Elements of Algebra

CHAP. VII. ADDITIONS. 563 we shall have, in general, $t \pm u \sqrt{A} = (T \pm v \sqrt{A})^m$, $m$ being any positive whole number; whence, on account of the ambiguity of the signs, we derive $t = \frac{(T+v\sqrt{A})^m + (T-v\sqrt{A})^m}{2}$ $u = \frac{(T+v\sqrt{A})^m - (T-v\sqrt{A})^m}{2\sqrt{A}}$ Though these expressions appear under an irrational form, it is easy to see that they will become rational, if we involve the powers of $T \pm v \sqrt{A}$; for it is well known that $(T \pm v \sqrt{A})^m = T^m \pm m T^{m-1} v \sqrt{A} + \frac{m(m-1)}{2} T^{m-2} v^2 A$ $+ \frac{m(m-1)\times(m-2)}{2\times3} T^{m-3} v^3 A \sqrt{A} +, \&c.$ Wherefore, $t = T^m + \frac{m(m-1)}{2} A T^{m-2} v^2$ $+ \frac{m(m-1)\times(m-2)\times(m-3)}{2\times3\times4} A^2 T^{m-4} v^4 +, \&c.$ $u = m T^{m-1} v + \frac{m(m-1)\times(m-2)}{2\times3} A T^{m-3} v^3$ $+ \frac{m(m-1)\times(m-2)\times(m-3)\times(m-4)}{2\times3\times4\times5} A^2 T^{m-5} v^5 +, \&c.$ Where we may take for $m$ any positive whole numbers whatever. It is evident that, by successively making $m = 1, 2, 3, 4,$ &c. we shall have values of $t$ and $u$, that will go on increasing. I shall now shew that, in this manner, we may obtain all the possible values of $t$ and $u$, provided $T$ and $v$ are the least of them. For this purpose, it is sufficient to prove, that, between the values of $t$ and $u$, which answer to $m$, any number whatever, and those which would answer to the number, $m + 1$, it is impossible to find any intermediate values, that will satisfy the equation $t^2 - A u^2 = 1$. For example, let us make the values $t'''$, $u'''$, which result from the supposition of $m = 3$, and the values $t^{iv}$, $u^{iv}$, which result from the supposition of $m = 4$, and let us suppose it possible that there are other intermediate values, $\theta$ and $\upsilon$, which would likewise satisfy the equation $t^2 - A u^2 = 1$. O o 2