Elements of Algebra

CHAP. IX. ADDITIONS. 585 the product sought would be $x''^2 + ax''y'' + by''^2$. In the same manner, we might find the product of four, or of a still greater number of formulæ similar to this, $x^2 + axy + by^2$, and these products likewise will always have the same form. 90. If we make $x' = x$, and $y' = y$, we shall have $X = x^2 - by^2$, $Y = 2xy + ay^2$; and, consequently, $(x^2 + axy + by^2)^2 = X^2 + aXY + bY^2$. Therefore, if we wish to find rational values of x and y, such, that the formula $x^2+axy+by^2$ may become a square, we shall only have to give the preceding values to x and y, and we shall have, for the root of the square, the formula, $x^2 + axy + by^2$; $x$ and $y$ being two indeterminate numbers. If we farther make $x'' = x' = x$, and $y'' = y' = y$, we shall have $x' = xx - byy$, $y' = xy + yx + ayy$; that is, by substituting the preceding values of x and y, $x' = x^3 - 3bxy^2 + aby^3$, $y' = 3x^2y + 3axy^2 + (a^2 - b)y^3$; wherefore, $(x^2 + axy + by^2)^3 = x'^2 + ax'y' + by'^2$. Thus, if we proposed to find the rational values of $x'$ and $y'$, such, that the formula $x'^2 + ax'y' + by'^2$ might become a cube, we should only have to give to $x$ and $y$ the foregoing values, by which means we should have a cube, whose root would be $x^2 + axy + by^2$; $x$ and $y$ being both indeterminate. In a similar manner, we may resolve questions, in which it is required to produce fourth, fifth powers, \ldots but we may, once for all, find general formulæ for any power whatever, $m$, without passing through the lower powers. Let it be proposed, therefore, to find rational values of x and y, such, that the formula, $x^2 + axy + by^2$, may become a power, $m$; that is, let it be required to solve the equation, $x^2 + axy + by^2 = z^m$. As the quantity $x^2 + axy + by^2$ is formed from the product of the two factors, $x + \alpha y$, and $x + \beta y$, in order that