Elements of Algebra

CHAP. VIII. OF ALGEBRA. 241 that it may be abridged. In order to which, let us begin with making the sum $x+y$ equal to the difference of the squares $x^2-y^2$; we shall then have $x+y=x^2-y^2$; and dividing by $x+y$, because $x^2-y^2 = (x+y) \times (x-y)$, we find $1=x-y$, and $x=y+1$. Consequently, $x+y = 2y+1$, and $x^2-y^2=2y+1$; farther, as the product $xy$, or $y^2+y$, must be equal to the same quantity, we have $y^2+y=2y+1$, or $y^2=y+1$, which gives, as before, $y=\frac{1+\sqrt{5}}{2}$. 687. The preceding question leads also to the solution of the following. 5. To find two numbers, such, that their sum, their product, and the sum of their squares, may be all equal. Let the numbers sought be represented by $x$ and $y$; then there must be an equality between $x+y$, $xy$, and $x^2+y^2$. Comparing the first and second quantities, we have $x+y=xy$, whence $x=\frac{y}{y-1}$; consequently, $xy$, and $x+y=\frac{y^2}{y-1}$. Now, the same quantity is equal to $x^2+y^2$; so that we have $\frac{y^2}{y^2-2y+1} + y^2 = \frac{y^2}{y-1}$. Multiplying by $y^2-2y+1$, the product is $y^4-2y^3+2y^2=y^3-y^2$, or $y^4=3y^3-3y^2$; and dividing by $y^2$, we have $y^2=3y-3$; which gives $y=\frac{3}{2}\pm\sqrt{(\frac{9}{4}-3)} = \frac{3+\sqrt{-3}}{2}$; consequently, $y-1=\frac{1+\sqrt{-3}}{2}$, whence results $x=\frac{3}{1+\sqrt{-3}}$; and multiplying both terms by $1-\sqrt{-3}$, the result is $x=\frac{6-2\sqrt{-3}}{4}$, or $\frac{3-\sqrt{-3}}{2}$. Therefore the numbers sought are $x=\frac{3-\sqrt{-3}}{2}$, and $y=\frac{3+\sqrt{-3}}{2}$, the sum of which is $x+y=3$, their product $xy=3$; and lastly, since $x^2=\frac{3-3\sqrt{-3}}{2}$, and R