Elements of Algebra

CHAP. VIII. OF ALGEBRA. 365 Let us also try the second method, and represent the root by $3+py+qy^2$; this gives $9+12y+6y^2+y^3=9+6py+6qy^2+p^2y^2+2pqy^3+q^2y^4$, in which we must first have $6p=12$, and $p=2$; then $6q+p^2=6q+4=6$, and $q=\frac{1}{3}$; farther, $1=2pq+q^2y=\frac{4}{3}+\frac{1}{9}y$; hence $y=-3$, and, consequently, $x=-1$, and $1+x^3=0$; from which we can draw no further conclusion, because, if we wished to make $x=-1+z$, we should find the formula, $3z-3z^2+z^3$, the first term of which vanishes; so that we cannot make use of either method. We have therefore sufficient grounds to suppose, after what has been attempted, that the formula $1+x^3$ can never become a square, except in these three cases; namely, when 1. $x=0$, 2. $x=-1$, and 3. $x=2$. But of this we may satisfy ourselves from other reasons. 122. Let us consider, for the sake of practice, the formula $1+3x^3$, which becomes a square in the following cases; when 1. $x=0$, 2. $x=-1$, and 3. $x=2$, and let us see whether we shall arrive at other similar values. Since $x=1$ is one of the satisfactory values, let us suppose $x=1+y$, and we shall thus have $1+3x^3=4+9y+9y^2+3y^3$. Now, let the root of this new formula be $2+py$, so that $4+9y+9y^2+3y^3=4+4py+p^2y^2$. We must have $9=4p$, and $p=\frac{9}{4}$, and the other terms will give $9+3y=p^2=\frac{81}{16}$, and $y=-\frac{21}{16}$; consequently, $x=-\frac{5}{16}$, and $1+3x^3$ becomes a square, namely, $\frac{3721}{4096}$, the root of which is $-\frac{61}{64}$, or $+\frac{61}{64}$: and, if we chose to proceed, by making $x=-\frac{5}{16}+z$, we should not fail to find new values. Let us also apply the second method to the same formula, and suppose the root to be $2+py+qy^2$; which supposition gives $4+9y+9y^2+3y^3 = 4+4py+4qy^2+p^2y^2+2pqy^3+q^2y^4$; therefore, we must have $4p=9$, or $p=\frac{9}{4}$, and $4q+p^2=9=4q+\frac{81}{16}$, or $q=\frac{63}{64}$: and the other terms will give $3=2pq+q^2y=\frac{567}{128}+q^2y$, or $567+128q^2y=384$, or $128q^2y=-183$; that is to say, $128 \times (\frac{63}{64})^2 y = -183$, or $\frac{63^2}{32}y=-183$. So that $y=-\frac{1952}{1323}$, and $x=-\frac{639}{1323}$; and these values