Elements of Algebra

330 ELEMENTS PART II. thus obtain $p^2 + qr = p^2 + \frac{2mpq}{n} + \frac{m^2q^2}{n^2}$, in which the terms $p^2$ vanish; after which we may divide by $q$, so that we find $r = \frac{2mp}{n} + \frac{m^2q}{n^2}$, or $n^2r = 2nmp + m^2q$, an equation from which $x$ is easily determined. This, therefore, is the fourth case in which our formula may be transformed into a square; the application of which is easy, and we shall illustrate it by a few examples. 55. Question 3. Required a number, $x$, such, that double its square, shall exceed some other square by unity; that is, if we subtract unity from this double square, the remainder may be a square. For instance, the case applies to the number 5, whose square 25, taken twice, gives the number 50, which is greater by 1 than the square 49. According to this enunciation, $2x^2 - 1$ must be a square; and as we have, by the formula, $a = -1$, $b = 0$, and $c = 2$, it is evident that neither $a$ nor $c$ is a square; and farther, that the given quantity cannot be resolved into two factors, since $b^2 - 4ac = 8$ which is not a square; so that none of the first three cases will apply. But, according to the fourth, this formula may be represented by $x^2 + (x^2 - 1) = x^2 + (x-1) \times (x+1)$. If, therefore, we suppose its root $= x + \frac{m(x+1)}{n}$, we shall have $x^2 + (x+1) \times (x-1) = x^2 + \frac{2mx(x+1)}{n} + \frac{m^2(x+1)^2}{n^2}$. This equation, after having expunged the terms $x^2$, and divided the other terms by $x+1$, gives $n^2x - n^2 = 2mnx + m^2x + m^2$; whence we find $x = \frac{m^2+n^2}{n^2-2mn-m^2}$; and, since in our formula, $2x^2 - 1$, the square $x^2$ alone is found, it is indifferent whether we take positive or negative values for $x$. We may at first even write $-m$, instead of $+m$, in order to have $x = \frac{m^2+n^2}{n^2+2mn-m^2}$. If we make $m=1$, and $n=1$, we find $x=1$, and