NOTES ON DIOPHANTINE APPROXIMATION

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1 NOTES ON DIOPHANTINE APPROXIMATION Jan-Hendrik Evertse December 11, Approximation of algebraic numbers Literature: W.M. Schmidt, Diophantine approximation, Lecture Notes in Mathematics 785, Springer Verlag 1980, Chap.II, 1,2, Chap. IV, 1 L.J. Mordell, Diophantine Equations, Pure and applied Mathematics series, vol. 30, Academic Press, reprint of the 1971 edition. 8.1 Liouville s Theorem and Roth s Theorem We are interested in the problem how well a given real algebraic number can be approximated by rational numbers. We define the height H(ξ) of a rational number ξ as follows: choose integers x, y such that ξ = x/y and gcd(x, y) = 1. Then H(ξ) := max( x, y ). More generally, let α be an algebraic number and f(x) = a 0 X n + a 1 X n a n Z[X] the minimal polynomial of α, that is, f(α) = 0, f is irreducible, gcd(a 0,..., a n ) = 1 and a 0 > 0. We may write f(x) = a 0 (X α 1 ) (X α n ) with α 1 = α. (1) Then the Mahler height of α (in the literature also called the Mahler measure of α and denoted by M(α)) is defined by H(α) := a 0 We start with a classic result of Liouville. n max(1, α i ). i=1 64

2 8.1 Liouville s Theorem and Roth s Theorem 65 Theorem (Liouville s inequality, 1844) Let α be a real algebraic number of degree n 1. There is a constant c(α) > 0 depending only on α such that ξ α c(α)h(ξ) n for every ξ Q with ξ α if n = 1. (2) One may take c(α) = 2 1 n H(α) 1. Proof. Let again f(x) = a 0 X n + a 1 X n a n Z[X] be the minimal polynomial of α and write f as in (1). Take ξ Q, let ξ = x/y where x, y Z, and consider the number F (x, y) := y n f(x/y) = a 0 x n + a 1 x n 1 y + + a n y n. By assumption, f(x/y) 0. Hence F (x, y) is a non-zero integer. So F (x, y) 1. To obtain an upper bound for F (x, y) we express F (x, y) as a 0 (x α 1 y) (x α n y) and estimate the factors as follows: Thus, x α 1 y = ξ α y ξ α H(ξ), x α i y 2 max(1, α i ) max( x, y ) = 2 max(1, α i )H(ξ) (i = 2,..., n). F (x, y) = a 0 n x α i y i=1 a 0 ξ α H(ξ) n 2 n 1 ξ α H(ξ) n 2 n 1 H(α). n max(1, α i ) i=2 Combined with F (x, y) 1 this gives (2). From Theorem we can derive a transcendence result which was already mentioned in Chapter 1. Corollary Let b be an integer with b 2. Then k=1 b k! is transcendental. Proof. Suppose α := k=1 b k! is algebraic of degree n. Let N be a positive integer, and define ξ N := N k=1 b k!. Notice that ξ N is a rational number with 0 < ξ N < 1 and such that b N! ξ N is an integer coprime with b. So H(ξ N ) = b N!. Further, 0 < α ξ N = b k! 2 b (N+1)! = 2H(ξ N ) (N+1). k=n+1

