Elements of Algebra

424 ELEMENTS PART II. $(b-a)p^2 = br^2 - as^2$, and, consequently, $p^2 = \frac{br^2 - as^2}{b-a}$; which formula must be a square: now, this happens when $r=s$. Let us, therefore, in order to remove the fractions, suppose $r = s + (b-a)t$, and we shall have $p^2 = \frac{br^2-as^2}{b-a} = \frac{bs^2+2b(b-a)st+b(b-a)^2t^2-as^2}{b-a} =$ $\frac{(b-a)s^2+2b(b-a)st+b(b-a)^2t^2}{b-a} =$ $s^2+2bst+b(b-a)t^2$. Let us now make $p = s + \frac{x}{y}t$, and we shall have $p^2 = s^2 + \frac{2x}{y}st + \frac{x^2}{y^2}t^2 = s^2 + 2bst + b(b-a)t^2$, in which the terms $s^2$ destroy each other; so that the other terms being divided by $t$, and multiplied by $y^2$, give $2sxy + tx^2 = 2bsy^2 + b(b-a)ty^2$; whence $t = \frac{2sxy-2sby^2}{b(b-a)y^2-x^2}$ and $\frac{t}{s} = \frac{2xy-2by^2}{b(b-a)y^2-x^2}$. So that $t = 2xy-2by^2$, and $s = b(b-a)y^2-x^2$. Farther, $r = 2(b-a)xy - b(b-a)y^2 - x^2$; and, consequently, $p = s + \frac{x}{y}t = b(b-a)y^2 + x^2 - 2bxy = (x-by)^2 - aby^2$. Having therefore found $p, r$, and $s$, it remains to determine $z$; and, for this purpose, let us subtract the first equation, $p^2+azq^2=r^2$, from the second, $p^2+bzq^2=s^2$; the remainder will be $zq^2(b-a) = s^2-r^2 = (s+r) \times (s-r)$. Now, $s+r = 2(b-a)xy-2x^2$, and $s-r = 2b(b-a)y^2-2(b-a)xy$, or $s+r = 2x((b-a)y-x)$, and $s-r = 2(b-a) \times (by-x)y$; so that $(b-a)zq^2 = 2x((b-a)y-x) \times 2(b-a) \times (by-x)y$, or $zq^2 = 2x((b-a)y-x) \times (by-x)2y$, or $zq^2 = 4xy((b-a)y-x) \times (by-x)$; consequently, $z = \frac{4xy((b-a)y-x) \times (by-x)}{q^2}$. We must therefore take the greatest square for $q^2$, that will divide the numerator; but let us observe, that we have already found $p = b(b-a)y^2+x^2-2bxy=(x-by)^2-aby^2$; and therefore we may simplify, by making $x = v+by$, or $x-by=v$; for then $p=v^2-aby^2$, and