Elements of Algebra

538 ADDITIONS. CHAP. V. Chap. IV.; we shall therefore abstract from those cases, or rather we shall endeavour to reduce all the rest to them. Farther, if the numbers $A$ and $B$ were divisible by any square numbers, we might likewise abstract from those divisors; that is to say, suppress them, only by taking for $A$ and $B$ the quotients, which we should have, after dividing the given values by the greatest squares possible; in fact, supposing $A = \alpha^2 A'$, and $B = \beta^2 B'$, we shall have to make the number, $A' \alpha^2 y^2 + B' \beta^2$ a square; therefore, dividing by $\beta^2$, and making $\frac{\alpha y}{\beta} = y'$; we shall have to determine the unknown quantity $y'$; so that $A' y'^2 + B'$ may be a square. Whence it follows that, when we have found a value of $y'$ that will make $A' y'^2 + B'$ become a square (rejecting in the given values of $A$ and $B$ the square factors $\alpha^2$ and $\beta^2$, which they might contain), we shall only have to multiply the value found for $y'$ by $\frac{\beta}{\alpha}$, in order to have that which answers to the quantity proposed. 51. Let us, therefore, consider the formula $Ay^2 + B$, in which $A$ and $B$ are given integers, not divisible by any square; and, as we suppose that $y$ may be a fraction, let us make $y = \frac{p}{q}$, $p$ and $q$ being integers prime to each other, in order that the fraction may be reduced to its least terms; we shall therefore have the quantity $\frac{Ap^2}{q^2} + B$, which must be a square; wherefore, $Ap^2 + Bq^2$ must be a square also; so that we shall have to resolve the equation, $Ap^2 + Bq^2 = z^2$, supposing $p$, $q$, and $z$, to be integer numbers. Now, I say that $q$ must be prime to $A$, and $p$ prime to $B$; for if $q$ and $A$ had a common divisor, it is evident that the term $Bq^2$ would be divisible by the square of that divisor; and the term $Ap^2$ would only be divisible by the first power of the same divisor, because $p$ and $q$ are prime to each other, and $A$ is supposed not to contain any square factor; wherefore the number $Ap^2 + Bq^2$ would only be once divisible by the common divisor of $q$ and $A$; consequently, it would be impossible for that number to be a square. In the same manner, it may be proved, that $p$ and $B$ can have no common divisor.