n= (k+1)! = 2 10 n! <

Transcription

1 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Notes taken by D. Wolczuk A complex number α is said to be algebraic if it is the root of a non-zero polynomial with integer coefficients. A complex number which is not algebraic is said to be transcendental. Recall: The degree of an algebraic numbers is the degree of its minimal polynomial. In 844, Liouville proved, Theorem. Let α be an algebraic number of degree d >. There is a positive number C which is effectively computable in terms of α, such that α p q > Cα), q d for p, q Z, q > 0. Suppose that fx) = a d x d + + a x + a 0 Z[x] is the minimal polynomial of α. Note that if α is not real, then we can take Cα) to be the absolute value of the imaginary part of α. Suppose α R. Then for p/q Q, fp/q) /q d since fp/q) 0 since f is the minimal polynomial of α and d >. By the mean value theorem, a real number θ, with θ between α and p/q such that /q d fp/q) = fα) fp/q) = α p/q f θ). ) but f x) = da d x d + + a. Observe that if α p/q >, we can take Cα) to be any positive number less than. Thus we may assume that α p/q. Therefore The result now follows from ). f θ) d a d α + ) d + + a ) = Cα)). Liouville proved the existence of transcendental numbers such as θ = 0 n! n= by using Thrm. For example, put p k /q k = k n= 0 n! thus q k = 0 k! and p k = 0 k! kn= 0 n!. Obeserve that θ p k /q k = n=k+ 0 n! < 0 k+)! = qk k+. ) Suppose that θ was algebraic of degree d. Note that d >, since θ is not a rational number. Then by Thrm, Cα) > 0 s.t. α p k /q k > Cα)/qk d. But this and ) imply qk+ d k < /Cα) but qk k+ d as k so α is not algebraic. In 873, Hermite proved that e is transcendental.

2 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART In 874, Cantor proved that the transcendental numbers are dense in R by making use of the fact that the algebraic numbers are countable. In 88, Lindemann proved that π is transcendental. The Hermite-Lindemann theorem states that if β C, β 0 then one at least of {β, e β } is transcendental. Consider {πi, e πi } = {πi, } since i is algebraic, it follows that π is transcendental. Lindemann stated and Weierstrass proved in 885: If β,..., β n are algebraic numbers which are linearly independent over Q then e β,..., e βn are algebraically independent. In 900, Hilbert proposed as the 7th of his problems: If α is algebraic and α 0, and β is algebraic and irrational, prove that α β is transcendental. This was proved by Gelfond and Schneider independentaly in 934. In 967, Baker proved that if α,..., α n are algebraic numbers different from 0 and and β,..., β n are algebraic numbers for which, β,..., β n are linearly independent over Q then α β α βn n is transcendental. He also proved that e β 0 α β αn βn is transcendental for all nonzero algebraic numbers β 0,..., β n, α,..., α n. Quantitative Results? One can ask for a measure of how small the quantity β 0 + β log α + + β n log α n log α n+ when α n+ is an algebraic number. For our applications we are interested in the degenerate case when β 0 = 0, and β i s are integers. We still get bounds from Baker s argument. Put Λ = b log α + + b n log α n 3) where b i Z, i =,..., n. One can prove that Λ = 0 or Λ is bounded away from 0 in terms of the size of the b i s, n, the degree of the α i s and the heights of the α i s. For any algebraic number α, we define the height of α, denoted by Hα), by Hα) = max{ a d, a d,, a 0 } where fx) = a d x d + a x + a 0 is the minimal polyomial of α. Hα) is known as the naive height. Aim: If Λ 0, then Λ is not too small in terms of b,..., b n, α,..., α n. Suppose, in 3) that the logs are always the principal branch. Put d = [Qα,..., α n ) : Q]. Suppose that A i = maxhα i ), e) for i =,..., n and that B = max b,..., b n, e). Theorem. 993, Baker and Wustholz) If Λ 0, then Λ > exp 6nd) n+4 log A log A n log B). Proposition 3. Suppse that α is an algebraic number, α 0 with minimal polynomial fx) = a d x d + + a 0 then α < Hα) +. a d If α, the result is immediate. So suppose that α >.

3 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 3 We have fα) = 0 and so Hence a d α = a d a d α a 0 α d+. a d α a d + a d α + + a 0 α d+ ) α < H a d H + α + + α d+ ) ) < H α Remark. Note that since α has minimal polynomial x d f/x), we see from Prop 3 that ) H α > a 0 +. Let b,..., b n Z with absolute value at most B and let α,..., α n be nonzero algebraic numbers with heights at most A. Proposition 4. If Λ 0 then Λ > 3A) dnb Let a j denote the leading coefficient in the minimal polynomial of α j when b j 0, and let a j denote the leading coefficient of the minimal polynomial of α j ) when b j < 0. Then we put w = a b a bn n αb αbn n ) = a b a bn n α ǫ b α ǫn bn n ) where ǫ i = b i / b i for i =,..., n. Notice that w is an algebraic integer of degree at most d as a b i i α ǫ i b i i is an algebraic integer. This is because a d α is a root of y d + a d y d + + a 0 a d d. Let σ be an embedding of Qα,..., α n ) in C which fixes Q. Each conjugate σw) of w is of the form σw) = a b a n bn σαǫ ) b σαn ǫn ) bn ). By Prop 3, a i σα ǫ i i ) < A and so σw) < A) Bn. ) If w = 0 and Λ 0 then Λ is a multiple of πi and the result holds. Suppose w 0, then N k/q w). Thus, by ), w σ id σw) A)nB ) d+. From the inequality e z z e z for all z C, and on setting z = Λ, we see that α b αbn n Λ e Λ. If Λ / we re done, so we may assume Λ < /, hence e Λ <. Thus α b αn bn Λ. Recall that a b a bn n αb αbn n ) A)nB ) d+

