Elements of Algebra

CHAP. VII. ADDITIONS. 565 $t'''u^{iv} - t^{iv}u''' = \frac{(T - v\sqrt{A})^{3} \times (T + v\sqrt{A})^{4} - (T - v\sqrt{A})^{4} \times (T + v\sqrt{A})^{3}}{2\sqrt{A}}$ Farther, $(T - v\sqrt{A})^3 \times (T + v\sqrt{A})^3 = (T^2 - Av^2)^3 = 1$, since $T^2 - Av^2 = 1$, by hypothesis; whence $(T - v\sqrt{A})^3 \times (T + v\sqrt{A})^4 = T + v\sqrt{A}$, and $(T - v\sqrt{A})^4 \times (T + v\sqrt{A})^3 = T - v\sqrt{A}$; so that the value of $t'''u^{iv} - u'''t^{iv}$ will be reduced to $\frac{2v\sqrt{A}}{2\sqrt{A}} = v$. It would follow from this, that we should have a value of $u < v$, which is contrary to the hypothesis; since $v$ is supposed to be the least possible value of $u$. There cannot, therefore, be any intermediate values of $t$ and $u$ between these, $t''', t^{iv}$, and $u''', u^{iv}$. And, as this reasoning may be applied, in general, to all the values of $t$ and $u$, which would result from the above formulæ, by making $m$ equal to any whole number, we may infer, that those formulæ actually contain all the possible values of $t$ and $u$. It is unnecessary to observe, that the values of $t$ and $u$ may be taken either positive, or negative; for this is evident from the equation itself, $t^2 - Au^2 = 1$. *Of the manner of finding all the possible Solutions, in whole numbers, of indeterminate Quadratic Equations of two unknown quantities.* 76. The methods, which we have just explained, are sufficient for the complete solution of equations of the form $Ay^2 + B = x^2$; but we may have to resolve equations of a more complicated form; for which reason, it is proper to shew how such solutions are to be obtained. Let there be proposed the equation $ar^2 + brs + cs^2 + dr + es + f = 0$, where $a, b, c, d, e, f$, are given whole numbers, and $r$ and $s$ are two unknown numbers, that must likewise be integer. I shall first have, by the common solution, $2ar + bs + d = \sqrt{((bs+d)^2 - 4a(cs^2 + es + d))}$, whence we see, that the difficulty is reduced to making $(bs+d)^2 - 4a(cs^2 + es + d)$ a square. In order to simplify, let us suppose *