Elements of Algebra

568 ADDITIONS. CHAP. VII. Now, I say that the values of $t$ and $u$, which answer to any exponent whatever, $m$, being divided by $\Delta$, will leave the same remainders as the values of $t$ and $u$, which would answer to the exponent $m + 2\mu$. For, denoting these last by $\theta$ and $\upsilon$, we shall have, $t \pm u \sqrt{\Delta} = (T \pm V \sqrt{\Delta})^m$, and $\theta \pm \upsilon \sqrt{\Delta} = (T \pm V \sqrt{\Delta})^{m+2\mu}$; wherefore, $\theta \pm \upsilon \sqrt{\Delta} = (t \pm u \sqrt{\Delta}) \times (T \pm V \sqrt{\Delta})^{2\mu}$, but we have just before found $T'' \pm V'' \sqrt{\Delta} = (T \pm V \sqrt{\Delta})^{2\mu}$; whence we shall have $\theta \pm \upsilon \sqrt{\Delta} = (t \pm u \sqrt{\Delta}) \times (T'' \pm V'' \sqrt{\Delta})$; then, by multiplying and comparing the rational parts, and the irrational parts, respectively, we derive $\theta = tT'' + \Delta uV''$, $\upsilon = tV'' + uT''$. Now, $V''$ is divisible by $\Delta$, and $T''$ leaves the remainder 1; therefore $\theta$ will leave the same remainder as $t$, and $\upsilon$ the same remainder as $u$. In general, therefore, the remainders of the values of $t$ and $u$, corresponding to the exponents $m + 2\mu$, $m + 4\mu$, $m + 6\mu$, &c. will be the same as those of the values, which correspond to any exponent whatever, $m$. Hence, therefore, we may conclude, that, if we wish to have the remainders arising from the division of the terms $t'$, $t''$, $t'''$, &c. and $u'$, $u''$, $u'''$, &c. which correspond to $m = 1$, $2$, $3$, &c. by the number $\Delta$, it will be sufficient to find these remainders as far as the terms $t^{2\mu}$ and $u^{2\mu}$ inclusive; for, after these terms, the same remainders will return in the same order; and so on to infinity. With regard to the terms $t^{2\mu}$ and $u^{2\mu}$, at which we may stop, one of them $u^{2\mu}$ will be exactly divisible by $\Delta$, and the other $t^{2\mu}$ will leave unity for a remainder; so that we shall only have to continue the divisions until we arrive at the remainders 1 and 0; we may then be sure that the succeeding terms will always give a repetition of the same remainders as those we have already found. We might also find the exponent, $2\mu$, a priori; for we should only have to perform the calculation pointed out, Art. 71, in the first place, for the number $\Delta$, and then for the number $\Delta^2$; and if $\pi$ be the rank of the term of the series $v'$, $v''$, $v'''$, &c. which, in the first case, will be $= 1$, and $\xi$ the rank of the term that will be $= 1$, in the second case, we shall only have to seek the smallest multiple of