Elements of Algebra

CHAP. IX. OF ALGEBRA. 375 other terms by $y^2$, we shall have $6k^2 + 4y(k^2-2a) = -2k^2 + p^2 - 2kpy$, or $y(4k^2 - 8a + 2kp) = p^2 - 8k^2$; or $p = \frac{2k^2-4a}{k}$, and $pk = 2k^2 - 4a$; so that $y(8k^2 - 16a) = \frac{-4k^4 - 16ak^2 + 16a^2}{k^2}$, and $y = \frac{-k^4 - 4ak^2 + 4a^2}{k^2(2k^2 - 4a)}$. If we now wish to find $x$, we have, first, $1+y = \frac{k^4 - 8ak^2 + 4a^2}{k^2(2k^2 - 4a)}$; and, in the second place, $1-y = \frac{3k^4 - 4a^2}{k^2(2k^2 - 4a)}$; so that $\frac{1+y}{1-y} = \frac{k^4 - 8ak^2 + 4a^2}{3k^4 - 4a^2}$; and, consequently, $x = \frac{h(k^4 - 8ak^2 + 4a^2)}{3k^4 - 4a^2}$; but this is just the same value that we found before, with regard to the even powers of $x$. 140. In order to apply this result to an example, let it be required to make the formula $2x^4 - 1$ a square. Here, we have $a = -1$, and $e=2$; and the known case when the formula becomes a square, is that in which $x=1$; so that $h=1$, and $k^2=1$; that is, $k=1$; therefore, we shall have the new value, $x = \frac{1+8+4}{3-4} = -13$; and since the fourth power of $x$ is found alone, we may also write $x = +13$, whence $2x^4 - 1 = 57121 = (239)^2$. If we now consider this as the known case, we have $h=13$, and $k=239$; and shall obtain a new value of $x$, namely, $\frac{13 \times (239^4 + 8 \times 239^2 + 4)}{3 \times 239^4 - 4} = \frac{42422452969}{9788425919}$. 141. We shall consider, in the same manner, a formula rather more general, $a+cx^2+ex^4$, and shall take for the known case, in which it becomes a square, $x=h$; so that $a+ch^2+eh^4 = k^2$. And, in order to find other values from this, let us