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Let a, b, c, x and y be positive integers. In this paper we sharpen a result of Le by showing that the Diophantine equation has at most two positive integer solutions (m,n) satisfying min (m,n)>1.
Let 𝒜={as(mod ns)}ks=0 be a system of residue classes. With the help of cyclotomic fields we obtain a theorem which unifies several previously known results related to the covering multiplicity of 𝒜. In particular, we show that if every integer lies in more than m0=⌊∑ ks=11/ns⌋ members of 𝒜, then for any a=0,1,2,… there are at least subsets I of {1,…,k} with ∑ s∈I1/ns=a/n0. We also characterize when any integer lies in at most m members of 𝒜, where m is a fixed positive integer.
Let u(n)=f(gn), where g > 1 is integer and f(X) ∈ ℤ[X] is non-constant and has no multiple roots. We use the theory of -unit equations as well as bounds for character sums to obtain a lower bound on the number of distinct fields among for n ∈ . Fields of this type include the Shanks fields and their generalizations.
Let a,b,c be relatively prime positive integers such that a2+b2=c2 with b even. In 1956 Jeśmanowicz conjectured that the equation ax+by=cz has no solution other than (x,y,z)=(2,2,2) in positive integers. Most of the known results of this conjecture were proved under the assumption that 4 exactly divides b. The main results of this paper include the case where 8 divides b. One of our results treats the case where a has no prime factor congruent to 1 modulo 4, which can be regarded as a relevant analogue of results due to Deng and Cohen concerning the prime factors of b. Furthermore, we examine parities of the three variables x,y,z, and give new triples a,b,c such that the conjecture holds for the case where b is divisible by 8. In particular, to prove our results, we shall show an important result which asserts that if x,y,z are all even, then x/2,y/2,z/2 are all odd. Our methods are based on elementary congruence and several strong results on generalized Fermat equations given by Darmon and Merel.
We prove that for any positive integers x,d and k with gcd (x,d)=1 and 3<k<35, the product x(x+d)⋯(x+(k−1)d) cannot be a perfect power. This yields a considerable extension of previous results of Győry et al. and Bennett et al., which covered the cases where k≤11. We also establish more general theorems for the case where x can also be a negative integer and where the product yields an almost perfect power. As in the proofs of the earlier theorems, for fixed k we reduce the problem to systems of ternary equations. However, our results do not follow as a mere computational sharpening of the approach utilized previously; instead, they require the introduction of fundamentally new ideas. For k>11, a large number of new ternary equations arise, which we solve by combining the Frey curve and Galois representation approach with local and cyclotomic considerations. Furthermore, the number of systems of equations grows so rapidly with k that, in contrast with the previous proofs, it is practically impossible to handle the various cases in the usual manner. The main novelty of this paper lies in the development of an algorithm for our proofs, which enables us to use a computer. We apply an efficient, iterated combination of our procedure for solving the new ternary equations that arise with several sieves based on the ternary equations already solved. In this way, we are able to exclude the solvability of the enormous number of systems of equations under consideration. Our general algorithm seems to work for larger values of k as well, although there is, of course, a computational time constraint.
An integer may be represented by a quadratic form over each ring of p-adic integers and over the reals without being represented by this quadratic form over the integers. More generally, such failure of a local-global principle may occur for the representation of one integral quadratic form by another integral quadratic form. We show that many such examples may be accounted for by a Brauer–Manin obstruction for the existence of integral points on schemes defined over the integers. For several types of homogeneous spaces of linear algebraic groups, this obstruction is shown to be the only obstruction to the existence of integral points.
A problem posed in the early eighteenth century asks for right-angled triangles, each of whose sides exceeds double the area by a perfect square. We summarize known results and find such triangles with the smallest possible standard generators.
We revisit recent work of Heath-Brown on the average order of the quantity r(L1(x))⋯r(L4(x)), for suitable binary linear forms L1,…,L4, as x=(x1,x2) ranges over quite general regions in ℤ2. In addition to improving the error term in Heath-Brown’s estimate, we generalise his result to cover a wider class of linear forms.
Generatingfunctions are used to derive formulas for the number of representations of a positive integer by each of the quadratic forms x12+x22+x32+2x42, x12+2x22+2x32+2x42, x12+x22+2x32+4x42 and x12+2x22+4x32+4x42. The formulas show that the number of representations by each form is always positive. Some of the analogous results involving sums of triangular numbers are also given.
