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A “class group” obstruction for the equation C y d = F ( x , z )

Denis Simon (2008)

Journal de Théorie des Nombres de Bordeaux

In this paper, we study equations of the form C y d = F ( x , z ) , where F [ x , z ] is a binary form, homogeneous of degree n , which is supposed to be primitive and irreducible, and d is any fixed integer. Using classical tools in algebraic number theory, we prove that the existence of a proper solution for this equation implies the existence of an integral ideal of given norm in some order in a number field, and also the existence of a specific relation in the class group involving this ideal. In some cases, this result...

A variety of Euler's sum of powers conjecture

Tianxin Cai, Yong Zhang (2021)

Czechoslovak Mathematical Journal

We consider a variety of Euler’s sum of powers conjecture, i.e., whether the Diophantine system n = a 1 + a 2 + + a s - 1 , a 1 a 2 a s - 1 ( a 1 + a 2 + + a s - 1 ) = b s has positive integer or rational solutions n , b , a i , i = 1 , 2 , , s - 1 , s 3 . Using the theory of elliptic curves, we prove that it has no positive integer solution for s = 3 , but there are infinitely many positive integers n such that it has a positive integer solution for s 4 . As a corollary, for s 4 and any positive integer n , the above Diophantine system has a positive rational solution. Meanwhile, we give conditions such that...

Almost powers in the Lucas sequence

Yann Bugeaud, Florian Luca, Maurice Mignotte, Samir Siksek (2008)

Journal de Théorie des Nombres de Bordeaux

The famous problem of determining all perfect powers in the Fibonacci sequence ( F n ) n 0 and in the Lucas sequence ( L n ) n 0 has recently been resolved [10]. The proofs of those results combine modular techniques from Wiles’ proof of Fermat’s Last Theorem with classical techniques from Baker’s theory and Diophantine approximation. In this paper, we solve the Diophantine equations L n = q a y p , with a > 0 and p 2 , for all primes q < 1087 and indeed for all but 13 primes q < 10 6 . Here the strategy of [10] is not sufficient due to the sizes of...

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