Elements of Algebra

322 ELEMENTS PART II. CHAP. IV. On the Method of rendering Surd Quantities of the form $\sqrt{(a + bx + cx^2)}$ Rational. 38. It is required in the present case to determine the values, which are to be adopted for $x$, in order that the formula $a + bx + cx^2$ may become a real square; and, consequently, that a rational root of it may be assigned. Now, the letters $a$, $b$, and $c$, represent given numbers; and the determination of the unknown quantity depends chiefly on the nature of these numbers; there being many cases in which the solution becomes impossible. But even when it is possible, we must content ourselves at first with being able to assign rational values for the letter $x$, without requiring those values also to be integer numbers; as this latter condition produces researches altogether peculiar. 39. We suppose here that the formula extends no farther than the second power of $x$; the higher dimensions require different methods, which will be explained in their proper places. We shall observe first, that if the second power were not in the formula, and $c$ were $= 0$, the problem would be attended with no difficulty; for if $\sqrt{(a + bx)}$ were the given formula, and it were required to determine $x$, so that $a + bx$ might be a square, we should only have to make $a + bx = y^2$, whence we should immediately obtain $x = \frac{y^2 - a}{b}$. Now, whatever number we substitute here for $y$, the value of $x$ would always be such, that $a + bx$ would be a square, and consequently, $\sqrt{(a + bx)}$ would be a rational quantity. 40. We shall therefore begin with the formula $\sqrt{(1 + x^2)}$; that is to say, we are to find such values of $x$, that, by adding unity to their squares, the sums may likewise be squares; and as it is evident that those values of $x$ cannot be integers, we must be satisfied with finding the fractions which express them. 41. If we supposed $1 + x^2 = y^2$, since $1 + x^2$ must be a square, we should have $x^2 = y^2 - 1$, and $x = \sqrt{(y^2 - 1)}$; so that in order to find $x$ we should have to seek numbers for $y$, whose squares, diminished by unity, would also leave squares; and, consequently, we should be led to a question as difficult as the former, without advancing a single step.