3 66 8 APPROXIMATION OF ALGEBRAIC NUMBERS On the other hand, Theorem implies that α ξ N c(α)h(ξ N ) n. For N sufficiently large this gives a contradiction. Let α be a real algebraic number of degree n 2. One of the central problems in Diophantine approximation is, to obtain improvements of (2) with in the right-hand side H(ξ) κ with κ < n instead of H(ξ) n, where as before, α is a real algebraic number of degree n 2. More precisely, the problem is, whether there exist κ < n and a constant c(α, κ) > 0 depending only on α, κ, such that ξ α c(α, κ)h(ξ) κ for every ξ Q. (3) Recall that by Dirichlet s Theorem, there exist infinitely many pairs of integers x, y such that x α y y 2, y 0. For such solutions we have x ( α + 1) y. Hence, writing ξ = x we infer that there is a constant c y 1(α) > 0 such that ξ α c 1 (α)h(ξ) 2 for infinitely many ξ Q. This shows that there can not exist an inequality of the shape (3) with κ < 2. In particular, for quadratic algebraic numbers α, Theorem gives the best possible result in terms of the exponent on H(ξ). Now let α be a real algebraic number of degree n 3. In 1909, the Norwegian mathematician A. Thue made an important breakthrough by showing that for every κ > n 2 +1 there exists a constant c(α, κ) > 0 such that (3) holds. In 1921, C.L. Siegel proved the same for every κ 2 n. In 1949, A.O. Gel fond and independently Freeman Dyson (the famous physicist) improved this to κ > 2n. Finally, in 1955, K.F. Roth proved the following: Theorem (Roth, 1955) Let α be a real algebraic number of degree n 3. Then for every κ > 2 there exists a constant c(α, κ) > 0 such that ξ α c(α, κ)h(ξ) κ for every ξ Q. (3) Remark. There are various equivalent formulations of Roth s Theorem which one may use depending on the application. 1) For every κ > 2 the inequality has only finitely many solutions. 2) For every C > 0, κ > 2 the inequality ξ α H(ξ) κ in ξ Q (4) ξ α CH(ξ) κ in ξ Q (5)

4 8.1 Liouville s Theorem and Roth s Theorem 67 has only finitely many solutions. It should be noted that Liouville s Theorem is effective, i.e., its proof gives a method to compute the constant c(α) in (2). In contrast, the results of Thue, Siegel, Gel fond, Dyson and Roth mentioned above are ineffective, i.e., with their methods of proof one can prove only the existence of a constant c(α, κ) > 0 as in (3), but one can not compute c(α, κ). Alternatively stated, the methods of proof of Thue,..., Roth show that the inequalities (4), (5) have only finitely many solutions, but they do not provide a method to determine these solutions. Thue used his result on the approximation of algebraic numbers stated above, to prove his famous theorem that the equation F (x, y) = m in x, y Z where F is a binary form in Z[X, Y ] such that F (X, 1) has at least three distinct roots and m is a non-zero integer, has at most finitely many solutions. We prove a more general result. Notice that every binary form is a product of linear forms. A binary form F (X, Y ) Z[X, Y ] is called square-free if it is not divisible by G(X, Y ) 2 for some non-constant binary form G. Theorem Let F (X, Y ) Z[X, Y ] be a square-free binary form of degree n 3. Then for every κ > 2 there is a constant c(f, κ) > 0 such that for every pair of integers (x, y) with F (x, y) 0 we have F (x, y) c(f, κ) max( x, y ) n κ. (6) Proof. Let F (X, Y ) = a 0 X n + a 1 X n 1 Y + + a n Y n. Distinguishing between a 0 0, a 0 = 0 we obtain that either F (X, Y ) = a 0 (X α 1 Y ) (X α n Y ) with α 1,..., α n distinct or F (X, Y ) = a 1 Y (X α 1 Y ) (X α n 1 Y ) with α 1,..., α n 1 distinct. The result holds in both cases, but we consider only the first case. Let (x, y) be a pair of integers with F (x, y) 0. First suppose that x is large compared with y, x C y with C := 2 max(1, α 1,..., α n ), say. Then F (x, y) a 0 n ( x α i y ) x n a 0 i=1 n i=1 ( 1 α ) i C 2 n a 0 max( x, y ) n C(F, κ) max( x, y ) n