4 4 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART and so since nb. Λ a b a n bn A) nb ) d > d 3A) ndb A) nbd We can rewrite Prop 4 as if Λ 0 then Λ > exp ndlog 3A)B). Suppose further that A e then 3A < A 3 and so Λ > exp 3ndlog A)B). Compare this with Thrm. Notice that Thrm gives a better lower bound for the modulus of Λ, Λ, when 3ndlog A)B > 6nd) n+4 log A log A n log B. WLOG, we may assume that A = A n, so α n has largest height. Thus Thrm improves on Prop 4 when B log B > 6nd)n+4 log A log A n. We get nontrivial information from Thrm when the b i s are large relative to n, d, A,..., A n. Basically, Thrm tells us that products of large powers of algebraic numbers can t be too close together. Simpler situation: Suppse a,..., a n Q, nonzero and let b,..., b n Z, nonzero. Put B j = b j, B = max j b j, A j = maxha j ), ) and Λ = b log a + + b n log a n. Conjecture. Lang + Waldschmidt) Let ǫ > 0. Cǫ) > 0 such that if Λ 0, then Notice if we take ǫ = / then Λ > Cǫ)) n B B B n A A n )+ǫ. Cǫ)) n Λ > B n+ǫ) A > exp 3nlog n+ǫ) Cǫ) ) + log B + log A). The rationale behind the conjecture: Let S be the set of linear combinations of the log a i s of the form b log a + + b n log a n where b j B j and Ha j ) A j for j =,..., n. S has cardinality at most B + ) B n + )A + ) A n + ). The numbers in S are contained in the interval [ nb log A, nb log A]. If the numbers are uniformly distributed in the interval we would expect that the distance to 0 from the smallest nonzero element of S in absolute value is about nb log A B + ) B n + )A + ) A n + ). This motivates their conjecture. Suppose that a,..., a n are positive integers of at most A and suppose that b,..., b n are postive integers of size at most B. Fix a,..., a n and suppose that log a,..., log a n are linearly independent over Q. Then the set T of linear combinations b log a + + b n log a n has cardinality at least B ) n due to linear independence). They all lie in the interval [D, Bn log A]. Thus there is a nonzero difference of two elements of R of the form b i log a + b n log a n where b j B for j =,..., n of size at most nb log A B ) n.

5 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 5 Thus, in general, the term log B which occurs in the lower bound for Λ as in Thrm can t be improved. Similarly, it can be shown that we need a factor of log A also. Suppose that Λ = b log α + + b n log α n = 0 with α,..., α n algebraic numbers and b,..., b n nonzero integers. Suppose that no subset of n elements from {log α,..., log α n } is Q-lineraly dependent. Then, up to ±, there is a unique nonzero n-tuple of coprime integers k,..., k n ) such that k log α k n log α n = 0. Claim: We can bound the k i s from above in terms of n, d and A,..., A n A i = Hα i )). Let α be an algebraic number with minimal polynomial fx) = a d x d + a x + a 0 so Hα) = max i a i ). Suppose that over C, fx) = a d x α ) x α d ) so wlog α = α. We put Mα) = a d d i= max, α i ). Mα) is known as the Mahler measure of α and it is a more natural height function for α than Hα). Jensen s Forumla. Let f be an analytic function in region containing the closed ball centered at the origin of radius r > 0. Suppose that α,..., α n are the zeros of f in the ball repeated with multiplicity. If f0) 0 then n log f0) = log r/α i + π log fre iθ ) dθ. π 0 i= Proposition 5. Landau) Let α be an algebraic number of degree d. Then Mα) d + ) / Hα). Let fz) = a d z d + +a z+a 0 be the minimal polynomial of α over Q. Note that by Parseval s equality π fe iθ ) dθ = a 0 π + + a d. 0 On the other hand, fe iθ ) = a d e iθ α e iθ α d. Apply Jensen s Formula with r = and let α,..., α n be the roots of f of modulus at most. Then log a 0 = log α α n + π log fe iθ ) dθ π 0 so a 0 log α α n = π log fe iθ ) dθ. π 0 Note that fz) = a d z α ) z α d ) so a d α α d = a 0. Thus Thus so hence a 0 α α n = a d α n+ α d = a d π d max, α i ) = Mα). i= log Mα) = log fe iθ ) dθ π 0 π ) Mα) = exp log fe iθ ) dθ π 0 π ) Mα) = exp log fe iθ ) dθ. π 0

6 6 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART By the arithmetic geometric mean inequality for functions, Thrm 84 of Inequalities by Polya, Hardy and Littlewood, π ) exp log fe iθ ) dθ π fe iθ ) dθ π 0 π 0 hence Mα) a a d d + )Hα), as required. Theorem 6. Let α,..., α n be nonzero algebraic numbers and suppose that log α,..., log α n are linearly dependent over Q. Suppose that A j = maxmα j ), e log α j d, e) for j =,..., n where d = [Qα,..., α n ) : Q]. Then there exist integers t,..., t n not all zero for which t log α + + t n log α n = 0 with for i =,..., n. t i n )d3 ) n log A log A n log A i For the proof of Thrm 6, we need some ideas from the geometry of numbers. A set S in R n is said to symmetric about the orgin 0 = 0,..., 0) if whenever x S, then x S. S is said to be convex if whenever x, y S and λ is a real number with 0 λ then λx + λ)y S. The volume of a set S in R n is the Riemann integral of the characteristic function of the set, when it is integrable. It can be shown that every bounded convex set in R n has a volume. An integer point x = x,..., x n ) in R n is a vector with x i Z for i =,..., n. Theorem 7. Minkowski, 896) Let A be a set in R n which is convex, bounded, symmetric about the origin and has volume MA). If MA) > n then A contains an integer point different from the origin. Let A m be the set of rational points in A all of whose coordinates have denominator m, so A m = { t m,..., tn m ) A t i Z, i =,..., n}. Let A m denote the cardinality of m. Then For m sufficiently large A m lim m m n = MA). A m > m) n. Thus there are two distinct points a = a,..., an) and b = m m b m,..., bn) A m m with a i b i mod m) for i =,..., n. Then a b) is an integer point claim) which isn t 0. a and b A m hence in A. b A since A is symmetric about 0. a+ b) = a b) A since A is convex. And the result follows. Remarks concerning Minkowski s Convex Body Thrm. ) n is sharp. Consider the set A in R n given by A = {x,..., x n ) R n x i < }.

