Elements of Algebra

518 ADDITIONS. CHAP. II. so that by continuing the series $P^0, P', P''$, \ldots the term, $P^0 = 1$, will necessarily return after a certain interval of terms; consequently, we may always find an infinite number of values for $p$ and $q$, which will render the formula $p^2 - \kappa q^2$ equal to unity. 38. Corollary 4. We may likewise demonstrate this theorem: If the equation $p^2 - \kappa q^2 = \pm H$ be resolvible in integer numbers, by supposing $\kappa$ a positive number, not square, and $H$ a positive number, less than $\sqrt{\kappa}$, the numbers $p$ and $q$ must be such, that $\frac{p}{q}$ may be one of the principal fractions converging to the value of $\sqrt{\kappa}$. Let us suppose that the upper sign must take place, so that $p^2 - \kappa q^2 = H$; wherefore, we shall have $p - q\sqrt{\kappa} = \frac{H}{p+q\sqrt{\kappa}}$, and $\frac{p}{q} - \sqrt{\kappa} = \frac{H}{q^2(\frac{p}{q} + \sqrt{\kappa})}$. Now, let us seek two integer positive numbers, $r$ and $s$, less than $p$ and $q$, and such, that $ps - qr = 1$, which is always possible, as we have demonstrated (Art. 23), and we shall have $\frac{p}{q} - \frac{r}{s} = \frac{1}{qs}$; subtracting this equation from the preceding, we shall have $\frac{r}{s} - \sqrt{\kappa} = \frac{H}{q^2(\frac{p}{q} + \sqrt{\kappa})} - \frac{1}{qs}$; so that we have $$ p - q\sqrt{\kappa} = \frac{H}{q(\frac{p}{q} + \sqrt{\kappa})}, $$ $$ r - s\sqrt{\kappa} = \frac{1}{q}(\frac{sH}{q(\frac{p}{q} + \sqrt{\kappa})} - 1). $$ Now, as $\frac{p}{q} > \sqrt{\kappa}$, and $H < \sqrt{\kappa}$, it is evident, that $\frac{H}{\frac{p}{q} + \sqrt{\kappa}}$ will be $<\frac{1}{2}$; whence $p - q\sqrt{\kappa}$ will be $<\frac{1}{2q}$; wherefore, $\frac{sH}{q(\frac{p}{q} + \sqrt{\kappa})}$ will much more be $<\frac{1}{2}$, since $s<q$; so that $r - s\sqrt{\kappa}$ will be a negative quantity, which taken