Chapter 6. Approximation of algebraic numbers by rationals. 6.1 Liouville s Theorem and Roth s Theorem

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1 Chapter 6 Approximation of algebraic numbers by rationals Literature: W.M. Schmidt, Diophantine approximation, Lecture Notes in Mathematics 785, Springer Verlag 1980, Chap.II, 1,, Chap. IV, 1 L.J. Mordell, Diophantine Equations, Pure and applied Mathematics series, vol. 30, Academic Press, reprint of the 1971 edition. 6.1 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. Recall that the height H(ξ) of a rational number ξ is given by H(ξ) := max( x, y ), where x, y are coprime integers such that ξ = x/y. In Homework exercise 10b, you were asked to prove the following inequality, which is a variation on a result of Liouville from 1844: Theorem 6.1. Let α be an algebraic number of degree d 1. Then there is an effectively computable number c(α) > 0 such that (6.1) α ξ c(α)h(ξ) d for every ξ Q with ξ α. Here we may take c(α) = den(α) d (1 + α ) 1 d. 87

2 Let α be an algebraic number of degree d. One of the central problems in Diophantine approximation is, to obtain improvements of (6.1) with in the righthand side H(ξ) κ with κ < d instead of H(ξ) d. More precisely, the problem is, whether there exist κ < d and a constant c(α, κ) > 0 depending only on α, κ, such that (6.) ξ α c(α, κ)h(ξ) κ for every ξ Q. Recall that by Dirichlet s Theorem, there exist infinitely many pairs of integers x, y such that x α y y, 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) for infinitely many ξ Q. This shows that there can not exist an inequality of the shape (6.) with κ <. In particular, for rational or quadratic algebraic numbers α, Theorem 6.1 gives the best possible result in terms of the exponent on H(ξ). Now let α be a real algebraic number of degree d 3. In 1909, the Norwegian mathematician A. Thue made an important breakthrough by showing that for every κ > d + 1 there exists a constant c(α, κ) > 0 such that (6.) holds. In 191, C.L. Siegel proved the same for every κ d. In 1949, A.O. Gel fond and independently Freeman Dyson (the famous physicist) improved this to κ > d. Finally, in 1955, K.F. Roth proved the following result, for which he was awarded the Fields medal. Theorem 6. (Roth, 1955). Let α be a real algebraic number of degree d 3. Then for every κ > there exists a constant c(α, κ) > 0 such that (6.) ξ α c(α, κ)h(ξ) κ for every ξ Q. As mentioned before, Roth s Theorem is valid also if α is a rational or quadratic number (with the proviso that ξ α if α Q) but then it is weaker than (6.1). Further, Roth s Theorem holds true also for complex, non-real algebraic numbers α; then we have in fact ξ α Im α for α Q, i.e., (6.) holds even with κ = 0. Exercise 6.1. Let α be a real algebraic number of degree d 3. Prove that the following three assertions are equivalent: (i) for every κ > there is a constant c(α, κ) > 0 with (6.); 88

3 (ii) for every κ >, the inequality (6.3) ξ α H(ξ) κ in ξ Q has only finitely many solutions; (iii) for every κ >, C > 0, the inequality (6.4) ξ α CH(ξ) κ in ξ Q has only finitely many solutions. It should be noted that Theorem 6.1 is effective, i.e., the constant c(α) in (6.1) can be computed. 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 (6.), but one can not compute such a constant. Equivalently, the methods of proof of Thue,..., Roth show that the inequalities (6.3), (6.4) 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. A binary form F (X, Y ) Z[X, Y ] is called square-free if it is not divisible in C[X, Y ] by (αx + βy ) for some α, β C, not both 0. Theorem 6.3. Let F (X, Y ) Z[X, Y ] be a square-free binary form of degree d 3. Then for every κ > there is a constant c(f, κ) > 0 such that for every pair of integers (x, y) with F (x, y) 0 we have (6.5) F (x, y) c(f, κ) max( x, y ) d κ. If F is a binary form of degree d the theorem holds true as well but then it is trivial since F (x, y) Z, hence 1. 89