5 68 8 APPROXIMATION OF ALGEBRAIC NUMBERS provided C(F, κ) 2 n a 0. Now suppose that x < C y. Then y 0. Let ξ := x/y. Notice that max( x, y ) H(ξ) (with equality if gcd(x, y) = 1). Assume without loss of generality that By Theorem we have For i = 2,..., n we have ξ α 1 = min i=1,...,n ξ α i. ξ α 1 c(α, κ)h(ξ) κ c(α, κ) max( x, y ) κ. α i α 1 α i ξ + ξ α 1 2 ξ α i, implying ξ α i 1 2 α i α 1 for i = 2,..., n. By combining these inequalities with max( x, y ) C y we obtain F (x, y) = y n a 0 ξ α i C (F, κ) y n max( x, y ) κ C(F, κ) max( x, y ) n κ. We deduce Thue s Theorem. Corollary Let F (X, Y ) be a binary form in Z[X, Y ] such that F (X, 1) has at least three distinct roots. Further, let m be a non-zero integer. Then the equation has at most finitely many solutions. F (x, y) = m in x, y Z Proof. We first make a reduction to the case that F (X, Y ) is square-free. We can factor the polynomial F (X, 1) as cg 1 (X) k1 g t (X) kt where c is a non-zero integer and g 1 (X),..., g t (X) are irreducible polynomials in Z[X] none of which is a constant multiple of the others. Let f (X) := cg 1 (X) g t (X) = a 0 X n + + a n with a 0 0; then n 3. Put F (X, Y ) = Y n f(x/y ) = a 0 X n + + a n Y n. Then F is square-free, so we can apply Theorem Let x, y be integers with F (x, y) = m. Then F (x, y) divides m. Take κ with 2 < κ < n. Then by the previous theorem, m F (x, y) c(f, κ) max( x, y ) n κ,

6 8.1 Liouville s Theorem and Roth s Theorem 69 implying that x, y are bounded. As mentioned before, the proof of Roth s Theorem is ineffective, and an effective proof of Roth s Theorem seems to be very far away. There are however effective improvements of Liouville s inequality, i.e., inequalities of the shape ξ α c(α, κ)h(ξ) κ for ξ Q where α is algebraic of degree n 3 and κ < n (but very close to n) and with some explicit expression for c(α, κ). We mention the following result of the Russian mathematician Fel dman, obtained using lower bound for linear forms in logarithms. Theorem (Fel dman, 1971) Let α be a real algebraic number of degree n 3. Then there exist effectively computable numbers c 1 (α), c 2 (α) > 0 depending on α such that ξ α c 1 (α)h(ξ) n+c 2(α) for ξ Q. (7) Idea of proof. Let f(x) be the minimal polynomial of α and F (X, Y ) = Y n f(x/y ). Following the arguments of the proof of Theorem in Section 7, using more precise lower bounds for linear forms in logarithms and making all computations precise, one shows that there are effectively computable constants A, B depending on F, such that if m is any non-zero integer, and x, y are any integers with then max( x, y ) A m B. F (x, y) = m Let ξ Q, let x, y be integers with ξ = x/y and gcd(x, y) = 1, and put m := F (x, y). In the proof of Theorem we saw that Hence F (x, y) 2 n 1 H(α)H(ξ) n ξ α. ξ α m 2 1 n H(α) 1 H(ξ) n (A 1 H(ξ)) 1/B 2 1 n H(α) 1 H(ξ) n = c 1 H(ξ) n+1/b with c 1 = A 1/B 2 1 n H(α) 1. The quantities c 1 (α), c 2 (α) are very small numbers for which one can find an explicit expression by going through the proof. For instance, Bugeaud proved in 1998, that (7)

7 70 8 APPROXIMATION OF ALGEBRAIC NUMBERS holds with ( c 1 (α) = (eh) n2 exp 10 27n n 14n D 1/2 ( log(ed)) n 1), ( ) 1 c 2 (α) = 10 26n n 14n D 1/2 ( log(ed)) where H = H(α) is the Mahler height of α and D = D Q(α) is the absolute value of the discriminant of the field Q(α) (see Chapter 3). One can obtain better results for certain special classes of algebraic numbers using other methods. For instance, M. Bennett obtained good effective improvements of Liouville s inequality for various numbers of the shape m a where m is a positive integer and a a positive rational number. For instance he showed that ξ H(ξ) 2.45 for ξ Q. The techniques used by Thue,..., Roth cannot be used in general to solve Diophantine equations, but together with suitable refinements, they allow to give explicit upper bounds for the number of solutions of Diophantine equations. For instance we have: Theorem (Bombieri, Schmidt, 1986) Let F (X, Y ) be a binary form in Z[X, Y ] such that F (X, 1) has precisely n 3 distinct roots. Then the equation F (x, y) = 1 in x, y Z has at most c n solutions where c is a positive constant not depending on F. The importance of the result is that the bound is uniform, i.e. for all binary forms F we get the upper bound cn. It is possible to compute c explicitly. Bombieri and Schmidt showed that for sufficiently large n, c can be taken equal to 430. Probably the constant c can be improved, but the dependence on n is optimal. For instance, let F (X, Y ) = (X a 1 Y ) (X a n Y ) + Y n, where a 1,..., a n are distinct integers. Then the equation F (x, y) = 1 has the n solutions (a 1, 1),..., (a n, 1). We finish with a remarkable result due to M. Bennett: Theorem (Bennett, 2002) Let n be an integer with n 3 and let a, b be positive integers. Then the equation ax n by n = 1 has at most one solution in positive integers x, y.