7 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 7 Note that µa) = n but the only integer point in A is the origin. ) Note that in the conclusion of Minkowski s Convex Body Thrm we can claim the existence of non-zero points in the set since if g is in the set then so is g by symmetry. Theorem 7. Minkowski s Linear Forms Thrm) Let B = β ij ) be an n x n matrix with real entries and non-zero determinant. Let c,..., c n be positive real numbers with c c n det B. Then there exists a non-zero integer point x = x,..., x n ) such that β i x + + β in x n < c i for i =,..., n and β n x + + β nn x n c n. Write L i x) = β i x + + β in x n < c i for i =,..., n and put L ix) = c i L i x) for i =,..., n. Thus we wish to solve L ix) < for i =,..., n and L nx). The determinant of the matrix determined by L ix) for i =,..., n is det B c c n. Therefore, WLOG we may suppose that c = = c n = and that detb. For each ǫ > 0 define A ǫ to be the subset of x R n for which L i x) < for i =,..., n and L n x) < + ǫ. Note that A ǫ is symmetric about the origin 0 and is bounded. Further A ǫ is convex since if x and y are in A ǫ and λ R with 0 λ then L i λx) + λ)y) λ L i x) + λ) L i y) λ + λ = for i =,..., n < λ + ǫ) + λ) + ǫ) = + ǫ for i = n Since µa ǫ ) = + ǫ) n > n and so by Minkowski s Convex Body Thrm there is a non-zero integer point x ǫ in A ǫ. Consider A /k for k =,,... and note that A A / A /3. We obtain a sequence x /k of non-zero integer points in A /k for k =,,.... Since A is bounded and contains only finitely many integer points there must be one point y of the form x /k for infinitely many k. Note that then L i y) < for i =,..., n and L n y). Suppose that α is an algebraic number with Mα). Note that if fx) = a d x d + + a 0 is the minimal polynomial of α, then, since Mα) = a di= d max, α i ) where α,..., α d are the roots of f, we see that a d =. Therefore if Mα) then α is an algebraic integer. Let α = α so α,..., α d are the conjugates of α. Note that αi k for i =,..., d and k =,,.... The elementary symmetric polynomials in the variables x,..., x d are bounded in absolute value by d when evaluated at points y,..., y d ) with y i, for i =,..., d. Thus when y,..., y d ) = α, k..., αd) k they are integers of size at most d. In particular, α k is the root of a non-zero polynomial with integer coefficients of absolute value at most d, hence of a finite set of non-zero polynomials. Therefore α k = α l for some k, l Z +. Thus either α = 0 or α is a root of unity. This was first proved by Kronecker in 857. S = {Mα) α algebraic }.

8 8 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Countable, is it dense? In 933, Lehmer asked if for each ǫ > 0 there exists an algebraic number α for which < Mα) < + ǫ. The smallest value of Mα) larger than which he found was Mα 0 ) = where α 0 is a root of x 0 + x 9 x 7 x 6 x 5 x 3 + x +. This is an example of an Salem number. One root inside the circle, one outside the circle and the rest on the unit circle. No smaller example has been found. Pisot numbers: Algebraic integers α all of whose conjugates, apart from α > lie strictly inside the unit circle. Let α = α,..., α d be the conjugates of a Pisot number α. Then for each k Z +, α k + + αk d is an integer. Thus if we let x denote the distance from x to the nearest integer for any x R then lim k αk = 0. Open Question: If θ R with θ > and lim k θ k = 0, is θ a Pisot number P.V. number, Pisot-Vijayaraghavan number)? There is a smallest Pisot number. In 979, Dobrowolski proved that if α is an algebraic number of degree d 3 and α is not a root of unity then Mα) > + 00 log log d log d ) 3. Theorem 8. Let d Z + and let α be an algebraic number of degree at most d, which is not a root of unity. Then log Mα) > d. We first need: Proposition 9. Let p be a prime number and let f Z[x,..., x k ]. Then there exists g Z[x,..., x k ] such that fx p,..., x p k ) fx,..., x k ) p = pgx,..., x k ). Put x = x,..., x k ) and x p = x p,..., x p k ). Observe that if fx) is a monomial, say fx) = ax i x i k k ), then where gx) = a ap p x i p x i kp k fx p ) fx) p = a a p )x i p x i kp k = pgx), Z[x,..., x k ] since p a a p by Fermat s little theorem. Suppose now that the result holds for f x) and f x). Thus f x p ) f x) p = pg x) and f x p ) f x) p = pg x), with g, g Z[x,..., x k ]. Then f + f )x p ) f + f )x) p = f x p ) + f x p ) f + f ) p x) = f x) p + pg x) + f x) p + pg x) f + f ) p x) But ) ) p p f + f ) p = f p + f p f + + f f p + f p p. Thus f + f )x p ) f + f )x) p = pg x) + g x) + ) p f p f + + ) p f f p ) p p p Z[x,..., x k ] since p ) p j for j p. The result now follows by induction on the number of monomials of f.

9 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 9 Proposition 0. Let α be a non-zero algebraic number. Suppose that h and l are distinct positive integers for which α h and α l are conjugates. Then α is a root of unity. Let α = α,..., α d be the conjugates of α and put k = Qα,..., α d ). Since α h and α l are conjugates there is an element σ of Galk/Q) for which We claim that for n =,,... σα h ) = α l. σ n α hn ) = α ln. True for n =. Suppose true for k n. Then σ n+ α hn+ ) = σσ n α hn ) h ) = σα ln ) h ) = σα h )) ln = α l ) ln = α ln+. The claim follows by induction. Since the Galois group is finite there is finite there is t Z + such that σ t is the identity. Then σ t α ht ) = α lt so α ht = α lt. Since α is non-zero α is a root of unity. Given an algebraic number α with conjugates α = α,..., α d over Q we define the house of α, denoted α, by α = max{ α,..., α d }. Theorem. Dobrowolski, 978) If α is a non-zero algebraic integer of degree d which is not a root of unity then α > + 4ed. Let α = α,..., α d be the conjugates of α over Q. Let f h x,..., x d ) = x h + + xh d for h =,,.... Put S h = f h α,..., α d ) for h =,,.... Note that S h is an integer for h =,... since it is fixed under elements of the Galois group of the splitting field over Q. Let p be a prime number. By Fermat s little thrm, On the other hand, by Prop 9, S p h S h mod p) for h =,,... S hp S p h = pgα,..., α d ) where g Z[x..., x d ]. Note that S hp and S p h are in Q so gα,..., α d ) is in Q. Since α,..., α d are algebraic integers and g Z[x,..., x d ], gα,..., α d ) Z. Thus Therefore S hp S p h mod p). S hp S h mod p), for h =,,... For each h Z + S h = α h + + αh d d α h. Suppose that α +. By Bertrand s Postulate there is a prime p with ed < p < 4ed. 4ed For h d, S h d + ) h de dlog+ 4ed 4ed ) de.

10 0 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART and Therefore, for h d, Thus, since S hp S h mod p), S hp d + 4ed ) hp de 4ed log+ 4ed ) de. S hp S h ed < p. S hp = S h for h =,..., d. But the Newton Sums, for h =,..., d determine the elementary symmetric polynomials in x,..., x d and so the minimal polynomials of α and of α p are the same. Thus α and α p are conjugates and so by Prop 0, α is a root of unity. Let c, k R + with k > c. Consider the function ft) = log + ) ct, for t > 0. kt f t) = + ct + ct kt and so f t) > 0 for + > 0 c+ k t so when > k c+ t i.e. when > k c so when t <. Futher f t) < 0 for t >. In addition log + ) t k c k c ct kt is positive for t sufficiently large. Take k = and c = 4e. Then Thus, for d 3, log + ) 4et Since log + 6e Therefore for d, Z > t for t > 4e = log + ) > 4ed d. ) >.073 > 44 =.077. log + ) > 4ed d. ) Theorem 8 continued) If α is not an algebraic integer and fx) = a d x d + +a 0 is the minimal polynomail of α then a d and so Mα) hence log Mα) log and the result holds. Thus we may suppose α is an algebraic integer. Note that the result holds if d =, so suppose d. Then, by Thrm, log Mα) log α log + ) 4ed and, since d, by ) log Mα) > d.

