2.II.20G

Number Fields | Part II, 2006

Let K=Q(26)K=\mathbb{Q}(\sqrt{26}) and let ε=5+26\varepsilon=5+\sqrt{26}. By Dedekind's theorem, or otherwise, show that the ideal equations

2=[2,ε+1]2,5=[5,ε+1][5,ε1],[ε+1]=[2,ε+1][5,ε+1]2=[2, \varepsilon+1]^{2}, \quad 5=[5, \varepsilon+1][5, \varepsilon-1], \quad[\varepsilon+1]=[2, \varepsilon+1][5, \varepsilon+1]

hold in KK. Deduce that KK has class number 2 .

Show that ε\varepsilon is the fundamental unit in KK. Hence verify that all solutions in integers x,yx, y of the equation x226y2=±10x^{2}-26 y^{2}=\pm 10 are given by

x+26y=±εn(ε±1)(n=0,±1,±2,).x+\sqrt{26} y=\pm \varepsilon^{n}(\varepsilon \pm 1) \quad(n=0, \pm 1, \pm 2, \ldots) .

[It may be assumed that the Minkowski constant for KK is 12\frac{1}{2}.]

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