Elements of Algebra

562 ADDITIONS. CHAP. VII. $t = Tt' + AV u'$, $u = Tu' + vt'$, and, for the determination of T and v, we shall have the equation $T^2 - AV^2 = 1$, which is similar to the one proposed. Thus, we may suppose $T = t'$, and $v = u'$, which will give $t = t'^2 + Au'^2, u = t'u' + u't'$. Calling $t'', u''$ the second values of $t$ and $u$, we shall have $t'' = t'^2 + Au'^2, u'' = 2t'u'$. Now, it is evident, that we may take these new values $t'', u''$, instead of the first $t', u'$; so that we shall have $t = Tt'' + AV u'',$ $u = Tu'' + vt'',$ where we may again suppose $T = t', v = u'$, which will give $t = t't'' + Au'u'', u = t'u'' + u't''.$ Thus, we shall have new values of $t$ and $u$, which will be $t''' = t't'' + Au'u'' = t'(t'^2 + 3Au'^2)$, $u''' = t'u'' + u't'' = u'(3t'^2 + Au'^2)$, and so on. 75. The foregoing method only enables us to find the values $t'', t''', \ldots$ $u'', u''', \ldots$ successively; let us now consider how this investigation may be generalised. We have first, $t = Tt' + AV u', u = Tu' + vt';$ whence this combination, $t \pm u \sqrt{A} = (t' \pm u' \sqrt{A}) \times (T \pm v \sqrt{A});$ then supposing $T = t'$, and $v = u'$, we shall have $t'' \pm u'' \sqrt{A} = (t' \pm u' \sqrt{A})^2$. Let us now substitute these values of $t''$ and $u''$, instead of those of $t'$ and $u'$, and we shall have $t \pm u \sqrt{A} = (t' \pm u' \sqrt{A})^2 \times (T \pm v \sqrt{A}),$ where, again making $T = t'$, and $v = u'$, and calling $t''', u'''$, the resulting values of $t$ and $u$, there will arise $t''' \pm u''' \sqrt{A} = (t' \pm u' \sqrt{A})^3$. In the same manner, we shall find $t^{iv} \pm u^{iv} \sqrt{A} = (t' \pm u' \sqrt{A})^4$, and so on. Hence, in order to simplify, if we now call T and v the first and the least values of $t, u$, which we before called $t', u'$,