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A complete determination of the complex quadratic fields of class-number one. (English) Zbl 0148.27802
Let $$h(d)$$ denote the class number of the quadratic field $$K(\sqrt d)$$. The old question as to whether we can have $$h(d)=1$$ with $$d<-163$$ is here answered in the negative. The main lines of the argument are as follows. We need consider only $$-d = p$$, where $$p$$ is a prime congruent to $$19\pmod{24}$$: we assume that $$h(-p) = 1$$. From a formula for the class number [E. Landau, Vorlesungen über Zahlentheorie, Satz 897] it is proved that $$h(-8p)= 4N + 2$$ and $$h(-12p) = 8M + 4$$ where $$M, N$$ are non-negative integers. Let $$r_1 = 2+\sqrt 3$$, $$R_1 = 1+\sqrt 2$$, $$w_n = \tfrac12 (r_1-1)r_1^n + \tfrac12 (r_1^{-1}-1)r_1^{-n}$$, $$y_n= (R_1^n-(-R_1)^{-n})/2\sqrt 2$$, $$z_n = (R_1^n+ (-R_1)^{-n})/2$$: then the $$w$$’s, $$y$$’s and $$z$$’s are integers. Using results on $$L$$-series it is shown that, for sufficiently large $$p$$, $$z_{2N+1}-4y_N=(w_M+1)^3+3 \quad\text{if }\;\frac{p+1}{4}\equiv 1\pmod 8 \tag1$$ with similar equations for the other residues of $$(p+1)/4$$ modulo 8. By detailed consideration of the error terms in the series used, it is shown that $$p\ge 200$$ is sufficiently large. Use of the properties of the sequences of $$y$$’s and $$z$$’s leads to one of the Diophantine equations $$8x^6 \pm 1 = y^2$$ or $$x^6 \pm 1 = 2y^2$$: the number of solutions of these is finite and we find that (1) and the equations similar to it have no solutions if $$p\ge 200$$.
Show Scanned Page ##### MSC:
 11R11 Quadratic extensions 11R29 Class numbers, class groups, discriminants
##### Keywords:
complex quadratic fields of class-number one
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