11 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Further remarks on heights. Definition. Let K be a field and let : K R. We say that is a valuation on K if ) a K, a 0 and a = 0 iff a = 0. ) a, b K, ab = a b. 3) a, b K, a + b a + b. Examples: ) The ordinary absolute value on C. ) Let K be any field. The map if a 0 a = 0 if a = 0 is known as the trivial valuation on K. 3) Let p be a prime and let K = Q. For any non-zero rational a/b we can define the p-adic order of a/b denoted by ord p a/b), by ord p a/b) = ord p a ord p b where the p-adic order of an integer is the exact power of p which divides it. i.e. ord 5 00) =. We now define p on Q by p ordpa/b) if a/b 0 a/b p = 0 if a/b = 0. One can check that p is a valuation. It is called the p-adic valuation on Q. 4) Let k be any field and let T be a transcendental element over k. Put K = kt). Let λ be a real number with 0 < λ <. Let pt) be an irreducible element in K. Then every non-zero element a of K can be written in the form pt) j ft) where j Z, f and g are coprime with gt) p. Define on K by λ j if a 0 a = 0 if a = 0. is a valuation on K. Definition. A valuation on a field on K is said to be non-archimedean if a, b K, a + b max a, b ). The last 3 examples are non-archimedean. For example p on Q is non-arch. since if a/b = p α a /b with a, p) = b, p) = and c/d = p β c /d with c, p) = d, p) = then a/b p = p α, c/d p = p β. WLOG let α = minα, β) pαa + p β c = b d p pαa d + p β α c b b d = p α p a d + p β α c b p = p α a d + p β α c b p p α = a/b p max a/b p, c/d p ). Definition. Let and be vaulations on a field K. We say that they are equivalent if there is a positive number γ such that a = a γ for all a K. We can define a metric and hence a topology on K by defining d : K X K R by da, b) = a b.

12 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Equivalent valuations induce the same topology. The trivial valuation induces the discrete topology. Ostrowski proved that every non-trivial valuation on Q is equivalent to either the ordinary absolute value or to a p-adic valuation for some prime p. Let us denote on Q by p and we put SQ) = {p, p a prime in Z}. By the Unique Factorization Thrm for Z, hence for Q, we have for α Q, if α 0 α v = ) 0 if α = 0 v SQ) ) is known as the product formula. Note that ) holds because we chose p from its equivalence class appropriately. Let K be a finite extension of Q. Then K = Qα) for some algebraic number α. Let α = α,..., α d be the conjugates of α. Let K be a finite extension of Q and let O K denote the ring of algebraic integers of K. Each prime p in Z is such that the ideal p) in O K splits into a product of prime ideals p) = p e pt et here p,...,p t are distinct prime ideals. e i is known as the ramification of p i for i =,..., t. Further if we put [O K /p i : Z/p] = f i, the residue class degree of p i, then e f + +e t f t = [K : Q]. If K is a Galois extension of Q then e = = e t and f = = f t. Since K is a finite extension of Q there is an β K such that K = Qβ). Suppose that β,..., β d be the conjugates of β over Q. Then the Q-isomorphisms σ of K into C are determined once we knwon σβ). Let σ i β) = β i for i =,..., d. We may suppose that β,..., β r are real and that β r +,..., β r +r are not real and that β r +i = β r +r +i for i =,..., r. Note that σ r +r +i = σ r +i for i =,..., r. We say that σ,..., σ r +r are the infinite primes of K and that the prime ideals of O K are the finite primes. Let SK) denote the union of the finite and the inifinite primes. We now define a valuation v on K for each prime in SK). Let α K. If v = p SK) then we put N K/Q p) ωpα)/d if α 0 α v = 0 if α = 0 where ω p α) is the order of p in the canonical decomposition of the fractional ideal α) as a product of prime ideals. Further if v = σ SK) we put σα) g/d if α 0 α v = 0 if α = 0 where if σ {σ,..., σ r } g = if σ {σ r +,..., σ r +r } Once again it is possible to check that the product formula holds: if α 0 = 0 if α = 0. v SK) Note that v SK) means the v for each prime in SK). We now introduce a new height function hα) on K. For α K we put hα) = max, α v ). v SK)

13 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 3 If L is a finite extension of K then for α K max, α v ) = v SL) v SK) max, α v ), so h is defined on the set of all algebraic numbers. The definition of h does not depend on the field containing α. What is the link with Mα)? Suppose that α is an algebraic number of degree d over Q and minimal polynomial fx) = a d x d + a 0 = d d di= x α i ). Put K = Qα) then hα) = Mα) /d. As before suppose that α,..., α r are the real conjugates of α and that α r+,..., α r +r are the conjugates which are not real and that α r +i = α r +r +i. Then so v=σ SK) It can be shown that Thus max, α v ) = p SK) hα) d = r i= r +r max, σ i α) /d ) max, σ i α) /d ) i=r + d max, α v ) d = max, α i ). v=σ SK) i= max, α d v) = v=p Sk) = Mα). p Sk) = p SK) max, α d v ) max, N K/Q p) ωpα) ) max,p fωpα) ) = a d. v=σ Sk) max, α d v ) We can now verify the following : For α an algebraic number and k Z +, hα k ) = max, α k v ) = max, α v ) k = hα) k. v SQα)) v SQα)) Recall that if α is a non-zero algebraic number with minimal polynomial fx) of degree d then gx) = x d f ) is the minimal polynomial of x α. But π ) π ) Mα) = exp log fe iθ ) dθ = exp log ge iθ ) dθ = Mα ). 0 Thus hα) = hα ) for α 0. Therefore hα k ) = hα) k for all k Z. Further if α, β are algebraic numbers then hαβ) = max, αβ v ) = v SQα,β)) v SQα,β)) v SQα,β)) = hα)hβ) max, α v ))max, β v )) max, α v )) 0 v SQα,β)) max, β v ))