8 8.1 Liouville s Theorem and Roth s Theorem 71 For instance, the equation (a + 1)x n ay n = 1 has (1, 1) as its only solution in positive integers. The proof of the above result uses various techniques (good lower bounds for linear forms in two logarithms, Diophantine approximation techniques based on so-called hypergeometric functions, and heavy computations). There are generalizations of Roth s Theorem, which state how well a given algebraic number can be approximated by an element from a given algebraic number field. Let K be an algebraic number field of degree d. The height of ξ K relative to K is defined by d H K (ξ) = a 0 max(1, σ i (ξ) ), i=1 where σ 1,..., σ d are the embeddings K C, and a 0 is an integer such that a 0 d i=1 (X σ i (ξ)) has coefficients in Z with gcd 1. The first generalization of Roth s Theorem states that for every algebraic number α and every κ > 2 there is a constant c(α, κ) > 0 such that α ξ c(α, κ)h K (ξ) κ for every ξ K with ξ α. This can be generalized further. Let α 1,..., α d be algebraic numbers in C. Then one can simultaneously approximate α 1 by σ 1 (ξ),..., α d by σ d (ξ). Theorem Let α 1,..., α d be algebraic numbers in C and κ > 2. Then there is a constant c > 0 depending on κ, α 1,..., α d such that d min(1, σ i (ξ) α i ) ch K (ξ) κ i=1 for every ξ K for which the left-hand side is non-zero. The theorem more or less implies that if one of the σ i (ξ) is very close to α i then for j i the number σ j (ξ) cannot be too close to α j.

9 72 8 APPROXIMATION OF ALGEBRAIC NUMBERS 8.2 Thue s approximation theorem We intend to prove the following result of Thue: Theorem Let α be a real algebraic number of degree n 3 and κ > n + 1. Then 2 the inequality ξ α H(ξ) κ in ξ Q (8) has only finitely many solutions. We introduce some notation. The norm of a polynomial P = D i=0 p ix i C[X] is given by P := D i=0 p i. It is easy to check that for P, Q C[X] we have P + Q P + Q, P Q P Q. (9) The k-th divided derivative of a polynomial P C[X] is defined by P {k} := 1 k! dk P/dX k. Thus, if P = D i=0 p ix i, then P {k} = D ( i ) i=0 pi X i k, where ( a k b) := 0 if b > a. We have the product rule (P Q) {k} = k j=0 P {k j} Q {j} for P, Q C[X]. Notice that if P Z[X] then also P {k} Z[X]. Further, since each binomial coefficient ( i k) can be estimated from above by 2 i 2 D, we have P {k} 2 D P. In the proof we need parameters θ, ε such that n+2 < θ < 2, 1 < θ < 2, (10) κ 1 0 < ε < 1, < κ 1. (11) κ(θ ε) (n+ε) n+ε Such a choice of θ is possible since κ > n + 1, and such a choice of ε is possible, since 2 1 κθ n < 1 2 < κ 1 n. It will turn out later that these choices for θ, ε are precisely what is needed to make all estimates work. Further, we use a positive integer r, the value of which will be specified later. C 1, C 2,... will be effectively computable constants depending on α, κ, θ, ε but (very important) independent of r. Like in many proofs in Diophantine approximation or transcendence theory, we start with the construction of a function with suitable properties, using Siegel s Lemma.

