Elements of Algebra

572 ADDITIONS. CHAP. VII. Since $n''' = 0$, and consequently $\angle \frac{D'''}{\mathfrak{D}}$, and $\angle \frac{D^{iv}}{\mathfrak{D}}$, we shall here stop, and make $D'''=M=1$, $D^{iv}=L=7$, $n'''=0=N$, and $y''' = \xi$, $y^{iv} = \psi$, because $D'''$ is $\angle D^{iv}$. I now observe, that A being $= -7$, and consequently negative, in order that the equation may be resolvible, we must have $M=1$, as we have just now found; so that we may conclude, that the resolution is possible. We shall therefore suppose $\xi = y''' = 0$, $\psi = y^{iv} = \pm 1$; and we shall have, from the foregoing formulæ, $y'' = \pm 1$, $y' = \mp 3 = z$, $y = \mp 12 \pm 1 = \mp 11$, the doubtful signs being arbitrary. Therefore, $x = 321y - 1171z = \mp 18$; and, consequently, $s = \frac{x+31}{7} = \frac{31 \mp 18}{7} = \frac{13}{7}$, or $= \frac{49}{7} = 7$. Now, as the value of $s$ is required to be a whole number, we can only take $s=7$. It is remarkable, that the other value of $s$, namely, $\frac{13}{7}$, although fractional, gives nevertheless a whole number for the value of the radical, $\sqrt{(30 + 62s - 7s^2)}$, and the same number, $11$, which the value $s=7$ gives; so that these two values of $s$ will be the roots of the equation, $30 + 62s - 7s^2 = 121$. We have supposed $n = 321$. Now, we may likewise make $n = -321$; but it is easy to foresee, that the whole change that would result from it, in the preceding formulæ, would be a change of the sign of the values of $m, m', m''$, and of $n', n''$, by which means the values of $y'$, and of $y$, will also have different signs; we should not therefore have any new result, since these values already have the doubtful sign $\pm$. It will be the same in all other cases; so that we need not take the value of $n$, successively, positive and negative. The value $s=7$, which we have just found, results from the value of $n = \pm 321$: and we may find other values of $s$, if we have found other values of $n$ having the requisite condition; but, as the divisor $B = 1171$, is a prime number, there can be no other values of $n$, with the same property, as we have elsewhere demonstrated*, whence we must conclude, that the number $7$ is the only one that satisfies the question. * Memoirs of Berlin, for the year 1767, page 194.