Elements of Algebra

CHAP. IV. OF ALGEBRA. 323 It is certain, however, that there are real fractions, which, being substituted for $x$, will make $1 + x^2$ a square; of which we may be satisfied from the following cases: 1. If $x = \frac{3}{4}$, we have $1 + x^2 = \frac{25}{16}$; and consequently $\sqrt{(1 + x^2)} = \frac{5}{4}$. 2. $1 + x^2$ becomes a square likewise, if $x = \frac{4}{3}$, which gives $\sqrt{(1 + x^2)} = \frac{5}{3}$. 3. If we make $x = \frac{5}{12}$, we obtain $1 + x^2 = \frac{169}{144}$, the square root of which is $\frac{13}{12}$. But it is required to shew how to find these values of $x$, and even all possible numbers of this kind. 42. There are two methods of doing this. The first requires us to make $\sqrt{(1 + x^2)} = x + p$; from which supposition we have $1 + x^2 = x^2 + 2px + p^2$, where the square $x^2$ destroys itself; so that we may express $x$ without a radical sign. For, cancelling $x^2$ on both sides of the equation, we obtain $2px + p^2 = 1$; whence we find $x = \frac{1 - p^2}{2p}$; a quantity in which we may substitute for $p$ any number whatever less than unity. Let us therefore suppose $p = \frac{m}{n}$; then we have $x = \frac{1 - \frac{m^2}{n^2}}{\frac{2m}{n}}$; and, if we multiply both terms of this fraction by $n^2$, we shall find $x = \frac{n^2 - m^2}{2mn}$. 43. In order, therefore, that $1 + x^2$ may become a square, we may take for $m$ and $n$ all possible integer numbers, and in this manner find an infinite number of values for $x$. Also, if we make, in general, $x = \frac{n^2 - m^2}{2mn}$, we find, by squaring, $1 + x^2 = 1 + \frac{n^4 - 2m^2n^2 + m^4}{4m^2n^2}$; or, by putting $1 = \frac{4m^2n^2}{4m^2n^2}$ in the numerator, $1 + x^2 = \frac{n^4 + 2m^2n^2 + m^4}{4m^2n^2}$; a fraction which is a square, and gives $\sqrt{(1 + x^2)} = \frac{n^2 + m^2}{2mn}$. We shall exhibit, according to this solution, some of the least values of $x$. If $n = 2, 3, 3, 4, 4, 5, 5, 5, 5.$ and $m = 1, 1, 2, 1, 3, 1, 2, 3, 4,$ Y 2