Elements of Algebra

550 ADDITIONS. CHAP. VII. CHAP. VII. *A direct and general method for finding all the values of $y$ expressed in Integer Numbers, by which we may render Quantities of the form $\sqrt{(Ay^2 + B)}$, rational; A and B being given Integer Numbers; and also for finding all the possible Solutions, in Integer Numbers, of indeterminate Quadratic Equations of two unknown Quantities.* [APPENDIX TO CHAP. VI.] 64. Though by the method of Art. 5, general formulæ may be found, containing all the rational values of $y$, by which $Ay^2 + B$ may be made equal to a square; yet those formulæ are of no use, when the values of $y$ are required to be expressed in integer numbers: for which reason, we must here give a particular method for resolving the question in the case of integer numbers. Let then $Ay^2 + B = x^2$; and as A and B are supposed to be integer numbers, and $y$ must also be integer, it is evident that $x$ ought likewise to be integer; so that we shall have to resolve, in integers, the equation $x^2 - Ay^2 = B$. Now, I begin by remarking, that if B is not divisible by a square number, $y$ must necessarily be prime to B; for suppose, if possible, that $y$ and B have a common divisor $\alpha$, so that $y = \alpha y'$, and $B = \alpha B'$; we shall then have $x^2 = A\alpha^2 y'^2 + \alpha B'$, whence it follows that $x^2$ must be divisible by $\alpha$; and as $\alpha$ is neither a square, nor divisible by any square (hyp.), because $\alpha$ is a factor of B, $x$ must be divisible by $\alpha$. Making then $x = \alpha x'$, we shall have $\alpha^2 x'^2 = A\alpha^2 y'^2 + \alpha B'$; or, dividing by $\alpha$, $\alpha x'^2 = A\alpha y'^2 + B'$; whence it is evident, that $B'$ must still be divisible by $\alpha$, which is contrary to the hypothesis. It is only, therefore, when B contains square factors, that $y$ can have a common measure with B; and it is easy to see, from the preceding demonstration, that this common measure of $y$ and B can only be the root of one of the square factors of B, and that the number $x$ must have the same common measure; so that the whole equation will be divisible by the square of this common divisor of $x$, $y$, and B.