Elements of Algebra

CHAP. XV. OF ALGEBRA. 459 5. Let us now remove the fractions, by making $u = f(f^2 + 3g^2) - h(h^2 + 3k^2)$; then $t = 3k(h^2 + 3k^2) - 3g(f^2 + 3g^2)$, where we may give any values whatever to the letters $f, g, h$, and $k$. 6. When therefore we have determined, from these four numbers, the values of $t$ and $u$, we shall have $p = ft + 3gu$, $q = gt - fu$, $r = kt - hu$, $s = ht + 3ku$; whence we shall at last arrive at the solution of the question, $x = p + q$, $y = p - q$, $z = r - s$, and $v = r + s$; and this solution is general, so far as to comprehend all the possible cases, since in the whole calculation we have admitted no arbitrary limitation. The whole artifice consisted in rendering our equation divisible by $t^2 + 3u^2$; for we have thus been able to determine the letters $t$ and $u$ by an equation of the first degree; and innumerable applications may be made of these formulæ, some of which we shall give for the sake of example. 1. Let $k = 0$, and $h = 1$, we shall have $t = -3g(f^2 + 3g^2)$, and $u = f(f^2 + 3g^2) - 1$; so that $p = -3fg(f^2 + 3g^2) + 3fg(f^2 + 3g^2) - 3g$, or $p = -3g$; $q = -(f^2 + 3g^2)^2 + f$; $s = -3g(f^2 + 3g^2)$; $r = -f(f^2 + 3g^2) + 1$; consequently, $x = -3g - (f^2 + 3g^2)^2 + f$, $y = -3g + (f^2 + 3g^2)^2 - f$, $z = (3g - f) \times (f^2 + 3g^2) + 1$; lastly, $v = -(3g + f) \times (f^2 + 3g^2) + 1$. If we also suppose $f = -1$, and $g = +1$, we shall have $x = -20$, $y = 14$, $z = 17$, and $v = -7$; and thence results the final equation, $-20^3 + 14^3 + 17^3 = -7^3$, or $14^3 + 17^3 + 7^3 = 20^3$. 2. Let $f = 2$, $g = 1$, and consequently $f^2 + 3g^2 = 7$; farther, $h = 0$, and $k = 1$; so that $h^2 + 3k^2 = 3$; we shall then have $t = -12$, and $u = 14$; so that $p = 2t + 3u = 18$, $q = t - 2u = -40$, $r = t = -12$, and $s = 3u = 42$. From this will result $x = p + q = -22$, $y = p - q = 58$, $z = r - s = -54$, and $v = r + s = 30$; therefore, $30^3 = 22^3 + 58^3 - 54^3$, or $58^3 = 30^3 + 54^3 + 22^3$; and as all these roots are divisible by $2$, we shall also have $29^3 = 15^3 + 27^3 + 11^3$.