Elements of Algebra

356 ELEMENTS PART II. $r = \frac{s + \sqrt{(13s^2+4)}}{4}$. But here $r > \frac{s+3s}{4}$, or $r > s$; wherefore let $r = s + t$, and we shall have $3s + 4t = \sqrt{(13s^2+4)}$, and $13s^2 + 4 = 9s^2 + 24st + 16t^2$; so that $4s^2 = 24st + 16t^2 - 4$, and $s^2 = 6ts + 4t^2 - 1$; therefore $s = 3t + \sqrt{(13t^2 - 1)}$. Here we have $s > 3t + 3t$, or $s > 6t$; we must therefore make $s = 6t + u$; whence $3t + u = \sqrt{(13t^2 - 1)}$, and $13t^2 - 1 = 9t^2 + 6tu + u^2$; then $4t^2 = 6tu + u^2 + 1$; and, lastly, $t = \frac{3u + \sqrt{(13u^2+4)}}{4}$, or $t > \frac{6u}{4}$, and $t > u$. If, therefore, we make $t = u + v$, we shall have $u + 4v = \sqrt{(13u^2 + 4)}$, and $13u^2 + 4 = u^2 + 8uv + 16v^2$; therefore $12u^2 = 8uv + 16v^2 - 4$, or $3u^2 = 2uv + 4v^2 - 1$; lastly, $u = \frac{v + \sqrt{(13v^2 - 3)}}{3}$, or $u > \frac{4v}{3}$, or $u > v$. Let us, therefore, make $u = v + x$, and we shall have $2v + 3x = \sqrt{(13v^2 - 3)}$, and $13v^2 - 3 = 4v^2 + 12vx + 9x^2$; or $9v^2 = 12vx + 9x^2 + 3$, or $3v^2 = 4vx + 3x^2 + 1$, and $v = \frac{2x + \sqrt{(13x^2+3)}}{3}$; so that $v > \frac{5}{3}x$, and $v > x$. Let us now suppose $v = x + y$, and we shall have $x + 3y = \sqrt{(13x^2 + 3)}$, and $13x^2 + 3 = x^2 + 6xy + 9y^2$, or $12x^2 = 6xy + 9y^2 - 3$, and $4x^2 = 2xy + 3y^2 - 1$; whence $x = \frac{y + \sqrt{(13y^2 - 4)}}{4}$, and, consequently, $x > y$. We shall, therefore, make $x = y + z$, which gives $3y + 4z = \sqrt{(13y^2 - 4)}$, and $13y^2 - 4 = 9y^2 + 24zy + 16z^2$, or $4y^2 = 24zy + 16z^2 + 4$; therefore $y^2 = 6yz + 4z^2 + 1$, and $y = 3z + \sqrt{(13z^2 + 1)}$. This formula being at length similar to the first, we may take $z = 0$, and go back as follows: