Elements of Algebra

372 ELEMENTS PART II. $d = 0$, we have, by both methods, $p = 0$; the former giving $x = \frac{c-2fg}{0}$, and the other giving $x = 0$; neither of which are proper for furnishing any further conclusions. 137. These then are the three formulæ, to which the methods hitherto explained may be applied; and, if in the formula proposed neither term be a square, no success can be expected, until we have found one such value of $x$ as will make the formula a square. Let us suppose, therefore, that our formula becomes a square in the case of $x = h$, or that $a + bh + ch^2 + dh^3 + eh^4 = k^2$; if we make $x = h + y$, we shall have a new formula, the first term of which will be $k^2$; that is to say, a square, which will, consequently, fall under the first case: and we may also use this transformation, after having determined by the preceding methods one value of $x$, for instance, $x = h$; for we have then only to make $x = h + y$, in order to obtain a new equation, with which we may proceed in the same manner. And the values of $x$, that may thus be found, will furnish new ones; which will also lead to others, and so on. 138. But it is to be particularly remarked, that we can in no way hope to resolve those formulæ in which the second and fourth terms are wanting, until we have found one solution; and, with regard to the process that must be followed after that, we shall explain it by applying it to the formula $a + ex^4$, which is one of those that most frequently occur. Suppose, therefore, we have found such a value of $x = h$, that $a + eh^4 = k^2$; then if we would find, from this, other values of $x$, we must make $x = h + y$, and the following formula, $a + eh^4 + 4eh^3y + 6eh^2y^2 + 4ehy^3 + ey^4$, must be a square. Now, this formula being reducible to $k^2 + 4eh^3y + 6eh^2y^2 + 4ehy^3 + ey^4$, it therefore belongs to the first of our three cases; so that we shall represent its square root by $k + py + qy^2$; and, consequently, the formula itself will be equal to the square $k^2 + 2kpy + p^2y^2 + 2kqy^2 + 2pqy^3 + q^2y^4$; from which we must first remove the second term by determining $p$, and consequently $q$; that is to say, by making $4eh^3 = 2kp$, or $p = \frac{2eh^3}{k}$; and $6eh^2 = 2kq + p^2$, or