Elements of Algebra

240 ELEMENTS SECT. IV. Let $x$ be the greater of the two numbers, and $y$ the less; then the three following expressions must be equal to one another: namely, the sum, $x+y$; the product, $xy$; and the difference of the squares, $x^2-y^2$. If we compare the first with the second, we have $x+y=xy$; which will give a value of $x$: for $y = xy-x = (y-1)x$ and $x = \frac{y}{y-1}$; consequently, $x+y = \frac{y}{y-1}+y = \frac{y^2}{y-1}$, and $xy = \frac{y^2}{y-1}$; that is to say, the sum is equal to the product; and to this also the difference of the squares ought to be equal. Now, we have $x^2-y^2 = \frac{y^2}{y^2-2y+1} - y^2 = \frac{-y^4+2y^3}{y^2-2y+1}$; so that making this equal to the quantity found $\frac{y^2}{y-1}$, we have $\frac{y^2}{y-1} = \frac{-y^4+2y^3}{y^2-2y+1}$; dividing by $y^2$, we have $\frac{1}{y-1} = \frac{-y^2+2y}{y^2-2y+1}$; and multiplying by $y^2-2y+1$, or $(y-1)^2$, we have $y-1 = -y^2+2y$; consequently, $y^2=y+1$; which gives $y = \frac{1}{2} \pm \sqrt{(\frac{1}{4}+1)} = \frac{1}{2} \pm \sqrt{\frac{5}{4}}$; or $y=\frac{1\pm\sqrt{5}}{2}$, and since $x=\frac{y}{y-1}$, we shall have, by substitution, and using the sign $+$, $x = \frac{\sqrt{5}+1}{\sqrt{5}-1}$. In order to remove the surd quantity from the denominator, multiply both terms by $\sqrt{5}+1$, and we obtain $x = \frac{6+2\sqrt{5}}{4} = \frac{3+\sqrt{5}}{2}$. Therefore the greater of the numbers sought, or $x$, $= \frac{3+\sqrt{5}}{2}$; and the less, $y, = \frac{1+\sqrt{5}}{2}$. Hence their sum $x+y=2+\sqrt{5}$; their product $xy = 2+\sqrt{5}$; and since $x^2 = \frac{7+3\sqrt{5}}{2}$, and $y^2 = \frac{3+\sqrt{5}}{2}$, we have also the difference of the squares $x^2-y^2 = 2+\sqrt{5}$, being all the same quantity. 686. As this solution is very long, it is proper to remark