4 Proof. We prove the inequality only for pairs of integers (x, y) with y x. Then the inequality can be deduced for pairs (x, y) with x > y by interchanging x, y and repeating the argument below. Next, we restrict to the case that y x and F is not divisible by Y. If F is divisible by Y we have F = Y F 1 where F 1 Z[X, Y ] is a square-free binary form of degree d 1 which is not divisible by Y. Then if the inequality holds for F 1 and with d 1 instead of d, it follows automatically for F. So assume that F is a square-free binary form of degree d that is not divisible by Y. Then F (X, Y ) = a 0 X d + a 1 X d 1 Y + + a d Y d with a 0 0, and so, F (X, Y ) = a 0 (X α 1 Y ) (X α d Y ) with α 1,..., α d distinct. Let (x, y) be a pair of integers with F (x, y) 0 and y x. Then y 0. Let ξ := x/y. Notice that y = max( x, y ) H(ξ) (with equality if gcd(x, y) = 1). Let i be the index with ξ α i = min j=1,...,d ξ α j. Let κ >. Theorem 6. says that if α i is real algebraic then there is a constant c(α i, κ) > 0 such that ξ α i c(α i, κ)h(ξ) κ c(α i, κ) max( x, y ) κ ; as has been observed above this is true as well if α i is not real. For j i we have α i α j α i ξ + ξ α j ξ α j, implying ξ α j 1 α i α j. Hence d d F (x, y) = y d a 0 ξ α j = max( x, y ) d a 0 ξ α j j=1 j=1 c(α i, κ) a 0 ( 1 α i α j ) max( x, y ) d κ. j i We deduce Thue s Theorem. 90

5 Corollary 6.4. 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, by showing that F is divisible in Z[X, Y ] by a square-free binary form F Z[X, Y ] of degree 3. We can factor the polynomial F (X, 1) as cg 1 (X) k1 g t (X) kt where c is a nonzero 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) := g 1 (X) g t (X). Then f Z[X], and deg f =: d 3 since F (X, 1) has at least three zeros in C. We have F (X, 1) = f (X)g(X) with g Z[X]. Put F (X, Y ) = Y d f(x/y ) and G(X, Y ) := Y deg F d g(x/y ). Then F = F G with G Z[X, Y ]. The polynomial f has degree d 3 and d distinct zeros, and it divides F (X, 1) in Z[X]. Hence F is square-free, F has degree d 3 and F divides F in Z[X, Y ]. Let x, y be integers with F (x, y) = m. Then F (x, y) divides m. Take κ with < κ < d. Then by Theorem 6.3, implying that x, y are bounded. m F (x, y) c(f, κ) max( x, y ) d κ, The total degree of a polynomial G = i a ix i 1 1 X ir r, notation totdeg G, is the maximum of all quantities i i r, taken over all tuples i = (i 1,..., i r ) with a i 0. For instance, 3X 7 1X 5 X 3 X 1 X 1 X 3 has total degree 15. Exercise 6.. Let F Z[X, Y ] be a square-free binary form of degree d 4, and let G Z[X, Y ] be a polynomial of total degree d 3. Prove that there are only finitely many pairs (x, y) Z with F (x, y) = G(x, y) and F (x, y) 0. 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 91

