. Carnegie Institution of Washington publication. 546 HISTORY OF THE THEORY OF NUMBERS. [CHAP. XXI are relatively prime integers, so that each is a cube. Since p2+3g2 is a cube, he stated without rigorous proof (cf. Ch. 12, Arts. 188-191) that it is the cube of a number t2+3u2 of like form and that ^S is the cube of . [Cf. papers 6, 10, 27-29, 72 of Ch. XX; also 30, 36 and 183 below.] Hence p = t(tz- 9u2) , q = 3u(t2 - u2) . But also p/4 shall be a cube. The same is true of the product 2p of 2t, f+3w, t — 3w, which are relatively prime since p and hence t is not divisible by 3. Thus the last t
. Carnegie Institution of Washington publication. 546 HISTORY OF THE THEORY OF NUMBERS. [CHAP. XXI are relatively prime integers, so that each is a cube. Since p2+3g2 is a cube, he stated without rigorous proof (cf. Ch. 12, Arts. 188-191) that it is the cube of a number t2+3u2 of like form and that ^S is the cube of . [Cf. papers 6, 10, 27-29, 72 of Ch. XX; also 30, 36 and 183 below.] Hence p = t(tz- 9u2) , q = 3u(t2 - u2) . But also p/4 shall be a cube. The same is true of the product 2p of 2t, f+3w, t — 3w, which are relatively prime since p and hence t is not divisible by 3. Thus the last two are cubes, /3 and 03, whence 2t=f3-\-g3. Thus we have two cubes /3, g3, much smaller than x3, y3, whose sum is a cube 2t. A similar method of descent is used in the remaining case p = 3r, when the product of the relatively prime numbers 9r/4 and 3r2+q2 is a cube. As before, r = 2>u(t2—u2). Since is a cube and is the product of three relatively prime factors, each factor is a cube: t-\-u=f3, t—u = g3, so that/3— g3 is a cube 2u. J. A. Euler9 noted that, if p3+g3+r3 = 0 is possible, x = p2q, y = q2r, z=r2p satisfy xly+y/z+z[x = Q or x2z+y2x+z2y = Q. Inattempting to prove the latter impossible, he stated that yx is divisible by z, but admitted in a note that one can only conclude that the denominator of the irreducible fraction equal to yjz is a divisor of xy. For v = xyfz, we get x/y+v/x+y/v = 0, v<x. Continuing, we get solutions in smaller integers. L. Euler10 noted that p*+q3 = r3 implies AB(A+B) = 1 for A = p2/(gr), B = q2l(pr). Set A = aB. Then 53a(o!+l) = l, whereas «(«+!) is not a cube. N. Fuss I11 noted that a3 = fr3+c3 implies that a6 — 463c3 = (V — c3) 2. Conversely, a6 — 4d3 = D im- plies a3 = p2+p<?3 (since the square root of A2 — dB2 is of the form p2—dq2), whence p = r3, p+<?3 = cube. J. Glenie12 constructed on a given right line BC as base a triangle ABC such that A£3+AC3 = £C3. Through the mid point G of BC draw a perpendicular
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