Elements of Algebra

436 ELEMENTS PART II. sary, may also be represented in a more general manner; for if $\frac{b}{c} = \frac{\beta}{1}$, $\frac{d}{e} = \frac{n\beta-1}{n}$, we shall have $\frac{f}{g} = \frac{n\beta+\beta-1}{n+1}$, and $\frac{g}{h} = \frac{2n\beta+\beta-2}{2n+1}$. If in this last fraction we suppose $2n + 1 = m$, it will become $\frac{\beta m - 2}{m}$; consequently, the first gives $x = \beta^2 - a$, and the last furnishes $v = (\beta m - 2)^2 - am^2$. The only question therefore is, to make $xv + a$ a square. Now, because $v = (\beta^2 - a)m^2 - 4\beta m + 4$, we have $xv + a = (\beta^2 - a)^2 m^2 - 4(\beta^2 - a)\beta m + 4\beta^2 - 3a$; and since this must be a square, let us suppose its root to be $(\beta^2 - a)m - p$; the square of which quantity being $(\beta^2 - a)^2 m^2 - 2(\beta^2 - a)mp + p^2$, we shall have $-4(\beta^2 - a)\beta m + 4\beta^2 - 3a = -2(\beta^2 - a)mp + p^2$; wherefore $m = \frac{p^2 - 4\beta^2 + 3a}{(\beta^2 - a) \times (2p - 4\beta)}$. If $p = 2\beta + q$, we shall find $m = \frac{4\beta q + q^2 + 3a}{2q(\beta^2 - a)}$; in which we may substitute any numbers whatever for $\beta$ and $q$. For example, if $a = 1$, let us make $\beta = 2$: we shall then have $m = \frac{4q + q^2 + 3}{6q}$; and making $q = 1$, we shall find $m = \frac{8}{6}$; farther, $m = 2n + 1$; but without dwelling any longer on this question, let us proceed to another. 235. Question 12. Required three such numbers, $x, y,$ and $z$, that the sums and differences of these numbers, taken two by two, may be squares. The question requiring us to transform the six following formulæ into squares, viz. $x + y, x + z, y + z,$ $x - y, x - z, y - z,$ let us begin with the last three, and suppose $x - y = p^2$, $x - z = q^2$, and $y - z = r^2$; the last two will furnish $x = q^2 + z$, and $y = r^2 + z$; so that we shall have $q^2 = p^2 + r^2$, because $x - y = q^2 - r^2 = p^2$; hence, $p^2 + r^2$, or the sum of two squares, must be equal to a square $q^2$; now, this happens, when $p = 2ab$, and $r = a^2 - b^2$, since then $q = a^2 + b^2$. But let us still preserve the letters $p, q,$ and $r$, and consider also the first three formulæ. We shall have,