6 where α is algebraic of degree d 3 and κ < d (but very close to d) 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 6.5 (Fel dman, 1971). Let α be a real algebraic number of degree d 3. Then there exist effectively computable numbers c 1 (α), c (α) > 0 depending on α such that (6.6) ξ α c 1 (α)h(ξ) d+c (α) for ξ Q. The proof is to complicated to be given here, but we can give a brief sketch. The hard core is the following effective result on Thue equations, given by Fel dman. The proof is by making explicit the arguments of the proof given in the previous chapter. Lemma 6.6. Let F Z[X, Y ] be a binary form such that F (X, 1) has at least three zeros in C. Then there are effectively computable numbers A, B depending only on F, such that for each non-zero integer m and each solution (x, y) Z of F (x, y) = m we have max( x, y ) A m B. Proof of Theorem 6.5 (assuming Lemma 6.6). Let f(x) = a 0 X d + a 1 X d a d = a 0 (X α (1) ) (X α (d) ) be the primitive minimal polynomial of α and F (X, Y ) := Y d f(x/y ). Then F (X, Y ) is a binary form in Z[X, Y ]. Let ξ = x/y with x, y Z coprime. Then f(ξ) 0 and this implies m := F (x, y) 0. By Lemma 6.6 we have (6.7) F (x, y) = m ( max( x, y )/A ) 1/B = ( H(ξ)/A ) 1/B, where A, B are effectively computable positive numbers depending on F, hence α. It remains to estimate from below ξ α in terms of F (x, y) and H(ξ). Assume α (1) = α. Notice that F (x, y) = a 0 d i=1 (x α(i) y). We estimate the factors as follows: x α (1) y = ξ α y ξ α H(ξ), x α (i) y x + α (i) y (1 + α (i) )H(ξ) (i =,..., d). 9

7 Thus, d F (x, y) = a 0 x α (i) y i=1 d a 0 ξ α H(ξ) d (1 + α (i) ) = ξ α C(α)H(ξ) d, i= say, where C(α) is effectively computable. Hence Combined with (6.7) this gives ξ α F (x, y) C(α) 1 H(ξ) d. ξ α A 1/B C(α) 1 H(ξ) d+1/b = c 1 (α)h(ξ) d+c (α) with c 1 (α) = A 1/B C(α) 1, c (α) = 1/B. The quantities c 1 (α), c (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 (6.6) holds with ( c 1 (α) = exp 10 7d d 16d H d 1( log(edh) ) ) d 1, c (α) = (10 7d d 16d H d 1( log(edh) ) ) d 1 1 where H = H(α) is the height of α. One can obtain better results for certain special classes of algebraic numbers using other methods. 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 (6.8) ξ H(ξ).45 for ξ Q. Exercise 6.3. Using (6.8), compute explicit constants A, B such that the following holds: for any solution x, y Z of x 3 y 3 = m we have max( x, y ) A m B. Hint. Go through the proof of Theorem 6.3 and compute a constant c such that x 3 y 3 c max( x, y ) 3.45 for all x, y Z. Notice that we may have x > y. 93

8 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 6.7 (Bombieri, Schmidt, 1986). Let F (X, Y ) be a binary form in Z[X, Y ] such that F (X, 1) has precisely d 3 distinct roots. Then the equation F (x, y) = 1 in x, y Z has at most c d solutions where c is a positive constant not depending on d or F. The importance of the result is that the bound is uniform, i.e. for all binary forms F we get the upper bound cd. It is possible to compute c explicitly. Bombieri and Schmidt showed that for sufficiently large d, c can be taken equal to 430. Probably the constant c can be improved, but the dependence on d is optimal. For instance, let F (X, Y ) = (X a 1 Y ) (X a d Y ) + Y d, where a 1,..., a d are distinct integers. Then the equation F (x, y) = 1 has the d solutions (a 1, 1),..., (a d, 1). M. Bennett proved the following remarkable result: Theorem 6.8 (Bennett, 00). Let d be an integer with d 3 and let a, b be positive integers. Then the equation ax d by d = 1 has at most one solution in positive integers x, y. For instance, the equation (a + 1)x d ay d = 1 has (1, 1) as its only solution in positive integers. In his proof, Bennett uses various techniques (good lower bounds for linear forms in two logarithms, Diophantine approximation techniques based on so-called hypergeometric functions, and heavy computations). We finish with an exercise related to the abc-conjecture, formulated by Masser and Oesterlé in The radical rad(n) of a non-zero 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. 94