14 4 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Theorem 6. Let α,..., α n be non-zero algebraic numbers which are multiplicatively dependent. Suppose that A i = maxhα i ), e) for i =,..., n and that A A A n. Let d = [Qα,..., α n ) : Q]. There exists integers t,..., t n, not all zero, with α t αn tn = and for k =,..., n t k n )d 3 ) n log A log A n For each m Z + let φm) denote Euler s phi function so φm) = m p m /p). Notice that Further φm) = m /p) = m p m p m m p m p p ) ) p p p p p ) m p m p m Therefore φm) m/. Thus if φm) = d we wee that m d. First suppose that n =. Then α is a root of unity. Since d = [Qα ) : Q] we see that α is a m-th root of unity with m d. In particular α t = with t d and the result follows. Next suppose that n > and we may suppose wlog that no subset of n elements from {α,..., α n } is multiplicatively dependent. Therefore there is a unique n-tuple of relatively prime integers k,..., k n ) with k > 0 for which α k αn kn =. Let j Z with j n for which k j k i for i =,..., n. Put c i = n )d 3 log A i ) for i j and c j = n )d 3 ) n log A log A n / log A j. Observe that c c n =. Consider the system of inequalities: x i k i x j k j c i for i =,..., n, i j ) Notice that the associated matrix B =. x j c j ) k k j k k j k n k j has determinant. By Minkowski s Linear Forms Theorem 7 ), there exists a non-zero integer point b,..., b n ) which satisfies ) and ). Put α = α b α bn. We claim that α is a root of unity. We have n α k j = α k jb α k jb n n

15 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 5 Since α k α kn n = we have α kj = α k jb k b j α k jb n k nb j n. Thus so Therefore Thus Accordingly, hα k j ) = hα k jb k b j α k jb n k nb j n ) hα) k j hα ) k jb k b j hα j ) 0 hα n ) k jb n k nb j hα ) c k j hα j ) 0 hα n ) cn k j hα) hα ) c hα j ) c j hα j+ ) c j+ hα n ) cn. logmα) /d ) loghα)) n log Mα) d i=,i j n i=,i j n i=,i j c i log hα i ) c i log A i. n )d 3 d 3. Hence log Mα). Therefore, by Theorem 8, α is a root of unity. Since the degree of α d is at most d we see that α is an m-th root of unity with m d. Therefore α m = α b m αn bnm =. Further by ), b j m c j d = d n )d 3 ) n log A log A n / log A j n )d 3 ) n log A log A n Furthermore, since k j k i for i =,..., n, and we see that, for i =,..., n, i j, b i k i k j b j c i b i c i + b j + b j + c j 3 c j. Since m d we see that b i m 3d c j n )d 3 ) n log A log A n. Note if α k αn kn = then k log α + + k n log α n = 0 for some choice of branches of the logarithms. In 970, Senge and Strauss proved that if a, b Z larger than with log a)/ log b irrational then the number of integers n for which the sum of the digits of n in base a plus the sum of the digits of n in base b lies below a fixed bound is finite. The proof was not effective in the sense that given a bound one could not determine the finite set of integers n. We can overcome this difficulty using Theorems 3 and 6.

16 6 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Let α, β Z with 0 α < a and 0 β < b. Denote the number of digits in the base a expansion of n which are different from α by L α,a n). Similarly denote the number of digits in the base b expansion of n which are different from β by L β,b n). Put L α,a,β,b n) = L α,a n) + L β,b n). Note that the sum of the digits of n in base a and base b is bounded from below by L 0,a,0,b n). Note that for all n Z with n > we have L α,a,β,b n) < c log n, where c is a positive constant. Further on average, L α,a,β,b n) < c c log n, for c > 0. To fix ideas 33 = 5 + = So L 0,,0,3 33) = 4. The sum of the digits is this case is 5. Similarily 63 = = So L,,,5 63) =. Note the condition that log a log a is irrational is necessary. For suppose that = r so that log b log b s a s = b r. Then for each integer k Z + L 0,a,0,b a sk ) = since n = a sk and n = b rk. Theorem. Stewart) Let a, b Z a, b > with log a/ log b irrational and let α and β be integers with 0 α < a and 0 β < b. Then there is a positive number C, which is effectively computable in terms of a and b, such that if n Z with n > 5 then log log n L α,a,β,b n) > log log log n + C. Suppose that n > a + b and consider the expansions n = a a m + α a m a + a a m + + a r a mr, where 0 a < a and α a i < a α for i =,..., r and n = b b l + β b l b + b b l + + b t b lt, where 0 b < b and β b i < b β for i =,..., t. Further m > m > > m r 0 and l > l >... > l t 0. We put θ = c log log n, ) where c is a positive number which is effectively computable in terms of a and b and c > 4. We shall assume that n > c 5, where c, c 3,... denote positive number which are effectively computable in terms of a and b and may be determined independently of c. Define k to be the integer satisfying θ k log n 4 log a < θk+ ) and put Θ =, θ], Θ = θ, θ ],..., Θ k = θ k, θ k ]. If each of the intervals Θ,..., Θ k contains at least one term either of the form m m s with s r or of the form l l j with j t then By ), L α,a,β,b n) r + t k. 3) k + ) log θ > log log n log4 log a).

17 so Thus by ) LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 7 k > k > log log n log θ log4 log a) log θ. log log n log log log n + log c log4 log a). 4) Since L α,a,β,b 0 our result now follows from 3) and 4) in this case. Therefore we may assume that there is an integer s with s k for which the interval Θ s contains no term of the form m m i or of the form l l j. Define integers p and q by the inequalities and with the convention that m r+ = 0 and l t+ = 0. Then where and Note that so Further so m m p θ s and m m p+ > θ s, 5) l l q θ s and l l q+ > θ s, 6) b )a )n = b )a )a a m + b )αa m + b )a )a a m + + b )a )a r a mr b )α = A a mp + A, A = b )a )a a m m p + b )αa m m p + + b )a )a p A = b )a )a mp+ a m p+ + + b )a )a r a mr b )α. 0 < A < b )a )a m m p+ + b )αa m m p 0 < A < b )a )a m m p+. 7) 0 A < b )a )a m p++ + b )α 0 A < b )a )a m p++ Similarily b )a )n = B b lq + B, where B and B are integers with We have 0 < B < b )a )b l l q+. 9) 0 B < b )a )b l q++ A a mp ) = A a mp + A = A a mp + A B b lq + B B b lq + B B ). b lq If x and y are real numbers with absolute values at most / then + x max + y, + y ) + 4 max x, y ). ) + x 8) 0)

