Assume that x 3 mod 4. Equality.x 1/.x C 1/ D y q imlies that there exist ositive integers a and b such that x C 1 D 2 q 1 a q ; x 1 D 2b q : Notice that a q D.b q C 1/=2 q 2 <b q, which imlies a<b. We have b 2 C 2a b 2q C.2a/ q b 2 C 2a x 1 2 x C 3 2 D b 2q C.2a/ q D C 2.x C 1/ D : 2 2 (2.6)

8 18 2 Even Exonents We again invoke Theorem 2.9, this time the statement q j x. Sinceq>3,this imlies that the right-hand side of (2.6) is not divisible by q. Now Lemma 2.8 yields that the factors on the left-hand side are corime. Hence they are comlete squares. Since b 2 C 2a is a comlete square, we have 2a.b C 1/ 2 b 2 D 2b C 1, which contradicts the reviously established inequality a<b. If x 1 mod 4 then x 1 D 2 q 1 a q and x C 1 D 2b q. We obtain x 3 2 b 2q.2a/ q D ; 2 and the rest of the argument is the same Some Historical Remarks Here we give some historical and bibliograhical remarks on the equation x 2 y q D 1 (2.7) with q 5. Before Ko Chao, the roblem attracted many mathematicians. In 1921 Nagell [100] observed that a theorem of Lebesgue [66] asserting that the equation x 5 C y 5 D 8z 5 has no solions imlies that x 2 y 5 D 1 has no solions. In the same article Nagell [100] resented several conditions involving the congruence class of q modulo 16 and the arithmetic of the number field Q. q/ under which the equation has no solion. In articular, he roved that there are no solions with q 101, excet maybe with q D 31; 59; 73; 83 or 89. In 1934, he [103] imroved uon the latter result, by showing that there is no solion if q 6 1 mod 8.Thus,below101 only 73 and 89 remained untreated. In 1932, Selberg [126] solved comletely the Diohantine equation x 4 y q D 1, answering a question osed by Nagell [98] in In 1940, Obláth [108, 109] showed that (2.7) has no solion excet, ossibly, when 2 q 1 1.mod q 2 /; 3 q 1 1.mod q 2 /: (2.8) His starting oint was the key observation that it is sufficient to solve the equation 2 q 2 a q b q D 1, as it is clear from the roof of Theorem Then, he combined the results of Lubelski [75] on the Diohantine equation x q C y q D cz q with Theorem 2.9 to get (2.8). The abovementioned work of Lubelski generalizes the famous results of Wieferich [138] and Mirimanoff [95] on the Fermat equation x q C y q C z q D 0. Wieferich showed that if the latter equation has a solion with q not dividing xyz (the first case of the Fermat theorem), then 2 q 1 1 mod q 2, and Mirimanoff

9 2.4 The Cubic Field Q. 2/ 19 showed that 3 q 1 1 mod q 2. Wieferich-tye conditions systematically occur in Catalan s roblem; see Sects. 6.2 and 6.5. Obláth [110] combined his conditions with Nagell s results to show that there is no solion with q < In 1961, Inkeri and Hyyrö [52] imroved uon Nagell s Theorem 2.9: they showed that existence of a nontrivial solion of (2.7) imlies that q 2 j x and q 3 j.y 1/. Using this result, they roved that there is no nontrivial solions with q < Equation x 2 y q D 1 continued to attract researchers even after the work of Ko Chao. We have already seen Chein s contribion. In 2004 Mignotte [88] suggested yet another roof of the theorem of Ko Chao. He adated the classical argument of Kummer to show that (2.7) has no solion when q 5 is a regular 2 rime. The first irregular rimes congruent to 1 modulo 8 are 233 and 257. This gives a weaker result than Obláth s; however, modern shar estimates [65] for binary logarithmic forms imly that (2.7) is imossible for q > 200. This gives an alternative roof of Ko Chao s theorem. One can also aly a dee theorem of Ribet [118] on the Diohantine equation a C 2 b C c D 0 which imlies that (2.7) has no nontrivial solions. This is not the easiest way to rove Ko Chao s theorem, since Ribet s work uses the advanced machinery of Galois reresentations. We are left with the equation x 2 y 3 D 1. It will be solved in Sect. 2.5, after some rearation in Sect The Cubic Field Q. 3 2/ In this auxiliary section we determine the ring of integers O K and the grou of units U K D O K of the cubic field K D Q. 3 2/. We shall use this in Sect Everywhere througho this section we use the notation D 1 C 3 2; D 2 1: Notice that is a unit of K. We start from a simle observation. Proosition The rincial ideal./ is a rime ideal of K. It satisfies./ 3 D.3/. Proof. One verifies that 3 D 3.Since is a unit, this imlies that.3/ D./ 3. Since a rational rime number cannot slit in K into more than 3 rimes, the ideal./ is rime. 2 An odd rime number ` is regular if it does not divide the class number of the `th cyclotomic field and irregular if it does.

