Elements of Algebra

CHAP. V. ADDITIONS. 545 on account of the indeterminate number $m$, that this expression of $x$ must comprehend all the values that can be given to $x$, in order to make the proposed formula a square; for whatever be the square number, to which this formula may be equal, it is evident, that the root of this number may always be represented by $g + m(x - f)$, giving to $m$ a suitable value. So that when we have found, by the method above explained, a single satisfactory value of $x$, we have only to take it for $f$, and the root of the square which results for $g$; and, by the preceding formula, we shall have all the other possible values of $x$. In the preceding example, we found $y = \frac{5}{3}$, and $x = -\frac{2}{3}$; so that, making $g = \frac{5}{3}$, and $f = -\frac{2}{3}$, we shall have $x = \frac{19-10m-2m^2}{3(m^2-13)}$ which is a general expression for the rational values of $x$, by which the quantity $7 + 15x + 13x^2$ may be made a square. 58. *Example* 2. Let it also be proposed to find a rational value of $y$, so that $23y^2 - 5$ may be a square. As $23$ and $5$ are not divisible by any square number, we shall have no reduction to make. So that making $y = \frac{p}{q}$, the formula $23p^2 - 5q^2$ must become a square, $z^2$; so that we shall have the equation $23p^2 = z^2 + 5q^2$. We shall therefore make $z = nq - 23q'$, and we must take for $n$ an integer number, not $7 \frac{2}{3}$, such, that $n^2 + 5$ may be divisible by $23$. I find $n = 8$, which gives $n^2 + 5 = 23 \times 3$, and this value of $n$ is the only one that has the requisite conditions. Substituting, therefore, $8q - 23q'$, in the room of $z$, and dividing the whole equation by $23$, we shall have $p^2 = 3q^2 - 2 \times 8qq' + 23q'^2$, in which we see that the coefficient $3$ is already less than the value of $B$, which is $5$, abstracting from the sign. Art. 52. Thus, we shall multiply the whole equation by $3$, and shall have $3p^2 = (3q - 8q')^2 + 5q'^2$; so that making $\frac{q'}{p} = y'$, the formula $-5y'^2 + 3$ must be a square, the coefficients $5$ and $3$ admitting of no reduction. Therefore, let $y' = \frac{r}{s}$ ($r$ and $s$ being supposed prime to NN