Elements of Algebra

460 ELEMENTS PART II. 3. Let $f = 3$, $g = 1$, $h = 1$, and $k = 1$; so that $f^2 + 3g^2 = 12$, $h^2 + 3k^2 = 4$; also $t = -24$, and $u = 32$. Here, these two values being divisible by 8, and as we consider only their ratios, we may make $t = -3$, and $u = 4$. Whence we obtain $p = 3t + 3u = +3$, $q = t - 3u = -15$, $r = t - u = -7$, and $s = t + 3u = +9$; consequently, $x = -12$, and $y = 18$, $z = -16$, and $v = 2$, whence $-12^3 + 18^3 - 16^3 = 2^3$, or $18^3 = 16^3 + 12^3 + 2^3$, or, dividing by the cube of 2, $9^3 = 8^3 + 6^3 + 1^3$. 4. Let us also suppose $g = 0$, and $k = h$, by which means we leave $f$ and $h$ undetermined. We shall thus have $f^2 + 3g^2 = f^2$, and $h^2 + 3k^2 = 4h^2$; so that $t = 12h^3$, and $u = f^3 - 4h^3$; also, $p = st = 12fh^3$, $q = -f^4 + 4fh^3$, $r = 12h^4 - hf^3 + 4h^4 = 16h^4 - hf^3$, and $s = 3hf^3$; lastly, $x = p + q = 16fh^3 - f^4$, $y = p - q = 8fh^3 + f^4$, $z = r - s = 16h^4 - 4hf^3$, and $v = r + s = 16h^4 + 2hf^3$. If we now make $f = h = 1$, we have $x = 15$, $y = 9$, $z = 12$, and $v = 18$; or, dividing all by 3, $x = 5$, $y = 3$, $z = 4$, and $v = 6$; so that $3^3 + 4^3 + 5^3 = 6^3$. The progression of these three roots, 3, 4, 5, increasing by unity, is worthy of attention; for which reason, we shall investigate whether there are not others of the same kind. 249. Question 4. Required three numbers, whose difference is 1, and forming such an arithmetical progression, that their cubes added together may make a cube. Let $x$ be the middle number, or term, then $x-1$ will be the least, and $x+1$ the greatest; the sum of the cubes of these three numbers is $3x^3 + 6x = 3x(x^2+2)$, which must be a cube. Here, we must previously have a case, in which this property exists, and we find, after some trials, that that case is $x=4$. So that, according to the rules already given, we may make $x = 4+y$; whence $x^2 = 16 + 8y + y^2$, and $x^3 = 64 + 48y + 12y^2 + y^3$, and by these means our formula becomes $216 + 150y + 36y^2 + 3y^3$, in which the first term is a cube, but the last is not. Let us, therefore, suppose the root to be $6+fy$, or the formula to be $216 + 108fy + 18f^2y^2 + f^3y^3$, and destroy the two second terms, by writing $108f = 150$, or $f = \frac{25}{18}$; the other terms, divided by $y^2$, will give $36 + 3y = 18f^2 + f^3y = 18\frac{25^2}{18^2} + \frac{25^3}{18^3}y$, or