10 20 2 Even Exonents In the sequel, we write Ord instead of Ord./. Proosition The ring of integers O K is ZŒ 3 2. i Proof. Denote by D the discriminant of K and d D ho K WZŒ 3 2.ThenDd 2 is the discriminant of the ring ZŒ 3 2. Since the conjugates of 2 are 2 and 2 2, where D. 1 C 3/=2 is a rimitive cubic root of unity, we have Dd 2 D ` 2 det k 3 2 D : 0k;`2 It follows that d j 6. Ifd is even, then D is odd, which is imossible because 2 ramifies in K. Thus, d divides 3. Since ZŒ 3 2 D ZŒ, every 2 OK can be written as D a 0 C a 1 C a 2 2, where a 0 ;a 1 ;a Z. If we show that 3 Ord 3.a k / 0.k D 0; 1; 2/; (2.9) it would follow that 2 ZŒ, roving the roosition. Observe that Ord ak k k mod 3. It follows that the numbers Ord ak k.k D 0; 1; 2/ are airwise distinct. Lemma A.1 imlies that Ord. / D min Ord ak k : 0k2 B Ord. / 0, because is an algebraic integer. Hence Ord ak k 0 for k D 0; 1; 2, which is only ossible if Ord 3.a k / 0 for all k. Thisroves(2.9) and the roosition. Proosition The unit grou U K is generated by 1 and. Proof. The field K has a real embedding and a air of comlex conjugate embeddings. We identify K with its real embedding and denote the comlex embeddings by and N, sothat j./j 2 D 1 D 1 C 3 2 C 4: The Dirichlet unit theorem (Aendix A.2) imlies that the rank of the unit grou is 1. Also, since K has a real embedding, it cannot contain roots of unity other than 1. Thus, U K is generated by 1 and a unit, where we may assume that 0<<1: (2.10)

11 2.5 The Equation x 2 y 3 D 1 21 Then D m,wherem 1. This imlies the inequality j./jdjn./jj./jd 1 C 3 2 C 4 1=2 <2: (2.11) Proosition 2.12 imlies that D a 0 C a 1 2 C a2 4,wherea0 ;a 1 ;a 2 2 Z. We have a k D 1 3 Tr 3 k 2.k D 0; 1; 2/ ; where Tr W K Q is the trace ma. Using inequalities (2.10)and(2.11), we obtain ja 0 j; ja 1 j; ja 2 j5=3. Thus, a 0 ;a 1 ;a 2 2f0; 1g.Also,a 0 0, since otherwise is not a unit. Among the 18 remaining ossibilities, only 1,,and 1 are units. Among the latter, only belongs to the interval.0; 1/. Thus, D. Remark The following more refined argument shows that a 2 D 0 (which means that only six ossibilities are to be verified instead of 18). Assume that is not a fundamental unit. Then D m with m 2. This imlies, instead of (2.11), the inequality j./jdjn./jj./j 1=2 D 1 C 3 2 C 4 1=4 <1:5: Hence ja 2 jd 1 ˇ 3 ˇTr 1 4 ˇˇˇˇ 0:9; that is, a 2 D The Equation x 2 y 3 D 1 This equation has a long history. Already Fermat stated (as usual, witho a roof) that it has no solions in ositive integers excet the obvious D 1. Euler [31] was the first to rove this. This roof, quite involved, is reroduced in Ribenboim s book [117, Sect. A.2]. Theorem 2.15 (Euler). The only solion of the equation x 2 y 3 D 1 in nonzero integers x;y is. 3/ D 1. Actually, Euler roved much more: the equation x 2 y 3 D 1 has no solions in rational numbers x;y other than. 1; 0/,.0; 1/ and. 3; 2/. A reader familiar with the notion of ellitic curve can exress this as the ellitic curve x 2 y 3 D 1 has rank 0 and torsion 6 over Q.

