Elements of Algebra

CHAP. V. ADDITIONS. 541 signs, and then it is evident, that $A''$ will become $< A'$, because $A'' = \frac{n'^2 - B}{A'}$, and $B =$, or $< A'$, and $n =$, or $< \frac{A'}{2}$. We shall therefore apply the same reasoning here that we did in the preceding Article; and if $A''$ is a square, we shall have the resolution of the equation: but if $A''$ be not a square, and $A'' < B$, or becomes so, when divided by a square, we must multiply the equation by $A'$, and shall thus have, by making $\frac{p}{q'} = y$, and $A'' = c$, the formula $By'^2 + c$, which must be a square, and in which the coefficients, $B$ and $c$, (after having suppressed in $c$ the square divisors, if there are any), will be less than those of the formula $Ay^2 + B$ of Art. 51. But if these cases do not take place, we shall, as before, make $q' = \nu q'' + q'''$, and the equation will be changed into this, $p''^2 = A''q''^2 - 2n''q''q''' + A'''q'''^2$, where $n''' = n'' - \nu A''$, and $A''' = A''\nu^2 - 2n''\nu + A' = \frac{n'''^2 - B}{A''}$. We shall therefore take for $\nu$ such an integer number, that $n''$ may not be $> \frac{A''}{2}$, abstracting from the signs; and, as $B$ is not $> A''$ (hyp.), it follows, from the equation, $A''' = \frac{n'''^2 - B}{A''}$, that $A'''$ will be $< A''$; so that we may go over the same reasoning as before, and shall draw from it similar conclusions. Now, as the numbers $A, A', A'', A''', \&c.$ form a decreasing series of integer numbers, it is evident, that, by continuing this series, we shall necessarily arrive at a term less than the given number $B$; and then calling this term $c$, we shall have, as we have already seen, the formula $By'^2 + c$ to make equal to a square. So that by the operations we have now explained, we may always be certain of reducing the formula, $Ay^2 + B$, to one more simple, such as $By'^2 + c$; at least, if the problem is resolvible. 55. Now, in the same manner as we have reduced the