Elements of Algebra

520 ADDITIONS. CHAP. II. first (abstracting from the signs), as in the preceding case; therefore, &c. So that when we have to resolve, in integer numbers, an equation, of the form, $p^2 - kq^2 = \pm H$, where $H < \sqrt{k}$, we have only to follow the same process as in Art. 33, making $A=1$, $B=0$, and $c = -k$; and, if in the series $p^0, p', p'', p''', \&c.$ we find a term $= \pm H$, we shall have the solution required; if not, we may be certain that the given equation admits of no solution in integer numbers. 39. *Scholium*. We have considered (Art. 33) only one root of the equation $Ax^2 + Bx + C = 0$, which we have supposed positive; if this equation have both its roots positive, we must take them successively for $a$, and perform the same operation with both; but if one of the two roots, or both, were negative, then we should first change them into positive, by only changing the sign of $B$, and should proceed as before: but then we should take the values of $p$ and $q$ with contrary signs; that is to say, the one positive, and the other negative (Art. 29). In general, therefore, we shall give the ambiguous sign $\pm$ to the value of $B$, as well as to $\sqrt{E}$; that is to say, we shall make $q' = \mp \frac{1}{2}B$, and let us put $\pm$ before $\sqrt{E}$, and we must take these signs, so that the root $a = \frac{\mp \frac{1}{2}B \pm \frac{1}{2}\sqrt{E}}{A}$ may be positive, which may always be done in two different ways: the upper sign of $B$ will indicate a positive root; in which case, we must take both $p$ and $q$ with the same signs; on the contrary, the lower sign of $B$ will indicate a negative root; in which case, the values of $p$ and $q$ must be taken with contrary signs. 40. *Example*. Required what integer numbers must be taken for $p$ and $q$, in order that the quantity, $9p^2 - 118pq + 378q^2$ may become the least possible. Comparing this quantity with the general formula of Problem 3, we shall have $A=9$, $B=-118$, $c=378$; wherefore, $B^2 - 4AC = 316$; whence we see that this case belongs to that of Art. 33. We shall therefore make $E=316$, and $\frac{1}{2}\sqrt{E} = \sqrt{79}$, where we at once observe, that $\sqrt{79} > 8$, and $<9$; so that in the formulæ of which we shall only have to find the approximate integer value, we may immediately take, instead of $\sqrt{79}$, the number $8$, or $9$, according as that radical shall be added, or subtracted, from the other numbers of the same formula.