Elements of Algebra

CHAP. XII. OF ALGEBRA. 399 $u^2 = \frac{g^2x^2 - 2fgxy + f^2y^2}{h^2}$, we have $t^2 + acu^2 = \frac{a^2f^2x^2 + c^2g^2y^2 + acg^2x^2 + acf^2y^2}{h^2} = \frac{ax^2(af^2 + cg^2) + cy^2(af^2 + cg^2)}{h^2}$; also, since $af^2 + cg^2 = h^2$, we have $t^2 + acu^2 = ax^2 + cy^2$. Thus, we have given easy rules for transforming the expression $t^2 + acu^2$ into a square, to which we have now reduced the formula proposed, $ax^2 + cy^2$. 187. Let us proceed farther, and see how the formula $ax^2 + cy^2$, in which $x$ and $y$ are supposed to have no common divisor, may be reduced to a cube. The rules already given are by no means sufficient for this; but the method which we have last explained applies here with the greatest success: and what is particularly worthy of observation, is, that the formula may be transformed into a cube, whatever numbers $a$ and $c$ are; which could not take place with regard to squares, unless we already knew a case, and which does not take place with regard to any of the other even powers; but, on the contrary, the solution is always possible for the odd powers, such as the third, the fifth, the seventh, &c. 188. Whenever, therefore, it is required to reduce the formula $ax^2 + cy^2$ to a cube, we may suppose, according to the method which we have already employed, that $x\sqrt{a} + y\sqrt{-c} = (p\sqrt{a} + q\sqrt{-c})^3$, and $x\sqrt{a} - y\sqrt{-c} = (p\sqrt{a} - q\sqrt{-c})^3$; the product $(ap^2 + cq^2)^3$, which is a cube, will be equal to the formula $ax^2 + cy^2$. But it is required, also, to determine rational values for $x$ and $y$, and fortunately we succeed. If we actually take the two cubes that have been pointed out, we have the two equations $x\sqrt{a} + y\sqrt{-c} = ap^3\sqrt{a} + 3ap^2q\sqrt{-c} - 3cpq^2\sqrt{a} - cq^3\sqrt{-c}$, and $x\sqrt{a} - y\sqrt{-c} = ap^3\sqrt{a} - 3ap^2q\sqrt{-c} - 3cpq^2\sqrt{a} + cq^3\sqrt{-c}$; from which it evidently follows, that $x = ap^3 - 3cpq^2$, and $y = 3ap^2q - cq^3$. For example, let two squares $x^2$, and $y^2$, be required, whose sum, $x^2 + y^2$, may make a cube. Here, since $a=1$, and $c=1$, we shall have $x = p^3 - 3pq^2$, and $y = 3p^2q - q^3$, which gives $x^2 + y^2 = (p^2 + q^2)^3$. Now, if $p=2$, and $q=1$, we find $x=2$, and $y=11$; wherefore $x^2 + y^2 = 125 = 5^3$.