18 8 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Notice that A b )a )amp++ a m +m p+ +. mp A a b )a )a m By 6), m m p+ θ s θ = c log log n and thus for n sufficiently large A A a mp. Similarly B B b lq. We put R = A a mp B b. lq From ) we conclude that maxr, R ) + 4 max A A a, B mp B b ) lq so maxr, R ) + 8 maxa m +m p+ +, b l l q+ + ). Since log + x) < x for x > 0 we have Thus if log R 0 then by 5) and 6), log R < 8ab maxa m +m p+, b l l q+ ). log log R < c 3 c 4 θ s. ) On the other hand log R = log A + m p log a l q log b, B and we can apply Prop 3 to give a lower bound for log R. We put n = 3, d = and α, α, α 3 to be A B, a and b respectively. Note that m p and l q are at most log n/ log and that the height of A B is at most the maximum of A and B. Thus, by Prop 3, if log R 0 then log R exp c 5 log4 max A, B )) log log n) so, by 7) and 9) log log R c 6 max, m m p, l l ) log log n. Suppose that log R = 0, hence that log A B + m p log a l q log b = 0. 3) By Theorem 6, there exists x, x, x 3 Z not all zero with and with By 5) and 7), and by 6) and 9), x log A B + x log a + x 3 log b = 0 4) max x, x, x 3 ) c 9 logmax A, B, e). log A c 0 θ s log B c θ s

19 Therefore LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 9 and so, by ), x c θ s c θ k x < log n 4 log a, for n sufficiently large. By 5) m m p θ s so m p m θ s. Since m log n and since log a θs θ k < log n 4 log a for n sufficiently large we see that m p > log n log a log n 4 log a = log n 4 log a. In particular m p > x. Recall that 4) and note that if x = 0 then log a/ log b is rational. Thus we may suppose that x 0. Eliminating log A B from 3) and 4) we find that x m p x ) log a + x 3 x l q ) log b) = 0. Since m p > x we see that x m p x 0 hence log a/ log b is rational. Let p be a prime with p 3 and let = n < n < be the sequence of positive integers which are composed of primes size at most p. In 898, Stormer proved that lim inf i n i+ n i ) >. In 908, Thue proved that lim n i+ n i ) =. i In 965, Erdos used a result of Mahler to prove the following. Let ǫ > 0 then there exists a positive number Np, ǫ), such that if n i > Np, ǫ), then n i+ n i > n ǫ i. In 973, Tijdeman used estimates for linear forms in logarithms to prove: Theorem 3. Let p be a prime and let = n < n < be the sequence of positive integers all of whose prime factors are at most p. There exists a positive number C, which is effectively computable in terms of p, such that if n i 3 then n i+ n i > n i /log n i ) C. Let p,..., p k be the primes of size at most p. Consider the prime decompositions of n i and n i+ for some i, say n i+ = p a pa k k with a j N and n i = p b p b k k with b j N Then 0 log n i+ = a b ) log p + + a k b k ) log p k n i and by Thrm with k = n, d =, and α,..., α k given by p,..., p k respectively. Note that max j k a j b j ) ) logn i+ )/ log + log n i log

20 0 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART since n i+ n i. Further observe that k since if k = the result is immediate. Therefore, by Theorem, log n i+ n i > exp k c k log p log p k log log n i ), where c is a positive number. Note that k c k log p log p k < e c k log k log p) k < e c k log k+k log log p < e c k log k since p < k and by the prime number theorem e c k log k < e c3p. Therefore log n i+ > exp 3 c3p log log n i ) = n i Since log + x) < x for x > 0, log n i ) ec 3 p. Therefore log n i+ n i = log + n i+ n i n i ) < n i+ n i n i. n i+ n i > n i log n i ) ec 3 p. Theorem 4. Tijdeman) Let p and q be distinct primes and let = n < n < be the sequence of positive integers whose only prime factors are p and q. There exists positive numbers c and N which are effectively computable in terms of p and q, such that if n i N, then n i n i+ n i < log n i ) c. Before we prove Thrm 4, we need some basic facts from Diophantine approximation. For any real number α we define an associated sequence α 0.α, ) where we put α 0 = α and α k =, for k =,,..., α k [α k ] provided that α k [α k ] 0; here [x] denotes the greatest integer x for x R. Next put a k = [α k ] for k = 0,,,.... Then α = a 0 + a + α k We put p k q k = a 0 + a for k = 0,,,... with p + + k, q k ) =. a k to α. The finite continued fraction is denote by [a 0, a,..., a k ]. We ll now show that for n =, 3,... p k q k is said to be a convergent p n = a n p n + p n and q n = a n q n + q n ) Note that p 0 = a 0 and q 0 = and p = a 0 a + and q = a. We have p = a 0a a + ) + a q a a + Observe that p = a 0 a a +)+a and q = a a +. Further p = a a a 0 +)+a 0 = a p +p 0 ) and q = a q + q 0. Thus ) holds for n =. Suppose ) holds for n k and we ll

21 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART prove by induction that ) holds for n = k. Define a sequence of rationals p j q j with p j and q j coprime integers with q j > 0 given by for j = 0,,,... p j q j = [a,..., a j+ ]. By the recuurence relations given by ) for n k we find that and But so Thus, on taking j = k, we find that In particular, and so by ) Similarly p k = a k p k + p k 3 q k = a kq k + q k 3. p j q j = a 0 + p j q j = a 0 + q j p j p j = a 0 p j + q j and q j = p j. ) p k = a 0 p k + q k and q k = p k. p k = a 0 a k p k + p k 3 ) + a kq k + q k 3 ) = a k a 0 + p k + q k ) + a 0p k 3 + q k 3 ) p k = a k p k + p k. q k = p k = a k p k + p k 3 = a k q k + q k. The result now follows by induction. Recall, α = [a 0, a,..., a k, α k+ ] and /α k+ /a k+ and so p 0 < p < < α < < p 3 < p. q 0 q q 3 q In particular α lies between p n /q n and p n+ /q n+ for n = 0,,,.... Proposition 5. p n q n+ q n p n+ = ) n+ for n = 0,,,.... Notice that the result holds for n = 0 since p 0 = a 0, q 0 =, p = a 0 a +, q = a hence p 0 q q 0 p = a 0 a a 0 a + =. Suppose its true for 0,..., n. Then p n q n+ q n p n+ = ) n+ = p n a n+ q n + q n ) q n a n+ p n + p n ) and the result follows by induction. = p n q n q n p n = )p n q n q n p n ) = ) n+

22 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART By Prop 5, p k p k+ = p kq k+ q k p k+ q k+ = )k+. q k q k+ q k q k q k+ Since α lies between p n /q n and p n+ /q n+ for n = 0,,,... we conclude that α p n <. q n q n q n+ qn It is not difficult to show that if α p/q < q then p/q = p n /q n for some n with n 0. Lemma 6. Tijdeman) Let p and q be distinct primes let h 0 k 0, h k, be the sequence of convergents to log p. There exists a positive number c which is effectively computable in terms of log q of p and q such that k j+ < kj c log q, for j = 0,,,.... We may suppose that j and so k j. Since log p log q h j k j < k j k j+. Hence k j log p h j log q < log q k j+. By Thrm with α = p, α = q, d = we find that k j log p h j log q > exp c logmax h j, k j )) where c, c,... denote positive numbers which are effectively computable in terms of p and q. Since log p h j log q k j < k j k j+ we see that h j < c k j hence Thus hence k j log p h j log q > exp c 3 log k j ) = k jc 3. k c 3 j Theorem 4 Put n i = n = p u q v. we may suppose that hence < log q k j + k j+ < k c 3 j log q. p u n, u log n log p. ) Let h 0 k 0, h k, be the sequence of convergents to log p. Choose j so that log q k j u < k j+. We may suppose that N is sufficiently large so that n 3 and j. We shall distinguish two cases according to whether h j k j is bigger or smaller than log p. log q Case h j k j > log p. log q

