Elements of Algebra

CHAP. VII. ADDITIONS. 557 Dρ of the series $D'$, $D''$, $D'''$, \ldots any more than the half of the following term $D_{\rho+1}$. Then we may make $D_\rho = L$, $n_\rho = N$, $D_{\rho+1} = M$, and $\theta_\rho = \psi$, $\theta_{\rho+1} = \xi$, or $D_\rho = M$, $D_{\rho+1} = L$, and $\theta_\rho = \xi$, $\theta_{\rho+1} = \psi$. We must always suppose, as we proceed, that we have taken, for $M$, the less of the two num- bers $D_\rho$, $D_{\rho+1}$. 71. The equation, $cy^2 - 2nyz + Dz^2 = 1$, will therefore be reduced to this, $L\xi^2 - 2N\xi\psi + M\psi^2 = 1$, where $N^2 - LM = A$, and where $2N$ is neither $> L$, nor $> M$, (abstracting from the signs). Now, $M$ being the less of the two coefficients $L$ and $M$, let us multiply the whole of the equation by the coefficient $M$; and making $v = M\psi - N\xi$, it is evident, that it will be changed into $v^2 - A\xi^2 = M$, in which we must make a distinction between the two cases of $A$ positive, and $A$ negative. 1st. Let $A$ be negative, and $=-a$ ($a$ being a positive number), the equation will then be $v^2 + a\xi^2 = M$. Now, as $N^2 - LM = A$, we shall have $a = LM - N^2$; whence we immediately perceive, that the numbers $L$ and $M$ must have the same signs; otherwise, $2N$ can neither be $> L$, nor $> M$; wherefore $N^2$ will not be $> \frac{LM}{4}$; therefore, $a$ =, or $> \frac{3}{4}LM$; and since $M$ is supposed to be less than $L$, or at least not greater than $L$, we shall have, *à fortiori*, $a$ =, or $> \frac{3}{4}M^2$; whence $M$ =, or $< \sqrt{\frac{4a}{3}}$; and $M < \frac{2}{\sqrt{3}}\sqrt{a}$. Hence, we see that the equation, $v^2 + a\xi^2 = M$, could not exist on the supposition of $v$ and $\xi$ being whole numbers, unless we made $\xi = 0$, and $v^2 = M$, which requires $M$ to be a square number. Let us, therefore, suppose $M = \mu^2$, and we shall have $\xi = 0$, $v = \pm \mu$; wherefore, from the equation, $v = M\psi - N\xi$, we shall have $\mu^2\psi = \pm \mu$, and, consequently, $\psi = \pm \frac{1}{\mu}$; so that $\psi$ cannot be a whole number, as it ought, by the hypothesis, unless $\mu$ be equal to unity, or $= \pm 1$, and, con- sequently, $M = 1$. Hence, therefore, we may infer, that the given equation is