The fermat equation over quadratic fields

Hao, Fred H.; Parry, Charles J.

doi:10.1016/0022-314x(84)90096-9

Cited by 18 publications

(22 citation statements)

References 9 publications

(8 reference statements)

Supporting

Mentioning

Contrasting

Unclassified

Order By: Relevance

“…L'entier q = 89 est un nombre premier décomposé dans K. Si q est un idéal premier de O K au-dessus de 89, on constate que l'on a a q (E) = −6. D'après les conditions (6) et (7), E 0 a donc bonne réduction en q, d'où v q (abc) = 0. On en déduit que 89 divise W 8 (cf.…”

Section: Le Théorèmeunclassified

Sur le théorème de Fermat sur $${\mathbb Q}\big (\sqrt{5}\big )$$ Q ( 5 )

Kraus

2015

Ann. Math. Québec

View full text Add to dashboard Cite

Soit p un nombre premier impair. En utilisant des arguments modulaires, on établit un critère permettant souvent de démontrer le théorème de Fermat sur le corps quadratique Q √ 5 pour l'exposant p. Il s'exprime en termes du résultant de Wendt des polynômes X n − 1 et (X + 1) n − 1. On en déduit le théorème de Fermat sur ce corps pour p si l'on a 5 ≤ p < 10 7 , et on obtient des résultats analogues au critère de Sophie Germain.Abstract Let p be an odd prime number. Using modular arguments, we give an easy testable condition which allows often to prove Fermat's Last Theorem over the quadratic field Q √ 5 for the exponent p. It is related to Wendt's resultant of the polynomials X n −1 and (X +1) n −1. We deduce Fermat's Last Theorem over this field for p in case one has 5 ≤ p < 10 7 , and we obtain results analogous to Sophie Germain type criteria.

show abstract

Section: Le Théorèmeunclassified

Sur le théorème de Fermat sur $${\mathbb Q}\big (\sqrt{5}\big )$$ Q ( 5 )

Kraus

2015

Ann. Math. Québec

View full text Add to dashboard Cite

show abstract

“…Démonstration : D'après (5), π divise l'un des α + βζ i et pour tout j on a . On a i = j, donc q r divise β.…”

Section: Lemme 1 L'ensemble S N'est Pas Videunclassified

Équation de Fermat et nombres premiers inertes

Kraus

2015

Int. J. Number Theory

View full text Add to dashboard Cite

Abstract. Let K be a number field and p a prime number ≥ 5. Let us denote by µ p the group of the pth roots of unity. We define p to be K-regular if p does not divide the class number of the field K(µ p ). Under the assumption that p is K-regular and inert in K, we establish the second case of Fermat's Last Theorem over K for the exponent p. We use in the proof classical arguments, as well as Faltings' theorem stating that a curve of genus at least two over K has a finite number of K-rational points. Moreover, if K is an imaginary quadratic field, other than Q( √ −3 , we deduce a statement which allows often in practice to prove Fermat's Last Theorem over K for the K-regular exponents.

show abstract

“…Set L=F(z). If A(L)=(0), Hao and Parry have proved that the equation x p +y p +z p =0 has only the trivial solutions in F (see [4]). …”

Section: Then A(l) Is a Z P [G]-module And We Havementioning

confidence: 99%