Elements of Algebra

376 ELEMENTS PART II. suppose $x=h+y$, and our formula will assume the following form: $a$ $ch^2+2chy+cy^2$ $eh^4+4eh^3y+6eh^2y^2+4ehy^3+cy^4$ $k^2+(2ch+4eh^3)y+(c+6eh^2)y^2+4ehy^3+cy^4$. The first term being a square, we shall suppose the root of this formula to be $k+py+qy^2$; and the formula itself will necessarily be equal to the square, $k^2+2kpy+p^2y^2+2kqy^2+2pqy^3+q^2y^4$; then determining $p$ and $q$, in order to expunge the second and third terms, we shall have for this purpose $2ch+4eh^3 = 2kp$; or $p = \frac{ch+2eh^3}{k}$; and $c+6eh^2 = 2kq+p^2$; or $q = \frac{c+6eh^2-p^2}{2k}$. Now, the last two terms of the general equation being divisible by $y^3$, they are reduced to $4eh+ey = 2pq+q^2y$; which gives $y = \frac{4eh-2pq}{q^2-e}$, and, consequently, the value also of $x=h+y$. If we now consider this new case as the given one, we shall find another new case, and may proceed, in the same manner, as far as we please. 142. Let us illustrate the preceding article, by applying it to the formula $1-x^2+x^4$, in which $a=1$, $c=-1$, and $e=1$. The known case is evidently $x=1$; and, therefore, $h=1$, and $k=1$. If we make $x=1+y$, and the square root of our formula $1+py+qy^2$, we must first have $p = \frac{ch+2eh^3}{k} = \frac{-1+2}{1} = 1$, and then $q = \frac{c+6eh^2-p^2}{2k} = \frac{-1+6-1}{2} = \frac{4}{2} = 2$. These values give $y=0$, and $x=1$. Now, this is the known case, and we have not arrived at a new one; but it is because we may prove, from other considerations, that the proposed formula can never become a square, except in the cases of $x=0$, and $x=\pm 1$. 143. Let there be given, also, for an example, the formula $2-3x^2+2x^4$; in which $a=2$, $c=-3$, and $e=2$. The known case is readily found; that is, $x=1$; so that $h=1$, and $k=1$: if, therefore, we make $x=1+y$, and the root $=1+py+qy^2$, we shall have $p=1$, and