23 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 3 Put n = p u k j q v+h j. Note that n Z and n > n since h j k j > log p log q. Thus n n i+. We have and so Therefore we have k j+ > u log n log p h j k j log p log q < k j k j+, h j log q k j log p < log q k j +. ) ) n q h j log = log n p k j and so ) n 0 < log < n For n sufficiently large in terms of p and q Since, for x <, 0 < n n < exp < log q k j+. log p log q. ) log n ) log p log q < log n. log + x) = x x + x3 3. Further, for 0 < x <, we have Thus for n sufficiently large. By ) log + x) > x x ) ) n log = log + n n n ) n n < 4 log p log q log n = x x/) > x/. so n < n + > n n ) 4 log p log qn log n Let c, c, denote positive constancts which are effectively compuatable in terms of p and q. Thus n < n + c n log n, for n > c. Since n i+ n, n i+ < n i + c n i log n i, for n i > c. Case h j k j < log p. log q Hence h j k j > log p. Put log q n = p u k j q v+h j. Again note n Z and n n i+. We have h j log q k j log p k j log q = h j log p k j log q < h j k j h j k j = k j k j.,

24 4 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Therefore, Accordingly We find from Lemma 6, that ) ) n log q h j log = log n p k j = h j log q k j log p log < k j log q k j k j. ) n k j > n < log q k j. ) kj+ c. log q By ) we have u log n and, since k log p j+ > u, we see that ) n log q log < ) n c log n log plog q = log q)+/c log p) /c log n) /c 3) Thus, for n sufficiently large, ) n log > ) n n n. Comparing this estimate with the upper bound 3), we find that n < n + log q) +/c log p) /c n log n) /c. Since n n i+ we see that Thus in cases and, Therefore n i+ n i + n i+ n i + n i+ n i + c 3 n i log n i ) /c, for n i > c 4 c 5n i log n i ) c 6, for n i > c 7. n i log n i ) c 8, for n i > c 9. Background from Algebraic Number Theory Let K be a finite extension of Q. Thus there exists some irreducible polynomial f Q[x]) such that K is Q[x]/fQ[x]. Suppose that fx) = x n + a n x n + + a 0 and that f factors over C as fx) = x α ) x α n ). By the Primitive Element Theorem there exists a θ such that K = Qθ). There are n field embeddings of K into C, that is, n isomorphisms of K into C which are the identity on Q. They are given by θ α i for i =,..., n and so determine n conjugate fields Qα ),..., Qα n ) in C, non necessarily distinct). Let α,..., α r be real numbers and let α r +,..., α n be complex numbers which are not real. We may suppose that α ri +i = α ri +r +i for i =,..., r.

25 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 5 Note that n = r + r. The ring of algebriac integers of K, denoted O K, consists of the elements of K which are roots of monic polynomials with integer coefficients. O K is a Dedekind domain. Definition. An integral domain O is a Dedekind domain if: ) O is a Noetherian ring. ) O is integrally closed in its field of fractions. 3) All non-zero prime ideals are maximal ideals. If O is a Dedekind domain then there is unique factorization of ideals into prime ideals up to reordering. Note that we do not always have unique factorization of elements of O K into prime elements in O K. Let K be an extension of Q of degree n with ring of algebraic integers O K. Then there exists a set {ω,..., ω n } of elements from O K such that every element of O K has a unique representation as an integral linear combination of ω,..., ω n. We call {ω,..., ω n } an integral basis for O K. Any two integral bases for O K for K) are related by a matrix with determinant ±. Definition. The discriminant D of K is D = detσ i ω j ))), where σ,..., σ n are the embeddings of K in C and {ω,..., ω n } is an integral basis. Note that D is a non-zero integer. Consider the set S of non-zero ideals in O K. We define a relation on S by saying that a b, for ideals a, b S, if there exists non-zero elements α, β in O K such that [α]a = [β]b; here [α] denote the principal ideal generated by α in O K and similarly for [β]. is an equivalence relation on S and if a [] then a is principal. We can define a multiplication on the equivalence classes of S by taking multiplication of representatives. This determines a finite abelian group called the ideal class group of K. The order of the group is called the class number and is denoted by h. Thus if a is an ideal of O K then a h []. Let K be a finite extension of Q. The group of units UK) of O K is the set of invertable elements in O K. It forms an abelian group under multiplication. Plainly the roots of unity of K are in UK). In 846, Dirichlet proved that UK) is isomorphic to µk) x Z r, where µk) is a finite torsion group, where r = r +r and where K has r real embeddings in C and R non-real embeddings. Let σ,..., σ r be the real embeddings of K in C and suppose that σ r +,..., σ r +r are the complex emebbedings and that σ r +i = σ r +r +i for i =,..., r. By Dirichlet s result there exists units u,..., u r such that every element x in UK) can be written in the form X = ζ u b u br r where b,..., b r Z, ζ a root of unity. The set {u,..., u r } is said to be a fundamental system of units. While the fundamental system of units need not be unique it is possible to attach a volume to the system which is independent of the choice of system. We make use of the logarithmic map L : K R r+ given by Lα) = log σ α),..., log σ r α), log σ r +α),..., log σ r +r α) ). L is an abelian group homomorphism.

