Elements of Algebra

CHAP. IV. OF ALGEBRA. 325 $1+x^2 = 1 + \frac{4m^2n^2}{n^4-2m^2n^2+m^4} = \frac{n^4+2m^2n^2+m^4}{n^4-2m^2n^2+m^4}$ the square of $\frac{n^2+m^2}{n^2-m^2}$. Now, as we obtain from that, the equation $1 + \frac{(2mn)^2}{(n^2-m^2)^2} = \frac{(n^2+m^2)^2}{(n^2-m^2)^2}$ we shall have, as before, $(n^2-m^2)^2 + (2mn)^2 = (n^2+m^2)^2$; that is, the same two squares, whose sum is also a square. 46. The case which we have just analysed furnishes two methods of transforming the general formula $a+bx+cx^2$ into a square. The first of these applies to all cases in which $c$ is a square; and the second to those in which $a$ is a square. We shall consider both these suppositions. First, let us suppose that $c$ is a square, or that the given formula is $a+bx+f^2x^2$. Since this must be a square, we shall make $\sqrt{(a+bx+f^2x^2)} = fx + \frac{m}{n}$, and shall thus have $a+bx+f^2x^2 = f^2x^2 + \frac{2mfx}{n} + \frac{m^2}{n^2}$, in which the terms containing $x^2$ destroy each other, so that $a+bx = \frac{2mfx}{n} + \frac{m^2}{n^2}$. If we multiply by $n^2$, we obtain $n^2a+n^2bx = 2mnfx+m^2$; hence we find $x = \frac{m^2-n^2a}{n^2b-2mnf}$; and, substituting this value for $x$, we shall have $\sqrt{(a+bx+f^2x^2)} = \frac{m^2f-n^2af}{n^2b-2mnf} + \frac{m}{n} = \frac{mnb-m^2f-n^2af}{n^2b-2mnf}$. 47. As we have got a fraction for $x$, namely, $\frac{m^2-n^2a}{n^2b-2mnf}$, let us make $x = \frac{p}{q}$, then $p = m^2-n^2a$, and $q = n^2b-2mnf$; so that the formula $a + \frac{bp}{q} + \frac{f^2p^2}{q^2}$ is a square; and as it continues a square, though it be multiplied by the square $q^2$, it follows, that the formula $aq^2+bpq+f^2p^2$ is also a square, by making $p=m^2-n^2a$, and $q=n^2b-2mnf$. Hence it is evident, that an infinite number of answers, in integer numbers, may result from this expression, because the values of the letters $m$ and $n$ are arbitrary. 48. The second case which we have to consider, is that in