Elements of Algebra

448 ELEMENTS PART II. for the two numbers sought $x = \frac{15}{23}$, and $y = \frac{12}{23}$, by which means $x^2 + y = \frac{225}{529} + \frac{12}{23} = \frac{961}{529} = (\frac{31}{23})^2$, and $y^2 + x = \frac{1024}{529} + \frac{15}{23} = \frac{1369}{529} = (\frac{37}{23})^2$. If we made $p=1$, and $q=3$, we should have $x = -\frac{8}{17}$, and $y = \frac{6}{17}$, an answer which is inadmissible, since one of the numbers sought is negative. But let $p=1$, and $q = \frac{3}{2}$, we shall then have $x = \frac{3}{20}$, and $y = \frac{7}{10}$, whence we derive $x^2 + y = \frac{9}{400} + \frac{7}{10} = \frac{389}{400} = (\frac{17}{20})^2$, and $y^2 + x = \frac{49}{100} + \frac{3}{20} = \frac{64}{100} = (\frac{8}{10})^2$. 240. Question 17. To find two numbers, whose sum may be a square, and whose squares added together may make a biquadrate. Let us call these numbers $x$ and $y$; and since $x^2 + y^2$ must become a biquadrate, let us begin with making it a square: in order to which, let us suppose $x = p^2 - q^2$, and $y = 2pq$, by which means, $x^2 + y^2 = (p^2 + q^2)^2$. But, in order that this square may become a biquadrate, $p^2 + q^2$ must be a square; let us therefore make $p = r^2 - s^2$, and $q = 2rs$, in order that $p^2 + q^2 = (r^2 + s^2)^2$; and we immediately have $x^2 + y^2 = (r^2 + s^2)^4$, which is a biquadrate. Now, according to these suppositions, we have $x = r^4 - 6r^2s^2 + s^4$, and $y = 4r^3s - 4rs^3$; it therefore remains to transform into a square the formula $x + y = r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4$. Supposing its root to be $r^2 + 2rs + s^2$, or the formula equal to the square of this, $r^4 + 4r^3s + 6r^2s^2 + 4rs^3 + s^4$, we may expunge from both the first two terms and also $s^4$, and divide the rest by $rs^2$, so that we shall have $6r + 4s = -6r - 4s$, or $12r + 8s = 0$; or $s = -\frac{12r}{8} = -\frac{3}{2}r$. We might also suppose the root to be $r^2 - 2rs + s^2$, and make the formula equal to its square $r^4 - 4r^3s + 6r^2s^2 - 4rs^3 + s^4$; the first and the last two terms being thus destroyed on both sides, we should have, by dividing the other terms by $r^2s$, $4r - 6s = -4r + 6s$, or $8r = 12s$; consequently, $r = \frac{12}{8}s$; so that by this second supposition, if $r = 3$, and $s = 2$, we shall find $x = -119$, or a negative value. But let us make $r = \frac{3}{2}s + t$, and we shall have for our formula, $r^2 = \frac{9}{4}s^2 + 3st + t^2$; $r^3 = \frac{27}{8}s^3 + \frac{27}{4}s^2t + \frac{9}{2}st^2 + t^3$.