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35
A kit on linear forms in three logarithms
"... Abstract. In this paper we give a general presentation of the results to be used to get a ‘good’ lower bound for a linear form in three logarithms of algebraic numbers in the socalled rational case. We recall the best existing general result — Matveev’s theorem — and we add a powerful new lower bou ..."
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Cited by 7 (1 self)
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Abstract. In this paper we give a general presentation of the results to be used to get a ‘good’ lower bound for a linear form in three logarithms of algebraic numbers in the socalled rational case. We recall the best existing general result — Matveev’s theorem — and we add a powerful new lower bound for linear forms in three logarithms. We treat in detail the ‘degenerate ’ case, i.e. the case when the conditions of the zerolemma are not satisfied. 1.
On Fibonacci numbers with few prime divisors
 Proc. Japan Acad
"... Abstract. If n is a positive integer, write Fn for the nth Fibonacci number, and ω(n) for the number of distinct prime divisors of n. We give a description of Fibonacci numbers satisfying ω(Fn) ≤ 2. Moreover, we prove that the inequality ω(Fn) ≥ (log n) log 2+o(1) holds for almost all n. We conjec ..."
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Cited by 6 (0 self)
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Abstract. If n is a positive integer, write Fn for the nth Fibonacci number, and ω(n) for the number of distinct prime divisors of n. We give a description of Fibonacci numbers satisfying ω(Fn) ≤ 2. Moreover, we prove that the inequality ω(Fn) ≥ (log n) log 2+o(1) holds for almost all n. We conjecture that ω(Fn) ≫ log n for composite n, and give a heuristic argument in support of this conjecture. 1.
COVERS OF THE INTEGERS WITH ODD MODULI AND THEIR APPLICATIONS TO THE FORMS x m − 2 n AND x 2 − F3n/2
"... Abstract. In this paper we construct a cover {as(mod ns)} k s=1 of Z with odd moduli such that there are distinct primes p1,...,pk dividing 2n1 −1,...,2nk − 1 respectively. Using this cover we show that for any positive integer m divisible by none of 3, 5, 7, 11, 13 there exists an infinite arithmet ..."
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Cited by 6 (2 self)
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Abstract. In this paper we construct a cover {as(mod ns)} k s=1 of Z with odd moduli such that there are distinct primes p1,...,pk dividing 2n1 −1,...,2nk − 1 respectively. Using this cover we show that for any positive integer m divisible by none of 3, 5, 7, 11, 13 there exists an infinite arithmetic progression of positive odd integers the mth powers of whose terms are never of the form 2n ± pa with a, n ∈{0, 1, 2,...} and p a prime. We also construct another cover of Z with odd moduli and use it to prove that x2 − F3n/2 has at least two distinct prime factors whenever n ∈{0, 1, 2,...} and x ≡ a (mod M), where {Fi}i�0 is the Fibonacci sequence, and a and M are suitable positive integers having 80 decimal digits. 1.
PERFECT POWERS EXPRESSIBLE AS SUMS OF TWO CUBES
, 2009
"... Let n ≥ 3. This paper is concerned with the equation a3 + b3 = cn, which we attack using a combination of the modular approach (via Frey curves and Galois representations) with obstructions to the solutions that are of Brauer–Manin type. We shall show that there are no solutions in coprime, nonzero ..."
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Cited by 5 (4 self)
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Let n ≥ 3. This paper is concerned with the equation a3 + b3 = cn, which we attack using a combination of the modular approach (via Frey curves and Galois representations) with obstructions to the solutions that are of Brauer–Manin type. We shall show that there are no solutions in coprime, nonzero integers a, b, c, for a set of prime exponents n having Dirichlet density 28219 ≈ 0.628, and for a set of exponents n having natural density 1.
A MULTIFREY APPROACH TO SOME MULTIPARAMETER FAMILIES OF DIOPHANTINE EQUATIONS
, 2006
"... We solve several multiparameter families of binomial Thue equations of arbitrary degree; for example, we solve the equation 5 u x n − 2 r 3 s y n = ±1, in nonzero integers x, y and positive integers u, r, s and n ≥ 3. Our approach uses several Frey curves simultaneously, Galois representations an ..."
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Cited by 4 (3 self)
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We solve several multiparameter families of binomial Thue equations of arbitrary degree; for example, we solve the equation 5 u x n − 2 r 3 s y n = ±1, in nonzero integers x, y and positive integers u, r, s and n ≥ 3. Our approach uses several Frey curves simultaneously, Galois representations and levellowering, new lower bounds for linear forms in 3 logarithms due to Mignotte and a famous theorem of Bennett on binomial Thue equations.
