Elements of Algebra

542 ADDITIONS. CHAP. V. formula, $AY^2 + B$, to $BY'^2 + C$, we might reduce this last to $CY''^2 + D$, where $D$ will be less than $C$, and so on; and as the numbers $A, B, C, D, \&c.$ form a decreasing series of integers, it is evident that this series cannot go on to infinity, and therefore the operation must always terminate. If the question admits of no solution in rational numbers, we shall arrive at an impossible condition; but, if the question be resolvible, we shall always be brought to an equation like that of Art. 53, in which one of the coefficients, as $A'$, will be a square; so that the known methods will be applicable to it: this equation being resolved, we may, by inverting the operation, successively resolve all the preceding equations, up to the first $AP^2 + BQ^2 = Z^2$. We will illustrate this method by some examples. 56. Example 1. Let it be proposed to find a rational value of $x$, such, that the formula, $7 + 15x + 13x^2$, may become a square*. Here, we shall have $a = 7, b = 15, c = 13$; and therefore $4c = 4 \times 13$, and $b^2 - 4ac = -139$; so that calling the root of the square in question $y$, we shall have the formula $4 \times 13y^2 - 139$, which must be a square. We shall also have $A = 4 \times 13$, and $B = -139$, where it will at once be observed, that $A$ is divisible by the square $4$; so that we must reject this square divisor, and simply suppose $A = 13$; but we must then divide the value found for $y$ by $2$, as is shewn, Art. 50. Making, therefore, $y = \frac{p}{q}$, we shall have the equation, $13p^2 - 139q^2 = z^2$; or, because $139$ is $> 13$, let us make $y = \frac{q}{p}$, in order to have $-139p^2 + 13q^2 = z^2$, an equation which we may write thus, $-139p^2 = z^2 - 13q^2$. We shall now make (Art. 52) $z = nq - 139q'$, and must take for $n$ an integer number not $\not> \frac{139}{2}$; that is to say, $\not> 70$ such, that $n^2 - 13$ may be divisible by $139$. Assuming now $n = 41$, we have $n^2 - 13 = 1668 = 139 \times 12$; so that by making the substitution, and then dividing by $-139$, we shall have the equation, $p^2 = -12q^2 + 2 \times 41qq' - 139q'^2$. Now, as $-12$ is not a square, this equation has not the \* See Chap. IV. Art. 57, of the preceding Treatise.