10 8.2 Thue s approximation theorem 73 Lemma For every positive integer r there exist polynomials P r, Q r degree at most nr 1, not both equal to 0, with the following properties: Z[X] of P r αq r is divisible by (X α) [θr], (12) P r C r 1, Q r C r 1. (13) Proof. Let K = Q(α) and σ 1,..., σ n the embeddings K C. Write P r = nr 1 i=0 p i X i, Q r = nr 1 i=0 q i X i, where p i, q i are unknowns, taken from the integers. The condition to be satisfied is, that P r {k} (α) αq {k} r (α) = 0 (k = 0,..., [θr] 1). Let d be a denominator of α, i.e., d Z >0, dα O K. By expanding the above expression and multiplying with d nr we obtain nr 1 i=0 ( ) i d nr α i p i k nr 1 i=0 ( ) i d nr α i+1 q i = 0 (k = 0,..., [θr] 1), k which is a system of [θr] linear equations in 2nr unknowns with coefficients in O K. Notice that 2nr > n[θr], i.e., the number of unknowns is larger than [K : Q] times the number of equations. So the condition of Siegel s Lemma (Theorem 4.1.2) is satisfied. According to this lemma, the above system has a non-trivial solution (p 0,..., p nr 1, q 0,..., q nr 1 ) Z 2nr such that max(max i p i, max q i ) ( 2A) n[θr] 2nr [θr] i ( 2A) θ 2 θ, where A = max h,k ( nr 1 j=0 ( ) j d nr σ h (α) j + k nr 1 j=0 d nr 2 nr 2nr max (1, max σ h (α) ) nr. h d nr ( j k ) σ h (α) j+1 ) So Lemma holds with C 1 = (12d max (1, max h σ h (α) )) nθ/(2 θ). An outline of the proof of Theorem Pick two solutions ξ 1, ξ 2 of (8) and consider the quantity P r (ξ 1 ) ξ 2 Q r (ξ 1 ).

11 74 8 APPROXIMATION OF ALGEBRAIC NUMBERS Let ξ i = x i /y i with x i, y i Z, gcd(x i, y i ) = 1 (i = 1, 2). Then since P r, Q r have degree at most nr 1, the number is an integer. On the other hand, writing A r := y nr 1 y 2 (P r (ξ 1 ) ξ 2 Q r (ξ 1 )) P r αq r = V r (X α) [θr] with V r C[X] as we may, we get ) A r = y1 nr y 2 (V r (ξ 1 )(ξ 1 α) [θr] (ξ 2 α)q r (ξ 1 ). Since ξ 1, ξ 2 are both very close to α, this quantity is very small, and in fact one can show that there are solutions ξ 1, ξ 2 of (8) and a positive integer r such that A r < 1. But then, A r = 0. On the other hand, we would like to prove that A r 0 and arrive at a contradiction. This argument has to be modified somewhat. We prove that if H(ξ 1 ) is sufficiently large, there is k {0,..., [rε] + 1} such that A rk := y1 nr y 2 (P r {k} (ξ 1 ) ξ 2 Q {k} r (ξ 1 )) 0. On the other hand we prove, similarly as above that there are solutions ξ 1, ξ 2 of (8) and a positive integer r such that A rk is an integer with absolute value < 1, hence 0. We work out the details in some lemmas. Lemma Let F Z[X], β an algebraic number such that (X β) m divides F, and g the minimal polynomial of β. Then there is G Z[X] such that F = g m G. Proof. By the unique factorization in Z[X], we can factor F in Z[X] as cg k 1 1 gs gs where c Z, g 1,..., g s are distinct primitive irreducible polynomials in Z[X] (i.e., whose coefficients have gcd 1) and k 1,..., k s > 0. One of the factors g 1,..., g s must be equal to g, since F (β) = 0, and then we have k 1 m since (X β) m divides f. The lemma readily follows. Lemma Let ξ 1, ξ 2 be two distinct solutions of (8), and r a positive integer, such that P r {k} (ξ 1 ) ξ 2 Q r {k} (ξ 1 ) = 0 for k = 0,..., [εr] + 1. Then H(ξ 1 ) C 2. Proof. By eliminating ξ 2 we obtain that P {k} r (ξ 1 )Q {l} r (ξ 1 ) P r {l} (ξ 1 )Q {k} r (ξ 1 ) = 0 for k, l = 0,..., [εr] + 1.

