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On some special quartic reciprocity laws. (English) Zbl 0215.06503

In a paper [J. Reine Angew. Math. 250, 42–48 (1971; Zbl 0222.12007)] I gave an elementary proof of a theorem of A. Scholz [Math. Z. 39, 95–111 (1934; Zbl 0009.29402)] to the effect that if \(p\equiv q\equiv 1\pmod 4\) are primes which are quadratic residues of each other and if \(\varepsilon_p\) and \(\varepsilon_q\) are fundamental units in the fields \(\mathbb Q(\sqrt p)\) and \(\mathbb Q(\sqrt q)\), then \[ \left(\frac{\varepsilon_p} q\right)=\left(\frac{\varepsilon_q} p\right)=\left(\frac p q\right)_4 \left(\frac q p\right)_4. \]
In the present paper this reciprocity law is expressed in terms of the quadratic partitions of \(p = c^2 + qd^2\), \(p = c^2 + 2qd^2\), \(p = 2c^2 + qd^2\) and \(p = c^2 -qd^2\). Some cases of special interest are those for which the class number \(h(\sqrt{-q})\) or \(h(\sqrt{-2q})\) is small so that the representations exists for a known class of primes. For example if \(q = 5, 13\) or \(37\), then \(p = c^2 + qd^2\) and we have \(\left(\frac{\varepsilon_q} p\right) = (-1)^4\) while for \(p =17, 73, 97\) and \(193\) we have \[ \left(\frac{\varepsilon_q} p\right) =\begin{cases} 1 &\text{ if }p = c^2 + qd^2 \\ -1 &\text{ if }2p = c^2 + qd^2\end{cases} . \] Application is also made to the problem of whether the Pell equation \(t^2 - pqu^2 = -1\) has a solution. It is proved that this equation has no solution if \(p = c^2+qd^2\) with \(d\) odd and \(p\equiv q\equiv 1\pmod 4\) with \(\left(\frac{\varepsilon_q} p\right) =1\).
All the proofs are extremely elementary. Some of the above results appear in a University of Arizona dissertation by J. Brandler, where the proofs involve the theory of quartic fields.
Reviewer: Emma Lehmer

MSC:

11A15 Power residues, reciprocity
11D09 Quadratic and bilinear Diophantine equations
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