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1
Content available remote On the number of solutions of simultaneous Pell equations
100%
|
|
tom 101
|
nr 3
215-221
2
Content available remote Simultaneous Pell equations
88%
Acta Arithmetica
|
2004
|
tom 115
|
nr 2
119-131
3
Content available remote Integers not of the form $c(2^a + 2^b)+p^{α}$
88%
Acta Arithmetica
|
2004
|
tom 115
|
nr 1
23-28
4
Content available remote Two conjectures on an addition theorem
63%
Acta Arithmetica
|
2011
|
tom 148
|
nr 4
395-411
5
Content available remote On the Diophantine equation (x² ± C)(y² ± D) = z⁴
63%
Acta Arithmetica
|
2010
|
tom 144
|
nr 1
69-95
6
Content available remote Squares in Lehmer sequences and the Diophantine equation Ax⁴ - By² = 2
63%
Acta Arithmetica
|
2009
|
tom 139
|
nr 3
275-302
7
Content available remote On the diophantine equation $x^y - y^x = c^z$
51%
EN
Applying results on linear forms in p-adic logarithms, we prove that if (x,y,z) is a positive integer solution to the equation $x^y - y^x = c^z$ with gcd(x,y) = 1 then (x,y,z) = (2,1,k), (3,2,k), k ≥ 1 if c = 1, and either $(x,y,z) = (c^k+1,1,k)$, k ≥ 1 or $2 ≤ x < y ≤ max{1.5×10^{10},c}$ if c ≥ 2.
8
Content available remote On a conjecture of Lemke and Kleitman
51%
|
|
nr 3
289-299
EN
Let G be a finite cyclic group of order n ≥ 2. Every sequence S over G can be written in the form $S = (n_1 g)· ... · (n_lg)$ where g ∈ G and $n_1, ..., n_l ∈ [1,ord(g)]$, and the index ind(S) of S is defined as the minimum of $(n_1 + ⋯ + n_l )/ord(g)$ over all g ∈ G with ord(g) = n. In this paper it is shown that any sequence S over G of length |S| ≥ n ≥ 5, 2 ∤ n, having an element with multiplicity at least n/3 has a subsequence T with ind(T) = 1. On the other hand, if n,d ≥ 2 are positive integers with d|n and $n > d²(d³-d²+d+1), we provide an example of a sequence S of length |S| ≥ n having an element with multiplicity l = n/d - d(d-1) - 1 such that S has no subsequence T with ind(T) = 1, giving a general counterexample to a conjecture of Lemke and Kleitman.
9
Content available remote On some special forms of simultaneous Pell equations
51%
10
Content available remote On the Diophantine equation $X^2 - (2^{2m}+1)Y^4 = -2^{2m}$
51%
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