12 8.2 Thue s approximation theorem 75 Define F (X) := P r (X)Q r(x) P r(x)q r (X). Then for k = 0,..., [εr], F {k} is a linear combination of P r {l} Q r {m} P r {m} Q r {l}, 0 l, m k+1. So F {k} (ξ 1 ) = 0 for k = 0,..., [εr]. Assume for the moment that F is not identically 0. Then F is divisible by (X ξ 1 ) [εr]+1. So by Lemma 8.2.3, F = (y 1 X x 1 ) [εr]+1 G for some G Z[X]. This implies that both y [εr]+1 1, x [εr]+1 1 divide coefficients of F. Hence H(ξ) [εr]+1 F. By (9) we have F C r for some constant C independent of r. Hence H(ξ) C 2 with C 2 = C 1/ε. It remains to show that F is not identically 0. Assume that F is identically 0. One of P r, Q r, say Q r, is not identically 0. Then (P r (X)/Q r (X)) = 0 hence P r (X)/Q r (X) is identically equal to some constant c Q. But then, Q r (X) = (c α) 1 ((P r (X) αq r (X)) is divisible by (X α) [θr]. This implies that Q r (X) is divisible by f(x) [θr], where f(x) is the minimal polynomial of α. But then comparing degrees, n[θr] nr 1 which is impossible since θ > 1. Lemma Let ξ 1, ξ 2 be two solutions of (8) with H(ξ 2 ) > H(ξ 1 ) and put Let 0 k [εr] + 1. Then A rk := y nr 1 y 2 ( λ := log H(ξ 2) log H(ξ 1 ). P r {k} A rk C r 4(H(ξ 1 ) u + H(ξ 1 ) v ) ) (ξ 1 ) ξ 2 Q {k} r (ξ 1 ) is an integer and with u = (n κ(θ ε))r + 2κ + λ, v = nr (κ 1)λ. Proof. Recall that P r {k} and Q r {k} are polynomials of degree at most nr 1 k with integral coefficients. This implies that A rk is an integer. We estimate A rk from above. We write H 1 := H(ξ 1 ), H 2 := H(ξ 2 ). Thus, H 2 = H λ 1. The polynomial P {k} r P {k} r αq {k} r αq {k} r is divisible by (X α) [θr] k. Hence = V (X α) [θr] k with V C[X]. This gives A rk = y nr 1 y 2 ( V (ξ 1 )(ξ 1 α) [θr] k (ξ 2 α)q {k} r (ξ 1 )) max( V (ξ 1 ), Q {k} r (ξ 1 ) ) ( y nr 1 y 2 ξ 1 α [θr] k + y nr 1 y 2 ξ 2 α ).