12 22 2 Even Exonents Later Euler [32, Vol. 2, Article 247] and Legendre [68, ] used an infinite descent argument to rove the following theorem. Theorem 2.16 (Euler, Legendre). The equation u 3 C 1 D 2v 3 has no solions in integers u; v with v 0; 1. Theorem 2.15 is an easy consequence of Theorem Proof of Theorem 2.15 (Assuming Theorem 2.16). Rewrite our equation as.x 1/.x C 1/ D y 3.Ifx is even then the factors on the left are corime. Hence they are comlete cubes, which is imossible, because two cubes cannot differ by 2. Now assume that x is odd. Then y is even, and, relacing x by x, wemay assume that x 3 mod 4. Rewrite the equation as x 1 x C 1 y 3 D : The factors on the left are corime rational integers, hence both cubes: x 1 D 2u 3 and x C 1 D 4v 3,whereu and v are nonzero integers. We obtain u 3 C 1 D 2v 3, whence v D 0 or 1 by Theorem It follows that x D 1or x D 3, which roves Theorem Here again, both Euler and Legendre roved much more: the equation u 3 C 1 D 2v 3 has no nontrivial solions in u; v 2 Q (and even in u; v 2 Q. 3/). This also imlies the rational version of Theorem 2.15mentionedabove. In 1922, Nagell [101, Sect. 10] suggested an alternative roof of Euler s theorem; see also [102]. Here we give a different roof, due to McCallum [80]. McCallum s argument is more transarent than the roofs of Euler, Euler-Legendre, and Nagell, b it does not extend to rational solions. (Recently Notari [107] suggested yet another roof, which is totally elementary and, like McCallum s roof, works only for integer solions.) Denote by K the cubic field Q. 3 2/. Recall that D 2 1 generates the grou of ositive units of K. Notice that integers u; v satisfy u 3 C 1 D 2v 3 if and only if v 3 2 u is a ositive unit of the field K. Hence Theorem 2.16 is equivalent to the following statement. Theorem The field K has no ositive units of the form a 0 C a 1 2 (with a 0 ;a 1 2 Z) other than 1 and. Theorem 2.17 is a articular case of the famous result of Delaunay [27]: given a cube-free integer d, the ring ZΠ3 d has at most one nontrivial ositive unit of the form a 0 C a 1 d; if such a unit exists, then it generates the grou of ositive units. Skolem [130, ] gave another roof of the first art of Delaunay s theorem using his local method. See also [96, Theorems 23.5 and 24.5]. The roof of McCallum, reroduced below in Sect , can be viewed as a simlified version of Skolem s local argument for d D 2. For the roof, we need a rearatory statement on the -adic convergence of binomial series.

13 2.5 The Equation x 2 y 3 D Binomial Series In this subsection K is an arbitrary number field, not just Q. 2/. If is a comlex number with j j <1then the binomial series P 1 n k kd0 k converges to.1 C / n. We need the -adic generalization of this: if Ord. / > 0 then the series P 1 n k kd0 -adically converges to.1 C / n. k Thus, let be a rime ideal of the field K,andletO be the local ring of : O Df 2 K W Ord. / 0g: Given ; ˇ 2 O, we say that ˇ mod N if Ord.ˇ / N. Proosition Let n be an integer, N a nonnegative integer, and arimeideal of a number field K. Then for any 2 K with Ord. / > 0 we have NX kd0 n k.1 C / n mod N C1 : k Proof. If n 0 then the assertion is an obvious consequence of the binomial formula. Now assume that n<0, and write n D m 1 with m 0. It will be convenient to relace by. Since. 1/ k m 1 k D mck m,wehavetorovethat NX m C k k.1 / m 1 mod N C1 : (2.12) m kd0 Deriving m times the identity mcn X.1 t/ 1 kd0 t k D t mcn C1.1 t/ 1 ; and dividing by mš, we obtain the identity NX.1 t/ m 1 m C k t k D t N C1 P.t/.1 t/ m 1 ; (2.13) m kd0 where P.t/ is a olynomial, deending on m and N. Notice that X N t N C1 P.t/ D 1.1 t/ mc1 m C k t k 2 ZŒt ; m kd0

