Elements of Algebra

CHAP. VII. OF ALGEBRA. 233 For example, let the hexagonal number be 1225, its root will be $x = \frac{1+\sqrt{9801}}{4} = \frac{1+99}{4} = 25$. 666. Suppose $a$ an heptagonal number, of which the root is required. Let this root be $x$, then we shall have $\frac{5x^2-3x}{2} = a$, or $x^2 = \frac{3}{5}x + \frac{2}{5}a$, which gives $x = \frac{3}{10} + \sqrt{(\frac{9}{100} + \frac{2}{5}a)} = \frac{3+\sqrt{(40a+9)}}{10}$; therefore the heptagonal numbers have this property, that if they be multiplied by 40, and 9 be added to the product, the sum will always be a square. Let the heptagon, for example, be 2059; its root will be found $x = \frac{3+\sqrt{(82369)}}{10} = \frac{3+287}{10} = 29$. 667. Let us suppose $a$ an octagonal number, of which the root $x$ is required. We shall here have $3x^2 - 2x = a$, or $x^2 = \frac{2}{3}x + \frac{1}{3}a$, whence results $x = \frac{1}{3} + \sqrt{(\frac{1}{9} + \frac{1}{3}a)} = \frac{1+\sqrt{(3a+1)}}{3}$. Consequently, all octagonal numbers are such, that if multiplied by 3, and unity be added to the product, the sum is constantly a square. For example, let 3816 be an octagon; its root will be $x = \frac{1+\sqrt{11449}}{3} = \frac{1+107}{3} = 36$. 668. Lastly, let $a$ be a given $n$-gonal number, the root of which it is required to assign; we shall then, by the last formula, have this equation: $\frac{(n-2)x^2-(n-4)x}{2} = a$, or $(n-2)x^2 - (n-4)x = 2a$; consequently, $x^2 = \frac{(n-4)x}{n-2} + \frac{2a}{n-2}$; whence, $x = \frac{n-4}{2(n-2)} + \sqrt{(\frac{(n-4)^2}{4(n-2)^2} + \frac{2a}{n-2})}$, or $x = \frac{n-4}{2(n-2)} + \sqrt{(\frac{(n-4)^2}{4(n-2)^2} + \frac{8(n-2)a}{4(n-2)^2})}$, or $x = \frac{n-4+\sqrt{(8(n-2)a+(n-4)^2)}}{2(n-2)}$.