Elements of Algebra

442 ELEMENTS PART II. that this value will lead us to others, if we write $p = 1 + s$; in consequence of which supposition, we have to transform the formula $(m^2 - 1) \times (m^2 - 1 + 4ms + 6m^2s^2 + 4m^2s^3 + m^2s^4)$ into a square; it will be no less a square, if we divide it by $(m^2 - 1)^2$; this division gives us $1 + \frac{4m^2s}{m^2-1} + \frac{6m^2s^2}{m^2-1} + \frac{4m^2s^3}{m^2-1} + \frac{m^2s^4}{m^2-1}$; and if to abridge we make $\frac{m^2}{m^2-1} = a$, we shall have to reduce the formula $1 + 4as + 6as^2 + 4as^3 + as^4$ to a square. Let its root be $1+fs+gs^2$, the square of which is $1 + 2fs + 2gs^2 + f^2s^2 + 2fgs^3 + g^2s^4$, and let us determine $f$ and $g$ in such a manner, that the first three terms may vanish; namely, by making $2f = 4a$, or $f = 2a$, and $6a = 2g + f^2$, or $g = \frac{6a-f^2}{2} = 3a - 2a^2$, the last two terms will furnish the equation $4a + as = 2fg + g^2s$; whence $s = \frac{4a-2fg}{g^2-a} = \frac{4a-12a+8a^3}{4a^2-12a^3+9a^2-a} = \frac{4-12a+8a^2}{4a^3-12a^2+9a-1}$; or, dividing by $a-1$, $s = \frac{4(2a-1)}{4a^2-8a+1}$. This value is already sufficient to give us an infinite number of answers, because the number $m$, in the value of $a$, $a = \frac{m^2}{m^2-1}$, may be taken at pleasure. It will be proper to illustrate this by some examples. 1. Let $m=2$, we shall have $a = \frac{4}{3}$; so that $s = 4 \times \frac{\frac{5}{3}}{-\frac{23}{9}} = -\frac{60}{23}$; whence $p = -\frac{37}{23}$, and $q = -\frac{74}{23}$; lastly, $\frac{x}{z} = \frac{949}{420}$, and $\frac{y}{z} = \frac{6005}{4947}$. 2. If $m=\frac{3}{2}$, we shall have $a = \frac{9}{5}$, and $s = 4 \times \frac{\frac{13}{5}}{-\frac{11}{25}} = -\frac{260}{11}$; consequently, $p = -\frac{249}{11}$, and $q = -\frac{747}{55}$, by which means we may determine the fractions $\frac{x}{z}$, and $\frac{y}{z}$. There is here a particular case that deserves to be at-