Diophantine approximations, Diophantine equations, Transcendence and Applications
 Indian J. Pure Appl. Math
"... This article centres around the contributions of the author and therefore, it is confined to topics where the author has worked. Between these topics there are connections and we explain them by a result of Liouville in 1844 that for an algebraic number α of degree n ≥ 2, there exists c> 0 depend ..."
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Cited by 4 (1 self)
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This article centres around the contributions of the author and therefore, it is confined to topics where the author has worked. Between these topics there are connections and we explain them by a result of Liouville in 1844 that for an algebraic number α of degree n ≥ 2, there exists c> 0 depending only on α such that  α − p c q qn for all rational numbers p q with q> 0. This inequality is from diophantine approximations. Any nontrivial improvement of this inequality shows that certain class of diophantine equations, known as Thue equations, has only finitely many integral solutions. Also, the above inequality can be applied to establish the transcendence of ∞ ∑ 1 numbers like. For an other example on connection between these topics, we 2n! n=1 refer to an account on equation (2) in this article.
Perfect powers from products of terms in Lucas sequences
, 2007
"... Abstract. Suppose that {Un} n≥0 is a Lucas sequence, and suppose that l1,..., lt are primes. We show that the equation Un1 · · · Unm = ±lx1 1 · · · lxt t yp, p prime, m < p, has only finitely many solutions. Moreover, we explain a practical method of solving these equations. For example, if { ..."
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Cited by 4 (3 self)
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Abstract. Suppose that {Un} n≥0 is a Lucas sequence, and suppose that l1,..., lt are primes. We show that the equation Un1 · · · Unm = ±lx1 1 · · · lxt t yp, p prime, m < p, has only finitely many solutions. Moreover, we explain a practical method of solving these equations. For example, if {Fn} n≥0 is the Fibonacci sequence, then we solve the equation Fn1 · · · Fnm = 2 x1 · 3 x2
The modular approach to Diophantine equations
 in Explicit Methods in Number Theory, Panoramas et Synthèses, Société Mathématique De
"... Abstract. The aim of these notes is to communicate Ribet’s Level–Lowering Theorem and related ideas in an explicit and simplified (but hopefully still precise) way, and to explain how these ideas are used to derive information about solutions to Diophantine equations. Contents ..."
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Cited by 3 (2 self)
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Abstract. The aim of these notes is to communicate Ribet’s Level–Lowering Theorem and related ideas in an explicit and simplified (but hopefully still precise) way, and to explain how these ideas are used to derive information about solutions to Diophantine equations. Contents
ON PERFECT POWERS IN LUCAS SEQUENCES
"... Abstract. Let (un)n≥0 be the binary recurrent sequence of integers given by u0 = 0, u1 = 1 and un+2 = 2(un+1 + un). We show that the only positive perfect powers in this sequence are u1 = 1 and u4 = 16. We also discuss the problem of determining perfect powers in Lucas sequences in general. ..."
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Cited by 2 (1 self)
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Abstract. Let (un)n≥0 be the binary recurrent sequence of integers given by u0 = 0, u1 = 1 and un+2 = 2(un+1 + un). We show that the only positive perfect powers in this sequence are u1 = 1 and u4 = 16. We also discuss the problem of determining perfect powers in Lucas sequences in general.
Integral points on hyperelliptic curves
 ALGEBRA NUMBER THEORY
, 2008
"... Let C: Y² = anX n + · · · + a0 be a hyperelliptic curve with the ai rational integers, n ≥ 5, and the polynomial on the right irreducible. Let J be its Jacobian. We give a completely explicit upper bound for the integral points on the model C, provided we know at least one rational point on C an ..."
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Cited by 2 (1 self)
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Let C: Y² = anX n + · · · + a0 be a hyperelliptic curve with the ai rational integers, n ≥ 5, and the polynomial on the right irreducible. Let J be its Jacobian. We give a completely explicit upper bound for the integral points on the model C, provided we know at least one rational point on C and a Mordell–Weil basis for J(Q). We also explain a powerful refinement of the Mordell–Weil sieve which, combined with the upper bound, is capable of determining all the integral points. Our method is illustrated by determining the integral points on the genus 2 hyperelliptic models Y² − Y = X5 − X and ` Y 2