13 76 8 APPROXIMATION OF ALGEBRAIC NUMBERS We take for granted for the moment that V (ξ 1 ) C r 4, Q(ξ 1 ) C r 4. Notice that y1 nr y 2 H1 nr H 2 = H1 nr+λ, ξ 1 α [θr] k H κ([θr] k) 1 H κ([θr] [εr] 1) 1 H κr(θ ε)+2κ 1, ξ 2 α H2 κ = H1 κλ. A combination of these estimates gives our lemma. We sketch how to estimate V (ξ 1 ). Notice that P r αq r = ( j [θr] P {k} r (α) αq r {k} (α) ) (X α) j. Hence P r {k} αq {k} r = ( ) ( ) P r {k} (α) αq {k} j r (α) (X α) j k. k j [θr] So V = ( ) j [θr] P r {k} (α) αq r {k} (j (α) k) (X α) j [θr]. Now one can easily estimate V (ξ 1 ) from above using estimates such as P {k} r (α) P r {k} max(1, α ) nr (2 n C 1 max(1, α ) n ) r (and likewise for Q {k} r (α)), ξ 1 α 1, ( j k) 2 nr, number of summands 2nr 4 nr, etc. In a similar fashion one can estimate Q {k} (ξ 1 ). Proof of Theorem We take a solution ξ 1 of (8) with H(ξ 1 ) > C 2 and then a solution ξ 2. Then by Lemma 8.2.4, for every positive integer r, there is k {0,..., [εr]+ 1} with A rk 0. We want to show that there are r, ξ 1, ξ 2, such that A rk < 1, hence A rk = 0 for k = 0,..., [εr] + 1 and obtain a contradiction. To this end, we want to choose r in such a way that u rε, v rε. This amounts to (n κ(θ ε))r + 2κ + λ rε, nr (κ 1)λ rε. This gives altogether the following conditions for r: r 2κ + λ (κ 1)λ, r κ(θ ε) (n + ε) n + ε. (14) We view the right-hand of both inequalities as linear functions in λ. By condition (11), 1 i.e., < κ 1, the second linear function increases faster than the first. This κ(θ ε) (n+ε) n+ε implies that there is λ 0 such that for every λ > λ 0 we have (κ 1)λ n + ε 2κ + λ κ(θ ε) (n + ε) > 1.

14 8.2 Thue s approximation theorem 77 Any interval of length > 1 contains an integer. So there is λ 0 such that for every λ > λ 0 there is an integer r satisfying all conditions in (14). We finish the proof. Assume that (8) has infinitely many solutions. So (8) has solutions ξ 1, ξ 2 such that } H(ξ 1 ) C 2, H(ξ 1 ) > (2C 4 ) 1/ε, log H(ξ 2 ) =: λ > λ (15) log H(ξ 1 ) 0. Then there is an integer r with (14). So by Lemma 8.2.4, there is k {0,..., [εr] + 1} with A rk 0. On the other hand, u εr, v εr, and so by Lemma 8.2.5, A rk C r 4 2H 1 (ξ 1 ) εr < 1, hence A rk = 0. This gives the contradiction we want. So (8) cannot have infinitely many solutions. Remark. To obtain a contradiction, we did not need the assumption that (8) has infinitely many solutions, but merely that there are solutions ξ 1, ξ 2 of (8) with (15). In other words, solutions ξ 1, ξ 2 of (8) with (15) cannot exist. The constants C 2, C 4 and λ 0 are effectively computable. So in fact we can prove the following sharpening of Theorem 8.2.1: Theorem There are effectively computable constants C, λ 0 depending on α, κ, such that if ξ 1 is a solution of (8) with H(ξ 1 ) C, then for any other solution ξ of (8) we have H(ξ) H(ξ 1 ) λ 0.

15 78 8 APPROXIMATION OF ALGEBRAIC NUMBERS 8.3 Exercises Exercise Let b be an integer 2. Prove that k=1 b 3k is transcendental. Exercise Let F (X, Y ) Z[X, Y ] be a square-free binary form of degree n 3 and let G(X, Y ) = a b i=0 j=0 g ijx i Y j Z[X, Y ] with a + b n 3. Prove that the Diophantine equation has only finitely many solutions. F (x, y) = G(x, y) in x, y Z with F (x, y) = 0 Exercise Let F (X, Y ) be a binary form in Z[X, Y ] of degree 2 such that all roots of F (X, 1) are non-real complex numbers, and let m be a non-zero integer. Give a direct proof that F (x, y) = m in x, y Z has at most finitely many solutions. Exercise Using Bennett s result 3 2 ξ 1 4 H(ξ) 2.45, give explicit constants A, B such that the following holds: for any solution x, y Z of x 3 2y 3 = m we have max( x, y ) A m B. Hint. Go through the proof of Theorem and compute a constant c such that x 3 2y 3 c max( x, y ) for all x, y Z. Exercise Prove that the following statements are equivalent: (1) Let α be a real algebraic number of degree n 3. Then for every κ > 2 there is a constant c(α, κ) > 0 such that α ξ c(α, κ)h(ξ) κ for ξ Q. (2) Let α be a real algebraic number of degree n 3. Then for every κ > 2 the inequality α ξ H(ξ) κ in ξ Q has only finitely many solutions. (3) Let α be a real algebraic number of degree n 3. Then for every C > 0, κ > 2 the inequality α ξ CH(ξ) κ in ξ Q has only finitely many solutions.

