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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 \(p\ge 200\).

MSC:
11R11 Quadratic extensions
11R29 Class numbers, class groups, discriminants
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