Elements of Algebra

574 ADDITIONS. CHAP. VII. $m'' = \frac{-1}{-1} = 1$, $n'' = -1 + 1 = 0$, $\text{D}''' = \frac{-13}{-1} = 13$, $y = y' + y''$. As $n'' = 0$, and consequently, $< \frac{\text{D}''}{2}$, and $< \frac{\text{D}'''}{2}$, we shall stop here, and shall have the transformed equation, $\text{D}''' y''^2 - 2n'' y'' y''' + \text{D}'' y'''^2 = 1$, or $13y''^2 - y'''^2 = 1$; which being reduced to the form, $y'''^2 - 13y''^2 = 1$, will admit of the method of Art. 71; and, as $A = 13$ is $< 100$, we may make use of the Table, Art. 41. Thus, we shall only have to see, whether, in the upper series of numbers belonging to $\sqrt{13}$, there be found the number 1 in an even place; for, in order that the preceding equation may be resolvible, we must find in the series $p^0, p', p'', \ldots$ a term $= -1$; but we have $p^0 = 1, p' = 4, p'' = 3, \ldots$ wherefore, $\ldots$. Now, in the series, $1, 4, 3, 3, 4, 1, \ldots$ we find 1 in the sixth place; so that $p^v = -1$; and hence we shall have a solution of the given equation, by taking $y''' = p^v$, and $y'' = q^v$, the numbers $p^v, q^v$, being calculated according to the formulæ of Article 25, giving to $\mu, \mu', \mu'', \ldots$ the values $3, 1, 1, 1, 1, 6, \ldots$ which form the lower series of numbers belonging to $\sqrt{13}$ in the same Table. We shall therefore have $p^0 = 1$ $p' = 3$ $p'' = p' + p^0 = 4$ $p''' = p'' + p' = 7$ $p^{iv} = p''' + p'' = 11$ $p^v = p^{iv} + p''' = 18$ $q^0 = 0$ $q' = 1$ $q'' = q' + q^0 = 1$ $q''' = q'' + q' = 2$ $q^{iv} = q''' + q'' = 3$ $q^v = q^{iv} + q''' = 5$. So that $y''' = 18$, and $y'' = 5$; therefore, $y' = y'' + y''' = 23$, and $y = 3y' + y'' = 74$. We have supposed $n = 35$; but we may also take $n = -35$. Let therefore $n = -35$, we shall make $m = \frac{-35}{12} = -3$, $n' = -35 + 3 \times 12 = 1$, $\text{D}'' = \frac{1-13}{12} = -1$, $y = -3y' + y''$,