Elements of Algebra

422 ELEMENTS PART II. their product, $p^4 - 4q^4$, to become a square; but this we have likewise shewn to be impossible. (Art. 209.) In general, let $b - a = m$, $c - a = n$, and $z = \frac{p}{q}$; then the formulæ $p^2 + mq^2$, and $p^2 + nq^2$, must become squares; but we have seen that this is impossible, both when $m = +1$, and $n = -1$, and when $m = +2$, and $n = -2$. It is also impossible, when $m = f^2$, and $n = -f^2$; for, in that case, we should have two formulæ, whose product would be $p^4 - f^4q^4$, that is to say, the difference of two biquadrates; and we know that such a difference can never become a square. Likewise, when $m = 2f^2$, and $n = -2f^2$, we have the two formulæ $p^2 + 2f^2q^2$, and $p^2 - 2f^2q^2$, which cannot both become squares, because their product $p^4 - 4f^4q^4$ must become a square. Now, if we make $fq = r$, this product is changed into $p^4 - 4r^4$, a formula, the impossibility of which has been already demonstrated. If we suppose $m = 1$, and $n = 2$, so that it is required to reduce to squares the formulæ $p^2 + q^2$, and $p^2 + 2q^2$, we shall make $p^2 + q^2 = r^2$, and $p^2 + 2q^2 = s^2$; the first equation will give $p^2 = r^2 - q^2$, and the second will give $r^2 + q^2 = s^2$; and therefore both $r^2 - q^2$, and $r^2 + q^2$, must be squares: but the impossibility of this is proved, since the product of these formulæ, or $r^4 - q^4$, cannot become a square. These examples are sufficient to shew, that it is not easy to choose such numbers for $m$ and $n$ as will render the solution possible. The only means of finding such values of $m$ and $n$, is to imagine them, or to determine them by the following method. Let us make $f^2 + mg^2 = h^2$, and $f^2 + ng^2 = k^2$; then we have, by the former equation, $m = \frac{h^2 - f^2}{g^2}$, and, by the latter, $n = \frac{k^2 - f^2}{g^2}$; this being done, we have only to take for $f, g, h$, and $k$, any numbers at pleasure, and we shall have values of $m$ and $n$ that will render the solution possible. For example, let $h = 3, k = 5, f = 1$, and $g = 2$, we shall have $m = 2$, and $n = 6$; and we may now be certain that it is possible to reduce the formulæ $p^2 + 2q^2$ and $p^2 + 6q^2$ to squares, since it takes place when $p = 1$, and $q = 2$. But the former formula generally becomes a square, if $p = r^2 - 2s^2$, and $q = 2rs$; for then $p^2 + 2q^2 =$