16 8.3 Exercises 79 (4) Let α 1,..., α m be distinct real algebraic numbers of degree n 3. Then for every κ > 2 there is a constant c > 0 such that m α i ξ ch(ξ) κ for ξ Q. i=1 The following exercise is related to the abc-conjecture, formulated by Masser and Oesterlé in The radical rad(n) of an integer N is the product of the primes dividing N. For instance, rad( ) = abc-conjecture. For every ε > 0 there is a constant C(ε) > 0 such that for all positive integers a, b, c with a + b = c, gcd(a, b, c) = 1 we have c C(ε)rad(abc) 1+ε. The abc-conjecture has many striking consequences. As an example we deduce Fermat s Last Theorem. Let x, y, z be positive coprime integers and n 4. Assume that x n +y n = z n. Apply the abc-conjecture with a = x n, b = y n, c = z n. Then rad(abc) xyz z 3. On the other hand, z n C(ε)(z 3 ) 1+ε. Taking ε < 1 it follows that z and n are bounded. 4 Granville and Langevin proved independently that the abc-conjecture is equivalent to the following: Granville-Langevin conjecture. Let F (X, Y ) Z[X, Y ] be a square-free binary form of degree n 3. Then for every κ > 2 there is a constant C(F, κ) > 0 such that rad(f (x, y)) C(F, κ) max( x, y ) n κ for every x, y Z with gcd(x, y) = 1, F (x, y) 0. Exercise In this exercise you are asked to deduce some consequences of the Granville-Langevin conjecture. (1) Prove that the Granville-Langevin conjecture implies Roth s Theorem. (2) Recall that an integer n 0 is called powerful if every prime in the prime factorization of n occurs with exponent at least 2. In other words, n is powerful if it can be expressed as a 2 b 3 for certain integers a, b with ( a, b ) (1, 1). Let F (X, Y ) Z[X, Y ] be a square-free binary form of degree at least 5 whose coefficients have gcd 1. Assuming the Granville-Langevin conjecture, prove that there

17 80 8 APPROXIMATION OF ALGEBRAIC NUMBERS are only finitely many pairs of integers x, y with gcd(x, y) = 1 such that F (x, y) is powerful. Hint. Estimate rad(f (x, y)) from above in terms of F (x, y) and then in terms of max( x, y ). (3) Assuming the Granville-Langevin conjecture, prove the following. Let f(x) Z[X] be a square-free polynomial of degree n 3. Then for every ε > 0 there is a constant C (f, ε) > 0 such that rad(f(x)) C(f, ε) x n 1 ε for all x Z with f(x) 0. Then deduce Schinzel s conjecture, that there are only finitely many integers x such that f(x) is powerful. Exercise This exercise provides some means to estimate the number of solutions ξ Q of α ξ H(ξ) κ. (1) Let ξ 1, ξ 2 be distinct rational numbers. Prove that ξ 1 ξ 2 (H(ξ 1 )H(ξ 2 )) 1. (2) Let α be a real number, and κ > 2, and consider the inequality α ξ H(ξ) κ in ξ Q with ξ > α. (16) Prove that if ξ 1, ξ 2 are two distinct solutions of (16) with H(ξ 2 ) H(ξ 1 ), then H(ξ 2 ) H(ξ 1 ) κ 1. (So the solutions of (16) grow very quickly.) Hint. Estimate from above ξ 1 ξ 2. (3) Let A 2, c > 1. Prove that the number of solutions ξ of (16) with A H(ξ) < A c is bounded above by 1 + log c log(κ 1). (4) Let α be a real algebraic number of degree n 3 and κ > n + 1. Compute an 2 explicit upper bound for the number of solutions of (16) in terms of the constants C and λ 0 from Theorem