Elements of Algebra

332 ELEMENTS PART II. But, as we have observed, it cannot be said that this analysis is easily found; and therefore we shall explain a general method for discovering, beforehand, whether the resolution of any such formula be possible or not; for there is an infinite number of them which cannot be resolved at all: such, for instance, as the formula $3x^2 + 2$, which can in no case whatever become a square. On the other hand, it is sufficient to know a single case, in which a formula is possible, to enable us to find all its answers; and this we shall explain at some length. 58. From what has been said, it may be observed, that all the advantage that can be expected on these occasions, is to determine, or suppose, any case in which such a formula as $a + bx + cx^2$, may be transformed into a square; and the method which naturally occurs for this, is to substitute small numbers successively for $x$, until we meet with a case which gives a square. Now, as $x$ may be a fraction, let us begin with substituting for $x$ the general fraction $\frac{t}{u}$; and, if the formula $a + \frac{bt}{u} + \frac{ct^2}{u^2}$ which results from it, be a square, it will be so also after having been multiplied by $u^2$; so that it only remains to try to find such integer values for $t$ and $u$, as will make the formula $au^2 + btu + ct^2$ a square; and it is evident, that after this, the supposition of $x = \frac{t}{u}$ cannot fail to give the formula $a + bx + cx^2$ equal to a square. But if, whatever we do, we cannot arrive at any satisfactory case, we have every reason to suppose that it is altogether impossible to transform the formula into a square; which, as we have already said, very frequently happens. 59. We shall now shew, on the other hand, that when one satisfactory case is determined, it will be easy to find all the other cases which likewise give a square; and it will be perceived, at the same time, that the number of those solutions is always infinitely great. Let us first consider the formula $2 + 7x^2$, in which $a = 2$, $b = 0$, and $c = 7$. This evidently becomes a square, if we suppose $x = 1$. Let us therefore make $x = 1 + y$; then, by substitution, we shall have $x^2 = 1 + 2y + y^2$, and our formula becomes $9 + 14y + 7y^2$, in which the first term is a square; so that we shall suppose, conformably to the second rule, the square root of the new formula to be