Elements of Algebra

CHAP. VII. ADDITIONS. 551 Hence I conclude, 1st. That if $B$ is not divisible by any square, $y$ and $B$ will be prime to each other. 2dly. That if $B$ is divisible by a single square $a^2$, $y$ may be either prime to $B$, or divisible by $a$, which makes two cases to be separately examined. In the first case, we shall resolve the equation $x^2 - Ay^2 = B$, supposing $y$ and $B$ prime to one another; in the second, we shall have to resolve the equation, $x^2 - Ay^2 = B'$, $B'$ being $= \frac{B}{a^2}$, supposing also $y$ and $B'$ prime to each other; but it will then be necessary to multiply by $a$ the values found for $y$ and $x$, in order to have values corresponding to the equation proposed. 3dly. If $B$ is divisible by two different squares, $a^2$ and $\beta^2$, we shall have three cases to consider. In the first, we shall resolve the equation $x^2 - Ay^2 = B$, considering $y$ and $B$ as prime to each other. In the second, we shall likewise resolve the equation, $x^2 - Ay^2 = B'$, $B'$ being $= \frac{B}{a^2}$, on the supposition of $y$ and $B'$ being prime to each other, and we shall then multiply the values of $x$ and $y$ by $a$. In the third, we shall resolve the equation $x^2 - Ay^2 = B''$, $B''$ being $= \frac{B}{\beta^2}$, on the supposition of $y$ and $B''$ being prime to each other, and we shall then multiply the values of $x$ and $y$ by $\beta$. 4thly, &c. Thus, we shall have as many different equations to resolve, as there may be different square divisors of $B$; but those equations will be all of the same form, $x^2 - Ay^2 = B$, and $y$ also will always be prime to $B$. 65. Let us therefore consider, generally, the equation $x^2 - Ay^2 = B$; where $y$ is prime to $B$; and, as $x$ and $y$ must be integers, $x^2 - Ay^2$ must be divisible by $B$. By the method, therefore, of Chap. IV. 48, we shall make $x = ny - Bz$, and shall have the equation, $(n^2 - A)y^2 - 2nByz + B^2z^2 = B$, from which we perceive, that the term, $(n^2 - A)y^2$, must be divisible by $B$, since all the others are so of themselves; wherefore, as $y$ is prime to $B$, (hyp.) $n^2 - A$ must be divisible by $B$; so that making $\frac{n^2 - A}{B} = c$, and dividing by $B$, we shall have, $cy^2 - 2nyz + Bz^2 = 1$. Now, this equation is simpler than the one proposed, because the second side is equal to unity.