Elements of Algebra

582 ADDITIONS. CHAP. VIII. way, we may run the risk of never arriving at the solution of the equation proposed. Let us take the example of Art. 101 of that chapter, in which it is required to resolve an equation of this form, $p^2 = 6q^2 + 1$, or $p^2 - 6q^2 = 1$. We have $p = \sqrt{}(6q^2+1)$; and, neglecting the constant term 1, $p = q\sqrt{6}$; wherefore $\frac{p}{q} = \sqrt{6} > 2, < 3$. Let us take the limit in minus, and make $\mu = 2$, and then $p = 2q+r$; substituting this value, therefore, we shall have $-2q^2 + 4qr + r^2 = 1$; whence, $q = \frac{2r + \sqrt{}(6r^2 - 2)}{2}$; or, rejecting the constant term $-2$, $q = \frac{2r+r\sqrt{6}}{2}$; whence, $\frac{q}{r} = \frac{2+\sqrt{6}}{2} > 2, < 3$. Let us again take the limit in minus, and make $q=2r+s$; the last equation will then become $r^2 - 4rs - 2s^2 = 1$; where we at once perceive, that we may suppose $s = 0$, and $r = 1$; so that we shall have $q = 2$, and $p = 5$. Let us now resume the former transformation, $-2q^2 + 4qr + r^2 = 1$, where we found $\frac{q}{r} > 2$, and $< 3$; and, instead of taking the limit in minus, let us take it in plus, that is to say, let us suppose $q = 3r + s$; or, since $s$ must then be a negative quantity, $q = 3r - s$, we shall then have the following transformation, $-5r^2 + 8rs - 2s^2 = 1$, which will give $r = \frac{4s + \sqrt{}(6s^2 - 5)}{5}$; wherefore, neglecting the constant term 5, $r = \frac{4s+s\sqrt{6}}{5}$, and $\frac{r}{s} = \frac{4+\sqrt{6}}{5} > 1$, and $< 2$. Let us again take the limit in plus, and make $r = 2s-t$, we shall now have $-6s^2 + 12st - 5t^2 = 1$; therefore $s = \frac{6t + \sqrt{}(6t^2 - 6)}{6}$; so that, rejecting the term $-6$, $s = \frac{6t+t\sqrt{6}}{6}$, and $\frac{s}{t} = 1 + \frac{\sqrt{6}}{6} > 1, < 2$. Let us continue taking the limits in plus, and make $s = 2t - u$, we shall next have $-5t^2 + 12tu - 6u^2 = 1$; wherefore, $t = \frac{6u + \sqrt{}(6u^2 - 5)}{5}$; and $\frac{t}{u} = \frac{6+\sqrt{6}}{5} > 1, < 2$.