Elements of Algebra

CHAP. II. ADDITIONS. 507 If we had, therefore, to resolve the equation $Ap^m + Bp^{m-1}q + cp^{m-2}q^2 + \text{&c.} + vq^m = \pm 1$, in whole numbers, we must seek for the values of $p$ and $q$ by the method of the preceding problem, except in the case where the equation $Ax^m + Bx^{m-1} + cx^{m-2} + \text{&c.} + v = 0$, had roots, or any divisors commensurable; for then, it is evident, that the quantity $Ap^m + Bp^{m-1}q + cp^{m-2}q^2 + \text{&c.}$ might be decomposed into two or more similar quantities of less degrees; so that it would be necessary for each of these partial formulæ to be separately equal to unity, which would give at least two equations that would serve to determine $p$ and $q$. We have elsewhere given a solution of this last problem (Memoires pour l'Academie de Berlin pour l'Année 1768); but the one we are going to explain is much more simple and direct, although both depend on the same theory of con- tinued fractions *. 31. Problem 3. Required the values of $p$ and $q$, which will render the quantity $Ap^2 + Bpq + Cq^2$ the least possible, supposing that whole numbers only are admitted for $p$ and $q$. This problem evidently is only a particular case of the preceding; but it may be proper to consider it separately, because it is capable of a very simple and elegant solution; and, besides, we shall have occasion afterwards to make use of it, in resolving quadratic equations for two unknown quantities in whole numbers. According to the general method, we must begin, there- fore, by seeking the roots of the equation $Ax^2 + Bx + C = 0$, which we know to be, $\frac{-B \pm \sqrt{(B^2 - 4AC)}}{2A}$. 1st, If $B^2 - 4AC$ be a square number, the two roots will be commensurable, and there will properly be no minimum, because the quantity $Ap^2 + Bpq + Cq^2$ will become $0$. 2d, If $B^2 - 4AC$ be not a square, then the two roots will be irrational, or imaginary, according as $B^2 - 4AC$ will be $> 0$, or $< 0$, which makes two cases that must be considered separately; we shall begin with the latter, which it is most easy to resolve. First case, when $B^2 - 4AC < 0$. 32. The two roots being in this case imaginary, we shall * See also Le Gendre's Essai sur la Theorie des Nombres, page 169.