14 24 2 Even Exonents which imlies that P.t/ 2 ZŒt. Hence Ord N C1 P. /.1 / m 1.N C 1/Ord. / N C 1; and (2.12) follows uon substiting t D in (2.13) Proof of Theorem 2.17 Now we are ready to rove Theorem We again use the notation K D QŒ 3 2, D 3 2 1, and D 2 C 1 from Sect Every element of the field K can be uniquely written as a 0 C a 1 2 C a2 4, where a 0 ;a 1 ;a 2 2 Q. This defines the Q-linear coefficient functions a 0 ;a 1 ;a 2 W K Q: Since 3 D 3, wehave 3m D 3 m m for any integer m. Hence for any ositive integer k the number k is divisible by 3 bk=3c in the ring ZŒ 3 2. It follows that Ord 3 a` k bk=3c.` D 0; 1; 2/; (2.14) for k 0, which will be used througho the roof. Let be a ositive unit of K with a 2./ D 0. We have to rove that D 1 or D. Proosition 2.13 imlies that D n,wheren 2 Z. We assume that n 0; 1 and obtain a contradiction. Fix a large ositive integer N, to be secified later. Since D 2 C, wehave. 2/ n D.1 =2/ n. Proosition 2.18 imlies that NX kd0 n k. 2/ n mod N C1 : (2.15) k 2 Alying the coefficient function a 2 to both sides of (2.15), and using (2.14), we obtain NX kd0 n a2 k. 2/ n a k. 2/ k 2./ mod 3 b.n C1/=3c : (2.16) Since a 2./ D 0, the right-hand side of congruence (2.16) vanishes. Since a 2.1/ D a 2./ D 0, so do the summands on the left of (2.16), corresonding to k D 0 and k D 1.Also,fork 2,wehave n n.n 1/ D k k.k 1/ n 2 : k 2

15 2.5 The Equation x 2 y 3 D 1 25 Hence, for k 2, thekth summand on the left of (2.16) is equal to n.n 1/A k, where 1 n 2 A k D a k.k 1/. 2/ k 2 k : k 2 Thus, (2.16) can be rewritten as n.n 1/.A 2 CCA N / 0 mod 3 b.n C1/=3c : Since n is distinct from 0 and 1, we may choose N so large that We obtain b.n C 1/=3c > Ord 3.n.n 1//: On the other hand, again using (2.14), we find A 2 CCA N 0 mod 3: (2.17) Ord 3.A k / bk=3c Ord 3.k.k 1// bk=3c log 3 k: (2.18) As one can easily verify, the right-hand side of (2.18) is ositive for k 6. Hence Ord 3.A k />0for k 6. Also,Ord 3.k.k 1// D 0 for k D 5, which imlies that Ord 3.A 5 />0. We have roved that Ord 3.A k />0for k 5. Combining this with (2.17), we obtain A 2 C A 3 C A 4 0 mod 3: Since a 2. 2 / D 1, a 2. 3 / D 3, anda 2. 4 / D 6,wehave A 2 C A 3 C A 4 D 1 8 n 2 16 C.n 2/.n 3/ 32 D n2 7n C 14 : 32 It follows that n 2 7n C 14 0 mod 3, which is imossible for n 2 Z. The theorem is roved. Remark The Diohantine equation y 2 x 3 D k, wherek is a nonzero integer, is usually called Mordell s equation.manyresults onit can befoundin Cha. 26 of Mordell s book [96]. Modern techniques based on logarithmic forms [11] or forms in ellitic logarithms [33] allow one to solve Mordell s equation comletely for small values of k, using electronic comations. For instance, this was done for jkj 10 4 in [33].

16 htt://