Elements of Algebra

554 ADDITIONS. CHAP. VII. $Q'' = \mu' P' + Q'$, $P'' = \frac{\overset{''}{Q}^2-A}{P'}$, $\mu'' \angle \frac{-Q'' \pm \sqrt{A}}{P''}$. $Q''' = \mu'' P'' + Q''$, $P''' = \frac{\overset{'''}{Q}^2-A}{P''}$, $\mu''' \angle \frac{-Q''' \mp \sqrt{A}}{P'''}$. $\ldots$ $\ldots$ $\ldots$ and we shall only continue these series until two corresponding terms of the first and the second series appear again together; then, if among the terms of the second series, $P^0, P', P'', \ldots$ there be found one positive, and equal to unity, this term will give a solution of the proposed equation; and the values of $y$ and $z$ will be the corresponding terms of the two series $p^0, p', p'', \ldots$ and $q^0, q', q'', \ldots$ calculated according to the formulæ of Art. 25; otherwise, we may immediately conclude, that the given equation is not resolvible in integer numbers. See the example of Art. 40. Third case, when A is a square. 69. In this case, the quantity $\sqrt{A}$ will become rational, and the quantity $cy^2 - 2nyz + bz^2$ will be resolvible into two rational factors. Indeed, this quantity is no other than $\frac{(cy-nz)^2 - Az^2}{c}$, which, supposing $A = a^2$, may be thrown into this form, $\frac{(cy + (n+a)z) \times (cy + (n-a)z)}{c}$. Now, as $n^2 - a^2 = Ac = (n+a) \times (n-a)$, the product of $n+a$ by $n-a$ must be divisible by $c$; and, consequently, one of these two numbers $n+a$, and $n-a$, must be divisible by one of the factors of $c$, and the other by the other factor. Let us, therefore, suppose $c=bc$, $n+a=fb$, and $n-a=gc$, $f$ and $b$ being whole numbers, and the preceding quantity will become the product of these two linear factors, $cy+fz$, and $by+gz$; therefore, since these two factors are both integers, it is evident that their product could not be $=1$, as the given equation requires, unless each of them were separately $=\pm 1$; we shall therefore make $cy+fz = \pm 1$, and $by+gz = \pm 1$, and by these means we shall determine the numbers $y$ and $z$. If we find these numbers integer, we shall have the solution of the equation proposed; otherwise, it will be irresolvible, at least in whole numbers.