Elements of Algebra

398 ELEMENTS PART II. This new solution is also obtained from the last method, in the following manner: If we make $x + y\sqrt{-ac} = (p\sqrt{a} + q\sqrt{-c})^2$, and $x - y\sqrt{-ac} = (p\sqrt{a} - q\sqrt{-c})^2$, we shall have $x^2 + acy^2 = (ap^2 + cq^2)^2$, and, consequently, equal to a square. Farther, because $x + y\sqrt{-ac} = ap^2 + 2pq\sqrt{-ac} - cq^2$, and $x - y\sqrt{-ac} = ap^2 - 2pq\sqrt{-ac} - cq^2$, we find $x = ap^2 - cq^2$, and $y = 2pq$. It is farther evident, that if the number $ac$ be resolvible into two factors, in a greater number of ways, we may also find a greater number of solutions. 185. Let us illustrate this by means of some determinate formulæ; and, first, if the formula $x^2 + y^2$ must become a square, we have $ac = 1$; so that $x = p^2 - q^2$, and $y = 2pq$; whence it follows that $x^2 + y^2 = (p^2 + q^2)^2$. If we would have $x^2 - y^2 = \square$; we have $ac = -1$; so that we shall take $x = p^2 + q^2$, and $y = 2pq$, and there will result $x^2 - y^2 = (p^2 - q^2)^2 = \square$. If we would have the formula $x^2 + 2y^2 = \square$, we have $ac = 2$; let us therefore take $x = p^2 - 2q^2$, or $x = 2p^2 - q^2$, and $y = 2pq$, and we shall have $x^2 + 2y^2 = (p^2 + q^2)^2$, or $x^2 + 2y^2 = (2p^2 + q^2)^2$. If, in the fourth place, we would have $x^2 - 2y^2 = \square$, in which $ac = -2$, we shall have $x = p^2 + 2q^2$, and $y = 2pq$; therefore $x^2 - 2y^2 = (p^2 - 2q^2)^2$. Lastly, let us make $x^2 + 6y^2 = \square$. Here we shall have $ac = 6$; and, consequently, either $a = 1$, and $c = 6$, or $a = 2$, and $c = 3$. In the first case, $x = p^2 - 6q^2$, and $y = 2pq$; so that $x^2 + 6y^2 = (p^2 + 6q^2)^2$; in the second, $x = 2p^2 - 3q^2$, and $y = 2pq$; whence $x^2 + 6y^2 = (2p^2 + 3q^2)^2$. 186. But let the formula $ax^2 + cy^2$ be proposed to be transformed into a square. We know beforehand, that this cannot be done, except we already know a case, in which this formula really becomes a square; but we shall find this given case to be, when $x = f$, and $y = g$; so that $af^2 + cg^2 = h^2$; and we may observe, that this formula can be transformed into another of the form $t^2 + acu^2$, by making $t = \frac{afx+cgy}{h}$, and $u = \frac{gx-fy}{h}$; for if $t^2 = \frac{a^2f^2x^2 + 2acfgxy + c^2g^2y^2}{h^2}$, and