Elements of Algebra

CHAP. XIV. OF ALGEBRA. 419 $z = \frac{24d^2+14cd-24c^2}{25}$; and this value being added to $a = c^2+d^2$, gives $\frac{c^2+14cd+49d^2}{25}$, the square root of which is $\frac{c+7d}{5}$. If we then subtract $z$ from $a$, there remains $\frac{49c^2-14cd+d^2}{25}$, which is the square of $\frac{7c-d}{5}$; the former of these two square roots being $x$, and the latter $y$. 222. Question 7. Required such a number, $x$, that whether we add unity to itself, or to its square, the result may be a square. It is here required to transform the two formulæ $x+1$, and $x^2+1$, into squares. Let us therefore suppose the first $x+1=p^2$; and, because $x=p^2-1$, the second, $x^2+1 = p^4-2p^2+2$, must be a square: which last formula is of such a nature as not to admit of a solution, unless we already know a satisfactory case; but such a case readily occurs, namely, that of $p=1$: therefore let $p=1+q$, and we shall have $x^2+1=1+4q^2+4q^3+q^4$, which may become a square in several ways. 1. If we suppose its root to be $1+q^2$, we shall have $1+4q^2+4q^3+q^4 = 1+2q^2+q^4$; so that $4q+4q^2=2q$, or $4+4q=2$, and $q=-\frac{1}{2}$; therefore $p=\frac{1}{2}$, and $x=-\frac{3}{4}$. 2. Let the root be $1-q^2$, and we shall find $1+4q^2+q^3+q^4 = 1-2q^2+q^4$; consequently, $q=-\frac{3}{2}$, and $p=-\frac{1}{2}$, which gives $x=-\frac{3}{4}$, as before. 3. If we represent the root by $1+2q+q^2$, in order to destroy the first, and the last two terms, we have $1+4q^2+4q^3+q^4 = 1+4q+6q^2+4q^3+q^4$, whence we get $q=-2$, and $p=-1$; and therefore $x=0$. 4. We may also adopt $1-2q-q^2$ for the root, and in this case we shall have $1+4q^2+4q^3+q^4 = 1-4q+2q^2+4q^3+q^4$; but we find, as before, $q=-2$. 5. We may, if we choose, destroy the first two terms, by making the root equal to $1+2q^2$; for we shall then have $1+4q^2+4q^3+q^4 = 1+4q^2+4q^4$; also, $q=\frac{4}{3}$, and $p=\frac{7}{3}$; consequently, $x=\frac{40}{9}$; lastly, $x+1=\frac{49}{9}=(\frac{7}{3})^2$, and $x^2+1=\frac{1681}{81}=(\frac{41}{9})^2$. EE2