9 The abc-conjecture has many striking consequences. As an example we deduce a weaker version of 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 Exercise 6.4. Assuming the abc-conjecture, prove that the equation x m + y n = z p has only finitely many solutions in positive integers x, y, z, m, n, p with x > 1, y > 1, z > 1, gcd(x, y, z) = 1 and 1 m + 1 n + 1 p < 1. 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 d 3. Then for every κ > there is a constant C(F, κ) > 0 such that rad(f (x, y)) C(F, κ) max( x, y ) d κ for every x, y Z with gcd(x, y) = 1, F (x, y) 0. Exercise 6.5. (i) Prove that the Granville-Langevin conjecture implies the abcconjecture (the converse is also true but this is much more difficult to prove). (ii) Prove that the Granville-Langevin conjecture implies Roth s Theorem. (iii) An integer n 0 is called powerful if every prime in the prime factorization of n occurs with exponent at least. In other words, n is powerful if it can be expressed as ±a b 3 for certain positive integers a, b not both equal to 1. Let F (X, Y ) Z[X, Y ] be a square-free binary form of degree at least 5. Assuming the Granville-Langevin conjecture, prove that there are only finitely many pairs of integers x, y with gcd(x, y) = 1 such that F (x, y) is powerful. (iv) Assuming the Granville-Langevin conjecture, prove the following. Let f(x) Z[X] be a polynomial of degree d with d distinct zeros in C. Then for every ε > 0 there is a constant C (f, ε) > 0 such that rad(f(x)) C(f, ε) x d 1 ε for all x Z with f(x) 0. 95

10 Hint. Construct from f a binary form F of degree d + 1. (v) Deduce the following conjecture of Schinzel: if f is any square-free polynomial in Z[X] of degree 3, then there are only finitely many integers x such that f(x) is powerful. 6. Thue s approximation theorem We intend to prove the following result of Thue: Theorem 6.9. Let α be a real algebraic number of degree d 3 and κ > d + 1. Then the inequality (6.9) ξ α H(ξ) κ in ξ Q has only finitely many solutions. Our basic tool will be Siegel s Lemma, which we recall here. We consider systems of linear equations (6.10) a 11 x a 1N x N = 0. a M1 x a MN x N = 0 with coefficients a ij from the ring of integers O K of a number field K. Proposition Let K be an algebraic number field of degree d. Then there is a constant C(K) > 0 for which the following holds. Let M, N be integers with N > dm > 0, let A be a real 1, and suppose that a ij O K, a ij A for i = 1,..., M, j = 1,..., N. Then (6.10) has a solution x = (x 1,..., x N ) Z N \ {0} such that (6.11) max 1 i N x i (C(K)NA) dm/(n dm). Proof. This is Corollary 4.4 from Chapter 4. 96

11 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 not difficult to check that (6.1) (6.13) P (α) P max(1, α ) deg P for P C[X], α C, P + Q P + Q, P Q P Q for P, Q C[X]. From these properties it can be deduced that if P C[X], α C, then for the polynomial P (X) := P (X + α) we have (6.14) P P (1 + α ) deg P. Exercise 6.6. Prove (6.1) (6.14). The k-th divided derivative of a polynomial P C[X] is defined by P ((k)) := P (k) /k!. Thus, if P = D i=0 p ix i, then P ((k)) = D i=0 ( ) i p i X i k k with ( ) a := 0 if b > a. b 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 i deg P, we have (6.15) P ((k)) deg P P. Lastly, we have the product rule (6.16) (P Q) ((k)) = k P ((k j)) Q ((j)) for P, Q C[X]. j=0 A brief outline of the proof of Theorem 6.9. We give a brief outline of the proof. More details and explanation are given later. We follow the usual procedure to assume that (6.9) has infinitely many solutions, and to construct a non-zero integer of absolute value < 1. The first step of the proof is to take, for any positive integer r, non-zero polynomials P r, Q r Z[X] of degree as small as possible such that P r αq r is divisible by (X α) r. Using Siegel s Lemma, one can prove the existence of such P r, Q r of degree at most m := [( 1 d + ε)r] for any ε > 0, where [x] denotes the largest integer x. 97

