Elements of Algebra

CHAP. XII. OF ALGEBRA. 397 Farther, in order to determine $x$ and $y$, we have the equations $x+y\sqrt{-c} = mp^2+2mpq\sqrt{-c}-mcq^2$, and $x-y\sqrt{-c} = mp^2-2mpq\sqrt{-c}-mcq^2$; in which $x$ is necessarily equal to the rational part, and $y\sqrt{-c}$ to the irrational part; so that $x = mp^2-mcq^2$, and $y\sqrt{-c} = 2mpq\sqrt{-c}$, or $y=2mpq$; and these are the values of $x$ and $y$ that will transform the expression $x^2+cy^2$ into a square, $m^2(p^2+cq^2)^2$, the root of which is $mp^2+mcq^2$. 183. If the numbers $x$ and $y$ have not a common divisor, we must make $m=1$. Then, in order that $x^2+cy^2$ may become a square, it will be sufficient to make $x=p^2-cq^2$, and $y=2pq$, which will render the formula equal to the square $(p^2+cq^2)^2$. Or, instead of making $x=p^2-cq^2$, we may also suppose $x=cq^2-p^2$, since the square $x^2$ is still left the same. Besides, the same formulæ having been already found by methods altogether different, there can be no doubt with regard to the accuracy of the method which we have now employed. In fact, if we wish to make $x^2+cy^2$ a square, we suppose, by the former method, the root to be $x+\frac{py}{q}$, and find $x^2+cy^2 = x^2+\frac{2pxy}{q}+\frac{p^2y^2}{q^2}$. Expunge the $x^2$, divide the other terms by $y$, multiply by $q^2$, and we shall have $cq^2y = 2pqx+p^2y$; or $cq^2y-p^2y = 2pqx$. Lastly, dividing by $2pq$, and also by $y$, there results $\frac{x}{y} = \frac{cq^2-p^2}{2pq}$. Now, as $x$ and $y$, as well as $p$ and $q$, are to have no common divisor, we must make $x$ equal to the numerator, and $y$ equal to the denominator, and hence we shall obtain the same results as we have already found, namely, $x=cq^2-p^2$, and $y=2pq$. 184. This solution will hold good, whether the number $c$ be positive or negative; but, farther, if this number itself had factors, as, for instance, the formula $x^2+acy^2$, we should not only have the preceding solution, which gives $x=acq^2-p^2$, and $y=2pq$, but this also, namely, $x=cq^2-ap^2$, and $y=2pq$; for, in this last case, we have, as in the other, $x^2+acq^2 = c^2q^4+2acp^2q^2+a^2p^4 = (cq^2+ap^2)^2$; which takes place also when we make $x=ap^2-cq^2$, because the square $x^2$ remains the same.