Elements of Algebra

CHAP. VII. ADDITIONS. 559 equation; for we shall only have to take, for $v$ and $\xi$, the corresponding terms of the series $p'$, $p''$, $p'''$, &c. $q'$, $q''$, $q'''$, &c. calculated according to the formulæ of Art. 25; and we may even find an infinite number of satisfactory values for $v$ and $\xi$, by continuing the same series to infinity. Now, as soon as we know two values of $v$ and $\xi$, we shall have, from the equation, $v = M\psi - N\xi$, that of $\psi$, which will also be a whole number; then we may go back from these values of $\xi$ and $\psi$, that is to say, of $\theta_{p+1}$, and $\theta_p$, to those of $\theta$ and $\theta'$, or of $y$ and $z$ (Art. 70). But if in the series $P'$, $P''$, $P'''$, &c. there is no term $= M$, we are sure that the equation proposed admits of no solution in whole numbers. It is proper to observe, that, as the series $p^0$, $p'$, $p''$, &c. as well as the two others, $q^0$, $q'$, $q''$, &c. and $\mu$, $\mu'$, $\mu''$, &c. depend only on the number A; the calculation, once made for a given value of A, will serve for all the equations in which A, or $n^2 - CB$, shall have the same value; and hence the foregoing method is preferable to that of Art. 68, which requires a new calculation for each equation. Lastly, so long as A does not exceed 100, we may make use of the Table given, Art. 41, which contains for each radical $\sqrt{A}$, the values of the terms of the two series $p^0$, $p'$, $p''$, $p'''$, &c. and $\mu$, $\mu'$, $\mu''$, &c. continued, until one of the terms $P'$, $P''$, $P'''$, &c. becomes $= 1$; after which, all the succeeding terms of both series return in the same order. So that, by means of this Table, we may judge, immediately, whether the equation, $v^2 - A\xi^2 = M$, be resolvible, or not. *Of the manner of finding all the possible solutions of the equation, $cy^2 - 2nyz + Bz^2 = 1$, when we know only one of them.* 72. Though, by the methods just given, we may successively find all the solutions of this equation, when it is resolvible in integer numbers; yet this may be done, in a manner still more simple, as follows: Call $p$ and $q$ the values found for $y$ and $z$; so that we have $cp^2 - 2npq + Bq^2 = 1$, and take two other whole numbers, $r$ and $s$, such, that $ps - qr = 1$; which is always possible, because $p$ and $q$ are necessarily prime to each other; then suppose $y = pt + ru$, and $z = qt + su$, $t$ and $u$ being two new indeterminate numbers; substituting these expressions in the equation, $cy^2 - 2nyz + Bz^2 = 1$,