12 To see this, view the coefficients of P r, Q r as a system of m + unknowns. The condition P r αq r divisible by (X α) r is equivalent to P r ((k)) (α) αq ((k)) r (α) = 0 for k = 0,..., r 1, and by expanding this, we get a system of r linear equations with coefficients in K := Q(α) in the m + unknown coefficients of P r, Q r. Now [K : Q] = d and m + > dr, hence by Proposition 6.10, this system has a non-trivial solution in integers. In the second step, we take two solutions of (6.9), say ξ 1 = x 1 /y 1, ξ = x /y with x i, y i Z, gcd(x i, y i ) = 1, y i > 0 for i = 1,, and consider the number A r := y [( 1 +ε)dr] 1 y (P r (ξ 1 ) ξ Q r (ξ 1 )). This number is clearly an integer. We want to show that we can choose the solutions ξ 1, ξ such that A r 0 and A r < 1, thus obtaining a contradiction. To prove A r < 1, we write and obtain P r αq r = V r (X α) r with V r C[X] A r = y [( 1 ) +ε)dr] 1 y (V r (ξ 1 )(ξ 1 α) r (ξ α)q r (ξ 1 ). To keep our discussion informal, we ignore ε, and just pretend that V r (ξ 1 ) 1, Q r (ξ 1 ) 1. Define λ := log H(ξ )/ log H(ξ 1 ), and write H 1 := H(ξ 1 ). Notice that we can make both H 1 and λ as large as we like, since we assumed that (6.9) has infinitely many solutions. Then y 1 H 1, y H λ 1, and thus, A r H (dr/)+λ 1 (H1 κr + H1 λκ ) = H λ+r( κ+d/) 1 + H (1 κ)λ+dr/ 1. We want to make both exponents negative, to make A r < 1. This amounts to r > λ κ 1 (κ 1)λ r < 1 d, d. Now our assumption κ > 1 d + 1 is equivalent to 0 < 1 κ 1d < κ d

13 and so the upper bound for r increases faster in terms of λ than the lower bound. For λ sufficiently large, the upper and lower bound differ by at least 1 and so there is a positive integer r between these bounds. That is, if we choose ξ 1, ξ such that H 1 and λ := log H(ξ )/ log H(ξ 1 ) are sufficiently large, we can choose r such that both exponents on H 1 are < 0 and hence A r < 1. Of course, we have to take into account ε, and estimate V r (ξ 1 ), Q r (ξ 1 ), but with some modifications in the above argument, we can still obtain A r < 1 for some r. What remains is to show that A r 0. Unfortunately, it is not all clear how to do this. Instead, we prove that for any two distinct solutions ξ 1, ξ of (6.9) and any positive integer r, there is a not too large value k such that P r ((k)) (ξ 1 ) ξ Q ((k)) r (ξ 1 ). Then A rk := y [( 1 +ε)dr] ( 1 y P ((k)) (ξ 1 ) ξ Q ((k)) r (ξ 1 ) ) r is a non-zero integer. Similarly as above we prove that there are solutions ξ 1, ξ of (6.9) and a positive integer r such that A rk < 1 and obtain a contradiction. We now give the precise proof of Theorem 6.9. We need a parameter ε with 0 < ε < 1. Later, ε will be chosen depending on d, κ. Further, r will be a positive integer, to be chosen later. We start with the construction of the polynomials P r, Q r. Lemma For every positive integer r there exist polynomials P r, Q r Z[X] of degree at most [( 1 + ε)dr], not both equal to 0, with the following properties: (6.17) (6.18) P r αq r is divisible by (X α) r, P r C r 1, Q r C r 1, where C 1 is an effectively computable number, depending only on α, ε. Proof. Let K = Q(α). Put m := [( 1 + ε)dr]. Write P r = m p i X i, Q r = i=0 m q i X i, i=0 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,..., r 1). 99

