Elements of Algebra

CHAP. V. ADDITIONS. 543 requisite conditions; since 12 is already less than 13, we shall multiply the whole equation by $-12$, and it will become $-12p^2 = (-12q+41q')^2 - 13q'^2$; so that $13q'^2 - 12p^2$ must be a square; or, making $\frac{q'}{p} = y'$, $13y'^2 - 12$ must be so too. Where, it is evident, we should only have to make $y' = 1$; but as we have got this value merely by chance, let us proceed in the calculation according to our method, until we arrive at a formula, to which the ordinary methods may be applied. As 12 is divisible by 4, we may reject this square divisor, remembering, however, that we must multiply the value of $y'$ by 2; we have therefore to make a square of the formula $13y'^2 - 3$; or making $y' = \frac{r}{s}$, (supposing $r$ and $s$ to be integers prime to each other; so that the fraction $\frac{r}{s}$ is already reduced to its least terms, as well as the fraction $\frac{q'}{p}$), the formula $13r^2 - 3s^2$ must be a square. Let the root be $z'$, which gives $13r^2 = z'^2 + 3s^2$; and, making $z' = ms - 13s'$, $m$ being an integer not $\not< 7\frac{1}{2}$, that is, $\not< 7$, and such, that $m^2+3$ may be divisible by 13. Assuming $m=6$, which gives $m^2+3 = 39 = 13 \times 3$, we have, by substituting the value of $z'$, and dividing the whole equation by 13, $r^2 = 3s^2 - 2 \times 6ss' + 13s'^2$. As the coefficient 3 of $s^2$ is neither a square, nor less than that of $s'^2$, in the preceding equation, let us make (Art. 54), $s = \mu s' + s''$, and substituting, we shall have the transformed equation, $r^2 = 3s''^2 - 2(6-3\mu)s''s' + (3\mu^2 - 2 \times 6\mu + 13)s'^2$; and here we must determine $\mu$ so, that $6-3\mu$ may not be $\not< 7\frac{1}{2}$, and it is clear that we must make $\mu=2$, which gives $6-3\mu=0$; and the equation will become $r^2 = 3s''^2 + s'^2$, which is evidently reduced to the form required, as the coefficient of the square of one of the two indeterminate quantities of the second side is also a square. In order to have the most simple solution, we shall make $s''=0, s'=1$,