Proof of some conjectures by Kaplansky (Q2746254)

The author completes several FREs (frequently requested exercises). In brief, set \(\omega_h=(\omega+P_h)/Q_h\) and recall that a typical line in the periodic continued fraction expansion of \(\omega=\sqrt D\) (\(D\) positive and not a square) is of the shape \(\omega_h=a_h-\overline\rho_h\), with \(\rho_h=(\omega+P_{h+1})/Q_h\), and that the tableaux of these lines is order reversed by conjugation, here denoted by the overline, since that yields a typical line \(\rho_h=a_h-\overline\omega_h\); here we have rewritten line zero as \(\omega+a_0=2a_0-(\overline\omega+a_0)\). The integer sequences \((P_h)\) and \((Q_h)\) are thus given by \(P_h+P_{h+1}=aQ_h\) and \(Q_hQ_{h+1}=D-P_{h+1}^2\). Plainly, both \(\omega_h>1\), since \(\omega_h\) is a complete quotient, and \(-1< \overline\rho_h<0\), since \(-\overline\rho_h\) is a remainder. The symmetry provided by conjugation implies that the expansion of \(\omega+a_0\) is pure periodic, that a period of even length \(r=2m\) has \(P_m=P_{m+1}\), and that one of odd length \(r=2m+1\) has \(Q_m=Q_{m+1}\). Finally, if \([a_0,a_1,\ldots,a_h]=x_h/y_h\) denotes the convergents then \(x_h^2-Dy_h^2=(-1)^{h+1}Q_{h+1}\). NEWLINENEWLINENEWLINEExercise 1 remarks that if \(D\equiv 3\pmod 4\) and is prime then the central partial quotient \(a_m\) is very large. Here the period length is even and then \(Q_m\bigm |2P_m\) with \(Q\bigm |(D-P_m^2)\) entails \(Q_m=2\). It follows from the inequalities cited above that \(P_m\) is the largest odd integer less than \(\sqrt D\), and \(a_m=P_m\). NEWLINENEWLINENEWLINEExercise 3 considers \(D=a^2+b^2\), \(D\) prime, and asks one to show there is a solution in integers \(x\) and \(y\) to \(a=x^2-Dy^2\). Then the period length is odd so \(D=P_{m+1}^2+Q_m^2\), for example, \(61=6^2+5^2\). However, contrary to the author, the ideal \(6\mathbb Z+(\sqrt{61}+5)\mathbb Z\) of the order \(\mathbb Z[\sqrt{61}]\) is not principal and there are no integers \(x\), \(y\) so that \(\pm 6=x^2-61y^2\). It is of course true that there is always a solution both to \(x^2-Dy^2=(-1)^hQ_h\), and to \(x^2-Dy^2=(-1)^{r-h}Q_{r-h}=(-1)^{h+1}Q_h\). But what was Kap's question? NEWLINENEWLINENEWLINEExercise 2, not strictly an FRE, asks about the parity of \(x_{r-1}\) in the smallest solution to \(x_{r-1}^2-Dy_{r-1}^2=1\), with \(D=RS\), where \(R>S>1\), and say given information on those \(V\) for which \(RX^2-SY^2=\pm V\) has a solution in integers \(X\), \(Y\). This scenario is the subject of a note of \textit{P. G. Walsh} and the reviewer [Am. Math. Mon. 106, 52-56 (1999; Zbl 0985.11004)]. Its remarks show that if \(V=1\) is admissible then the primitive period length \(r=2m\) of the expansion of \(\sqrt D\) is even and \(x_{m-1}^2-Dy_{m-1}^2=(-1)^m S\). That readily yields \(y_{r-1}=2x_{m-1}y_{m-1}/S\). Thus with \(R\) and \(S\) odd primes, solvability of \(RX^2-SY^2=\pm 1\) entails \(y_{r-1}\) is even and \(x_{r-1}\) is odd. Just so, \(x_{m-1}y_{m-1}/S\) is odd and \(x_{r-1}\) is even if \(RX^2-SY^2=\pm 2\) has a solution. The author contrives conditions on \(R\) and \(S\) exemplified by \(S=3\), \(R=73\) so the prime \(S\equiv 3\pmod 4\) is a square but is not a fourth power modulo the prime \(R\equiv 1\pmod 8\). Then \(73\cdot 1^2-3\cdot 5^2=-2\) entails \(x_{r-1}\) must be even.