14 Let b be a denominator of α, i.e., b Z >0, bα O K. expression and multiplying with b m we obtain m i=0 ( ) i b m α i p i k m i=0 By expanding the above ( ) i b m α i+1 q i = 0 (k = 0,..., r 1), k which is a system of r linear equations in m + > (1 + ε)dr unknowns with coefficients in O K. Thus, the number of unknowns is larger than [K : Q] times the number of equations, and the condition of Proposition 6.10 is satisfied. As a consequence, the above system has a non-trivial solution (p 0,..., p m, q 0,..., q m ) Z m+ such that max(max i p i, max i where (with 0 i m, 0 k r), dr q i ) (C(K)(m + )A) m+ dr (C(K)( + ε)dra) 1/ε, ( A = max max i,k ( ) i b m α i, max k k,i ( i ) )b m α i+1 k m b m max(1, α ) m+1 ( b max(1, α ) ) (+ε)dr. So using C(K)( + ε) (C(K)( + ε)) (+ε)dr, dr (+ε)dr, we see that Lemma 6.11 holds with C 1 = ( 4C(K)( + ε)b max(1, α ) ) d(1+ε 1 ). We now take two solutions ξ 1, ξ of (6.9) and show that for every r there is a not too large k such that P r ((k)) (ξ 1 ) ξ Q ((k)) r (ξ 1 ) 0. We start with a simple lemma. Lemma 6.1. Let F Q[X], β an algebraic number such that (X β) m divides F, and f Q[X] the minimal polynomial of β. Then f m divides F. Proof. By the unique factorization in Q[X], we can factor F in Q[X] as cf k 1 1 fs ks where c Q, f 1,..., f s are distinct monic irreducible polynomials in Q[X] and k 1,..., k s > 0. Since two monic irreducible polynomials cannot have a common zero, β is a zero of precisely one of f 1,..., f s, say f 1. Then f 1 = f and k 1 m since (X β) m divides F and f has no double zeros. Lemma Let ξ 1, ξ be two rational numbers, and r a positive integer. Then there is k d(εr + 1) such that P r ((k)) (ξ 1 ) ξ Q ((k)) r (ξ 1 ). 100

15 Proof. The proof rests upon an analysis of the polynomial F := P r Q r P rq r. We first show that F is not identically 0. Assume the contrary. One of P r, Q r, say Q r, is not identically 0. Then (P r /Q r ) = 0 hence P r /Q r is identically equal to some constant c Q. But then, Q r = (c α) 1( ) P r αq r is divisible by (X α) r. By Lemma 6.1, Q r is divisible by f r, where f is the minimal polynomial of α. But this is impossible, since by our assumption ε < 1 we have r deg f = rd > ( 1 + ε)dr deg Q r. We now prove our lemma. Assume there exists an integer t 1 such that P ((k)) r (ξ 1 ) = ξ Q ((k)) r (ξ 1 ) for k = 0,..., t (if not, we are done). By eliminating ξ we obtain P ((k)) r (ξ 1 )Q ((l)) r (ξ 1 ) P r ((l)) (ξ 1 )Q ((k)) r (ξ 1 ) = 0 for k, l t. For each k 0, F ((k)) is a linear combination of P r ((l)) Q ((m)) r P r ((m)) Q ((l)) r, 0 l, m k + 1. Hence F ((k)) (ξ 1 ) = 0 for k t 1, and therefore, F is divisible by (X ξ 1 ) t. By construction, P r αq r is divisible by (X α) r, hence P r αq r is divisible by (X α) r 1. So F = P r (Q r αp r) P r(q r αp r ) is divisible by (X α) r 1. But F Q[X] hence by Lemma 6.1 it is divisible by f r 1. So F is in fact divisible by (X ξ 1 ) t f r 1. Since deg F max(deg P r + deg Q r, deg P r + deg Q r ) (1 + ε)dr 1, deg f = d, it follows that This proves our lemma. t (1 + ε)dr 1 d(r 1) = d(εr + 1) 1. Take two solutions ξ 1, ξ of (6.9). Write ξ i = x i /y i with x i, y i Z, gcd(x i, y i ) = 1 and y i > 0 for i = 1,. For integers r > 0, k 0 consider the number A rk := y [( 1 ( ) +ε)dr] 1 y P ((k)) (ξ 1 ) ξ Q ((k)) r (ξ 1 ). r This is clearly an integer, and by Lemma 6.13 there is k < d(εr + 1) such that A rk 0. We proceed to estimate A rk. 101

