Elements of Algebra

560 ADDITIONS. CHAP. VII. and, in order to abridge, making $P = cp^2 - 2npq + bq^2$, $Q = cpr - n(ps + qr) + bqs$, $R = cr^2 - 2nrs + bs^2$, we shall have the equation transformed into this, $Pt^2 + 2Qtu + Ru^2 = 1$. Now we have, by hypothesis, $p = 1$; farther, if we call $\rho$ and $\sigma$, two values of $r$ and $s$ that satisfy the equation, $ps - qr = 1$, we shall have, in general, (Art. 42), $r = \rho + mp, s = \sigma + mq$, $m$ being any whole number; therefore, putting these values into the expression of $Q$, it will become $Q = cp\rho - n(p\sigma + q\rho) + bq\sigma + mP$; so that, as $P = 1$, we may make $Q = 0$, by taking $m = -cp\rho + n(p\sigma + q\rho) - bq\sigma$. We now observe, that the value of $Q^2 - PR$ is reduced (after the above substitutions and reductions), to this; $(n^2 - cb) \times (ps - qr)^2$; so that as $ps - qr = 1$, we shall have $Q^2 - PR = n^2 - cb = A$; therefore, making $P = 1$, and $Q = 0$, we shall have $-R = A$, that is, $R = -A$; so that the equation before transformed will become $t^2 - Au^2 = 1$. Now, as $y, z, p, q, r$, and $s$ are whole numbers, by the hypothesis, it is easy to perceive, that $t$ and $u$ will also be whole numbers; for, deducing their values from the equa- tions, $y = pt + ru$, and $z = qt + su$, we have $t = \frac{sy-rz}{ps-qr}$, and $u = \frac{qy-pz}{qr-ps}$; that is to say, (because $ps - qr = 1$), $t = sy - rs$, and $u = pz - qy$. We shall therefore only have to resolve, in whole numbers, the equation $t^2 - Au^2 = 1$, and each value of $t$ and $u$ will give new values of $y$ and $z$. For, substituting the value of the number $m$, already found, in the general values of $r$ and $s$, we shall have $r = \rho(1 - cp^2) - bpq\sigma + np(p\sigma + q\rho)$, $s = \sigma(1 - bq^2) - cpq\rho + nq(p\sigma + q\rho)$; or, because $cp^2 - 2npq + bq^2 = 1$, $r = (bq - np) \times (q\rho - p\sigma) = -bq + np$, $s = (cp - nq) \times (p\sigma - q\rho) = cp - nq$. Therefore, putting these values of $r$ and $s$ in the fore- going expressions of $y$ and $z$, we shall have, in general,