For a numerical semigroup S, a positive integer a and anonzero element m of S, we define a new numerical semigroup R(S,a,m) and call it the (a,m)-rotation of S. In this paper we study the Frobenius number and the singularity degree of R(S,a,m). Moreover, we observe that the rotations of ℕ are precisely the modular numerical semigroups.
In this paper we consider systems of diagonal forms with integer coefficients in which each form has a different degree. Every such system has a nontrivial zero in every p-adic field Qp provided that the number of variables is sufficiently large in terms of the degrees. While the number of variables required grows at least exponentially as the degrees and number of forms increase, it is known that if p is sufficiently large then only a small polynomial bound is required to ensure zeros in Qp. In this paper we explore the question of how small we can make the prime p and still have a polynomial bound. In particular, we show that we may allow p to be smaller than the largest of the degrees.
§1. Introduction, By a remarkable result of Erdos and Selfridge [3] in 1975. the diophantine equation
with integers k≥2 and m≥2, has only the trivial solutions. x = −j(j = i, …, m), y = 0. This put an end to the old question whether the product of consecutive positive integers could ever be a perfect power; for a brief account of its history see [7].
Erdős (see [4]) asked if there are infinitely many integers k which are not a difference or a sum of two powers, i.e., if there are infinitely many positive integers k with k≠|um ± vn| for u, v, m, n ε ℤ. This is certainly a very difficult problem. For example, it is known that the Catalan equation, i.e., the equation um − vn = 1 with uv≠0 and min (m, n)≥2 has only finitely many solutions (u, v, m, n), but there is no other positive integer k≥1 for which it is known that the equation
has only finitely many solution (u, v, m, n) with min (m, n)≥2.
Fermat gave the first example of a set of four positive integers {a1, a2, a3, a4} with the property that aiaj+1 is a square for 1≤i<j≤4. His example was {1, 3, 8, 120}. Baker and Davenport [1] proved that the example could not be extended to a set of 5 positive integers such that the product of any two of them plus one is a square. Kangasabapathy and Ponnudurai [6], Sansone [9] and Grinstead [4] gave alternative proofs. The construction of such sets originated with Diophantus who studied the problem when the ai are rational numbers. It is conjectured that there do not exist five positive integers whose pairwise products are all one less than the square of an integer. Recently Dujella [3] proved that there do not exist nine such integers. In this note we address the following related problem. Let V denote the set of pure powers, that is, the set of positive integers of the form xk with x and k positive integers and k>1. How large can a set of positive integers A be if aa′ + 1 is in V whenever a and a′ are distinct integers from A? We expect that there is an absolute bound for |A|, the cardinality of A. While we have not been able to establish this result, we have been able to prove that such sets cannot be very dense.
In this paper we continue the investigation begun in [11]. Let λ1…., λs and μ1, …, μs be real numbers, and define the forms
Further, let τ be a positive real number. Our goal is to determine conditions under which the system of inequalities
has a non-trivial integral solution. As has frequently been the case in work on systems of diophantine inequalities (see, for example, Brüdern and Cook [6] and Cook [7]), we were forced in [11] to impose a condition requiring certain coefficient ratios to be algebraic. A recent paper of Bentkus and Gotze [4] introduced a method for avoiding such a restriction in the study of positivede finite quadratic forms, and these ideas are in fact flexible enough to be applied to other problems. In particular, Freeman [10] was able to adapt the method to obtain an asymptotic lower bound for the number of solutions of a single diophantine inequality, thus finally providing the expected strengthening of a classical theorem of Davenport and Heilbronn [9]. The purpose of the present note is to apply these new ideas to the system of inequalities (1.1).
§1. Introduction. In 1946, Davenport and Heilbronn [9] proved a result which opened up the study of Diophantine inequalities. Suppose that Q(x) is a diagonal quadratic form with non-zero real coefficients in s variables. We write
A new criterion on Catalan's equation is proved by elementary means
This shows, without appealing either to the theory of linear forms in logarithms, or to any computation, that (C) has no solution (x, y, p, q) with min {p, q}≤41, except (3,2, 2, 3).
It is known that a system of r additive equations of degree k with greater than 2rk variables has a non-trivial p-adic solution for all p > k2r + 2. In this paper we consider the same system with more than crk variables, c > 2, and show the existence of a non-trivial solution for all p > r2k2+(2/(c − 2)) if r ≠ 1 and p > k2+(2/(c − 1)) if r = 1.