16 Lemma Let ξ 1, ξ be two solutions of (6.9) with H(ξ ) > H(ξ 1 ) and put Then λ := log H(ξ ) log H(ξ 1 ). ( A rk C r H(ξ1 ) u + H(ξ 1 ) v) where C is an effectively computable number depending only on α, κ, ε, and where u = ( 1 d κ + (κ + 1)εd) r + dκ + λ, v = ( 1 + ε)dr (κ 1)λ. Proof. We write H 1 := H(ξ 1 ), H := H(ξ ). Thus, H = H λ 1. The polynomial P ((k)) r P ((k)) r 1 y[( +ε)dr] αq ((k)) r αq ((k)) r is divisible by (X α) r k. Hence = V (X α) r k with V C[X]. This gives ( A rk = 1 y V (ξ 1 )(ξ 1 α) r k (ξ α)q {k} r (ξ 1 )) max( V (ξ 1 ), Q ((k)) r (ξ 1 ) ) ( y ( 1 +ε)dr 1 y ξ 1 α r k + y ( 1 ) +ε)dr 1 y ξ α. We take for granted for the moment that V (ξ 1 ) C r, Q(ξ 1 ) C r. Notice that y ( 1 +ε)dr 1 y H ( 1 +ε)dr 1 H = H ( 1 +ε)dr+λ 1, ξ α H κ = H κλ 1, ξ 1 α r k (H κ 1 ) r k (H κ 1 ) r dεr d = H κr(1 dε)+dκ 1. A combination of these estimates gives our lemma. We finish with estimating Q ((k)) r (ξ 1 ), V (ξ 1 ). Here, we use (6.1) (6.15). Define P (X) := P r ((k)) (X + α), Q(X) := Q ((k)) r (X + α), Ṽ (X) := V (X + α) and ξ := ξ 1 α. Then P P ((k)) (1 + α ) ( 1 +ε)dr ( (1 + α ) ) ( 1 +ε)dr P ( (1 + α ) ) ( 1 +ε)dr C r 1 10

17 and likewise Q ( (1 + α ) ) ( 1 +ε)dr C r 1. Since P α Q = X r k Ṽ we have Ṽ P + α Q (1 + α ) ( (1 + α ) ) ( 1 +ε)dr C r 1. Together with ξ = ξ α 1, this leads to Q ((k)) r (ξ 1 ) = Q( ξ) Q C r, V (ξ 1 ) = Ṽ ( ξ) Ṽ Cr, with C := ( 1 +ε)d (1 + α ) 1+( 1 +ε)d C 1. This completes our proof. Proof of Theorem 6.9. We take solutions ξ 1, ξ of (6.9) with H(ξ ) > H(ξ 1 ). Then by Lemma 6.13, for every integer r 1, there is k d(εr + 1) such that A rk is a non-zero integer. We want to show that there are r 1 and ξ 1, ξ, such that A rk < 1 for all k d(εr + 1) and obtain a contradiction. To this end, we want to choose r in such a way that u εdr, v εdr. Then A rk C r H(ξ 1 ) εdr < 1 if we assume in addition that H(ξ 1 ) > (C ) 1/dε. The conditions u = ( 1 d κ + (κ + 1)εd) r + dκ + λ εdr, v = ( 1 + ε)dr (κ 1)λ εdr can be reworked into a lower and an upper bound for r in terms of λ, in fact, these conditions are equivalent to (6.19) dκ + λ (κ 1)λ κ 1 r d (κ + )εd ( 1 + ε)d. We now have to use our condition κ > 1 d + 1. This is equivalent to 0 < 1 κ 1d < κ 1 1 d. Hence there is ε with 0 < ε < 1/d such that κ 1 ( 1 + ε)d > 1 κ 1 > 0; d (κ + )εd we can choose such ε depending only on d, κ. With this choice of ε, both the upper and lower bound in (6.19) are increasing functions of λ and the upper bound 103

