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Displaying similar documents to “The method of infinite ascent applied on A 4 ± n B 3 = C 2

On the Diophantine equation q n - 1 q - 1 = y

Amir Khosravi, Behrooz Khosravi (2003)

Commentationes Mathematicae Universitatis Carolinae

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There exist many results about the Diophantine equation ( q n - 1 ) / ( q - 1 ) = y m , where m 2 and n 3 . In this paper, we suppose that m = 1 , n is an odd integer and q a power of a prime number. Also let y be an integer such that the number of prime divisors of y - 1 is less than or equal to 3 . Then we solve completely the Diophantine equation ( q n - 1 ) / ( q - 1 ) = y for infinitely many values of y . This result finds frequent applications in the theory of finite groups.

A ternary Diophantine inequality over primes

Roger Baker, Andreas Weingartner (2014)

Acta Arithmetica

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Let 1 < c < 10/9. For large real numbers R > 0, and a small constant η > 0, the inequality | p c + p c + p c - R | < R - η holds for many prime triples. This improves work of Kumchev [Acta Arith. 89 (1999)].

On a ternary Diophantine problem with mixed powers of primes

Alessandro Languasco, Alessandro Zaccagnini (2013)

Acta Arithmetica

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Let 1 < k < 33/29. We prove that if λ₁, λ₂ and λ₃ are non-zero real numbers, not all of the same sign and such that λ₁/λ₂ is irrational, and ϖ is any real number, then for any ε > 0 the inequality | λ p + λ p ² + λ p k + ϖ | ( m a x j p j ) - ( 33 - 29 k ) / ( 72 k ) + ε has infinitely many solutions in prime variables p₁, p₂, p₃.

Further remarks on Diophantine quintuples

Mihai Cipu (2015)

Acta Arithmetica

Similarity:

A set of m positive integers with the property that the product of any two of them is the predecessor of a perfect square is called a Diophantine m-tuple. Much work has been done attempting to prove that there exist no Diophantine quintuples. In this paper we give stringent conditions that should be met by a putative Diophantine quintuple. Among others, we show that any Diophantine quintuple a,b,c,d,e with a < b < c < d < e s a t i s f i e s d < 1.55·1072 a n d b < 6.21·1035 w h e n 4 a < b , w h i l e f o r b < 4 a o n e h a s e i t h e r c = a + b + 2√(ab+1)...

The Diophantine equation ( b n ) x + ( 2 n ) y = ( ( b + 2 ) n ) z

Min Tang, Quan-Hui Yang (2013)

Colloquium Mathematicae

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Recently, Miyazaki and Togbé proved that for any fixed odd integer b ≥ 5 with b ≠ 89, the Diophantine equation b x + 2 y = ( b + 2 ) z has only the solution (x,y,z) = (1,1,1). We give an extension of this result.