Elements of Algebra

CHAP. VII. ADDITIONS. 569 and $\rho$, which being divided by $\pi$, will give the required value of $\mu$. Thus, for example, if we have A $= 6$, and $\Delta = 3$, we shall find for the radical $\sqrt{6}$, in the Table of Art. 41, $p^0 = 1$, $p' = -2$, $p'' = 1$; therefore, $\pi = 2$. Then we shall find, in the same Table, for the radical $\sqrt{(6 \times 9)} = \sqrt{54}$, $p^0 = 1$, $p' = -5$, $p'' = 9$, $p''' = -2$, $p^{\text{iv}} = 9$, $p^{\text{v}} = -5$, $p^{\text{vi}} = 1$; and hence $\rho = 6$. Now, the least multiple of 2 and 6 is 6, which being divided by 2 gives the remainder 3; so that we shall here have $\mu = 3$, and $2\mu = 6$. Therefore, in order to have, in this case, all the remainders of the division of the terms $t'$, $t''$, $t'''$, &c. and $u'$, $u''$, $u'''$, &c. by 3, it will be sufficient to find those of the six leading terms of each series; for the succeeding terms will always give a repetition of the same remainders: that is to say, the seventh terms will give the same remainders as the first, the eighth terms, the same as the second; and so on to infinity. Lastly, the terms $t^\mu$ and $u^\mu$ may sometimes happen to have the same properties as the terms $t^{2\mu}$ and $u^{2\mu}$; that is to say, $u^\mu$ may be divisible by $\Delta$, and $t^\mu$ may leave unity for a remainder. In such cases, we may stop at these very terms; for the remainders of the succeeding terms, $t^{\mu+1}$, $t^{\mu+2}$, &c. $u^{\mu+1}$, $u^{\mu+2}$, &c. will be the same as those of the terms $t'$, $t''$, &c. $u'$, $u''$, &c. and so of the others. In general, we shall denote by M the least value of the exponent $m$, that will render $t - 1$, and $u$, divisible by $\Delta$. 79. Let us now suppose that we have any expression whatever, composed of $t$ and $u$, and given whole numbers, so that it may always represent whole numbers; and that it is required to find the values, which must be given to the exponent $m$, in order that this expression may become divisible by any given number whatever, $\Delta$: we shall only have to make, successively, $m = 1, 2, 3$, &c. as far as M; and if none of these suppositions render the given expression divisible by $\Delta$, we may conclude, with certainty, that it can never become so, whatever values we give to $m$. But if in this manner we find one, or more values of $m$, which render the given expression divisible by $\Delta$, then calling each of these values N, all the values of $m$ that can possibly do the same, will be N, N + M, N + 2M, N + 3M, &c. and, in general, N + $\lambda$M; $\lambda$ being any whole number whatever. In the same manner, if we had another expression composed likewise of $t$, $u$, and given whole numbers, and, at the same time, divisible by any other given number whatever,