18 increases faster than the lower bound. So there is λ 0 such that for every λ > λ 0, the upper and lower bound in (6.19) differ by at least 1 and there is an integer r 1 with (6.19). We finish the proof. Assume that (6.9) has infinitely many solutions. So (6.9) has solutions ξ 1, ξ such that (6.0) H(ξ 1 ) > (C ) 1/dε, log H(ξ ) log H(ξ 1 ) =: λ > λ 0. Then there is an integer r 1 with (6.19), hence with u εdr, v εdr. Together with Lemma 6.14 this implies A rk C r H 1 (ξ 1 ) εdr < 1 for k d(εr + 1). On the other hand, by Lemma 6.13, there is k d(εr + 1) such that A rk is a nonzero integer. This gives the contradiction we want. So (6.9) cannot have infinitely many solutions. Remark. To obtain a contradiction, we did not need the assumption that (6.9) has infinitely many solutions, but merely that there are solutions ξ 1, ξ of (6.9) with (6.0). In other words, solutions ξ 1, ξ of (6.9) with (6.0) cannot exist. The constants C and λ 0 are effectively computable. So in fact we can prove the following sharpening of Theorem 6.9: Theorem Let α be an algebraic number of degree d and κ > 1 d+1. There are effectively computable constants C, λ 0 depending on α, κ, such that if ξ 1 is a solution of (6.9) ξ α H(ξ) κ in ξ Q with H(ξ 1 ) C, then for any other solution ξ of (6.9) we have H(ξ) H(ξ 1 ) λ 0. It should be noted that Theorem 6.15 would give an effective proof of Thue s Theorem in case we were extremely lucky and knew a solution ξ 1 of (6.9) with H(ξ 1 ) C. However, to find such a solution seems quite hopeless, since the constant C is very large. It is very likely that such a solution ξ 1 does not even exist. However, there are variations on Thue s method, which work only for special algebraic numbers α of the shape d a with a Q, where the constant C is much smaller and where a solution ξ 1 of (6.9) with H(ξ 1 ) C is known. For such α one can derive very strong 104

19 effective approximation results, for instance Bennett s estimate (6.8) mentioned in Section 6.1. On the other hand Theorem 6.15 can be used to estimate the number of solutions of (6.9). This is worked out in the exercise below. Exercise 6.7. (i) Let ξ 1, ξ be distinct rational numbers. Prove that ξ 1 ξ ( H(ξ 1 )H(ξ ) ) 1. (ii) Let α be a real number, and κ >, and consider the inequality (6.1) ξ α H(ξ) κ in ξ Q with ξ > α. Prove that if ξ 1, ξ are two distinct solutions of (6.1) with H(ξ ) H(ξ 1 ), then H(ξ ) H(ξ 1 ) κ 1. (So there are large gaps between the solutions of (6.1); we call such an inequality a gap principle.) Hint. Estimate from above ξ 1 ξ. (iii) Let A, c > 1. Prove that the number of solutions ξ of (6.1) with A H(ξ) < A c is bounded above by 1 + log c log(κ 1). (iv) Let α be a real algebraic number of degree d 3 and κ > d + 1. Compute an explicit upper bound for the number of solutions of (6.1) in terms of the constants C and λ 0 from Theorem