26 6 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART For any α K we let N K/Q α) denote the norm of α and we put N K/Q α) = r +r i= σ i α). Note that the norm is multiplicative, in other words, N K/Q αβ) = N K/Q α)n K/Q β). We see that if x UK) then N K/Q x) = ±. In particular, if α UK) then Lα) lies in the r dimensional subspace of R r+ given by H = {x,..., x r ) x + + x r = 0}. Thus L : UK) H. The image of UK) under L is a lattice in H and the kernel of L is just the set of roots of unity of K. The volume of a fundamental region for the lattice is called the regulator R K of K. Let {u,..., u r } be a fundamental system of units. We have where R K = det δ i log σ i u j ) i=,...,r//j=,...,r, if i r δ i =. if r < i In 98, Zimmert, sharpening work of Remak, proved that R K >.056. In 98, Landau proved that there is positive number C, which depends on d = [K : Q], such that loghr) < C D / log D ) d. Since the class number h of K is always at least this yields an upper bound for R in terms of the discriminant D and the degree d of K over Q. Let I be an ideal in O K. The norm of I, denoted NI), is the cardinality of O K /I. It can be shown that if α in O K and [α] denotes the principal ideal of O K generated by α then N[α]) = N K/Q α). Dedekind introduced a generalization of the Riemann zeta function ζs). Let [K : Q] <. He defined ζ K s) for Res) > by ζ K s) = I NI) s, where the sum is taken over all non-zero ideals of O K. This converges uniformly on compact subsets to an analytic function for Res) >. Further ζ Q s) = ζs). It can be shown that ζ K s) can be analytically continued to all of C with the exception of a simple pole at s =. Further there is a functional equation which relates ζ K s) with ζ K s). The Generalized Riemann Hypothesis GRH) is that the only zeros of ζ K s) with 0 Res) have Res) = /. There is an Euler product representation for Res) > given by ζ K s) = p Np) s ) where the product is taken over all prime ideals p of O K. Let ωk) denote the number of roots of unity in K. Then lim s )ζ π) Ks) = r r hr K. s ωk) D

27 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 7 Thue Equations Let Fx, y) = a n x n + a n x n y + + a 0 y n be a binary form with integer coefficients, and suppose n 3. Let m be a non-zero integer. Suppose that F has non-zero discriminant. The equation Fx, y) = m 4) is known as a Thue equation. Thue proved in 909 that if a n 0 and Fx, ) is irreducible then 4) has only finitely many solutions in integers x and y. Eg x 3 y 3 = 6. Thue - only finitely many. In fact,) is the only solution. By contrast x y = has infinitely many solutions in integers x and y. Let K be a finite extension of Q. It is possible to choose a fundamental system of units {u,..., u r } such that max i r, j r log uj) i < CR, where R is the regulator of K and C is a number which depends on [K : Q]. Theorem 7. Let F be an irreducible binary form with integer coefficients and degree n 3. Let m be a non-zero integer. There exists a positive number C, which is effectively computable in terms of F, such that all solutions in integers x and y of the Diophantine equation Fx, y) = m satisfy max{ x, y } < m C log log 4m. Let c, c,... denote positive numbers which are effectively computable in terms of F. Let Fx, y) = a n x n + + a xy n + a 0 y n. By considering an n Fx, y) = an n m and letting an n Fx, y) = fx, y) with X = a n x we see that we may suppose without loss of generality a n =. Next let Fx, y) = x α ) y) x α n) y) be the factorization of F over C. Suppose, as usual, that α ),..., α r) are real and that α r+i) = α r +r +i) for i =,..., r, hence that n = r + r. Suppose that x and y are integers for which Fx, y) = m, and put Then β i) = x α i) y, for i =,..., n. β ) β n) = m. Put K = Qα ) ) and let η ),..., η r ) be a fundamental system of units for which max log η i) j < c R K. i r j r Every point p in R r is within c of the lattice generated by the vectors log η i),..., log η r i) ) for i =,..., r. Take p = log m /n β ) ),..., log m /n β r) )). Then we can find integers b,..., b r such that b log η ) i + b log η ) i + + b r log η r) i + log m /n β i) ) < c, for i =,..., r. Now put, for j =,..., n, γ j) = β j) η j) b η j) r )br. )

28 8 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART Observe that, for j =,..., r, log m /n γ j) < c. Since γ r+i) = γ r +r +i) for i =,..., r we see that log m /n γ j) < c holds for j =,..., n except for j = n when r = n or for j = r + r and j = r + r when r < n. But γ ) γ n) = β ) β n) = m and so n log m /n γ j) = 0. Therefore j= log m /n γ j) ) < c 3 for j =,...n. ) Note that since β j) is an algebraic integer so is γ j) for j =,..., n. Further the height Hγ j) ) is at most c 4 m since the coefficients in the minimal polynomial are elementary symmetric polynomials in the γ i) s, hence are at most c 4 m in absolute value. Consider now the equations obtained from ) by taking logarithms: b log η j) + + b r log η r j) = log γ j), for j =,..., r. By Cramer s Rule, for i =,..., r: log η ) log γ) log η β ) r ) det... log η r) log γr) log η r) β b i = r) r log η ) log η ) 3) r det.. log η r) log η r r) Our aim is to bound b i in terms of m hence to bound x, y in terms of m. Let B = max b,..., b r ) and suppose that b i = B. Notice that, the determinant in the denominator, satisfies = R r. Expanding the determinant in the numerator of 3) along the i-th column we see that max log γ j) > c 5B. j r Let the maximum occur for j = J. Then log m /n log β J) = log βj) γ J) + log m /n γ J) ) β j) > c 5 B c 3. Since n j= log m /n β j) = 0 it follows that for some l with l n In particular, log m /n β l) ) < c 3 c 5 B n. β j) β l) < m /n c 7 e c 6B. 4)

29 LINEAR FORMS IN LOGARITHMS AND DIOPHANTINE EQUATIONS C.L. STEWART 9 Since β ) β n) = m, there exists an integer k with k l such that β k) > m /n c /n ) 7 e c 6B/n ). 5) Let j be an integer with j n and with j k and j l. Note that j exists since n 3. We have the identity Note Coef of y: α k) α l) )β j) α j) α l) )β k) = α k) α j) )β l). α k) α l) )x α j) y) α j) α l) )x α k) y) = α k) α j) )x α l) y). α k) α l) )α j) + α j) α l) )α k) = α l) α k) α l) α j) ) Divide by α k) α l) )β k) γ j) /γ k) to get = α k) α j) )α l) where and ηk) β j) γ k) γ j) β αj) α l) γ k) k) α k) α l) γ = αk) α j) )βl) γ k) j) α k) α l) β k) γ j) ) b η k) br r α η r j) r+ = λ η j) λ = α r+ = αj) α l) γ k) α k) α l) γ j) α k) α j) ) β l) γ k) α k) α l) β k) γ j) Put α i = ηk) i η j) i Thus α b αbr r α r+ = + λ Thus taking the principal branch of the logarithm so logα b α br r α r+) = log + λ ) α r+ b log α + + b r α r log α r+ b r+ log ) = log + λ ) α r+ and here b r+ b + + b r+ + r + ) max i=,...,r b i r + )B and we have introduced this factor since we have taken the principal branch of the logarithm. Put Λ = b log α + + b r log α r log α r+ b r+ log ). Since λ 0 we see that Λ 0. Put A r+ = maxhα r+ ), e). Put K = Qα k), α j), α l) ). Notice that d = [K : Q] n 3. By Thrm, Λ > exp c 6 log A r+ logr + )B). 6)