546 



HISTORY OF THE THEORY OF NUMBERS. 



[CHAP. XXI 



are relatively prime integers, so that each is a cube. Since p 2 +3g 2 is a cube, 

 he stated without rigorous proof (cf. Ch. 12, Arts. 188-191) that it is the 

 cube of a number t 2 +3u 2 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(t z - 9u 2 ) , q = 3u(t 2 - u 2 ) . 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 3 , whence 2t=f 3 -\-g 3 . Thus we have two cubes / 3 , g 3 , much smaller 

 than x 3 , y 3 , 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 3r 2 +q 2 is a cube. As before, r = 2>u(t 2 u 2 ). Since 



is a cube and is the product of three relatively prime factors, each factor is 

 a cube: t-\-u=f 3 , tu = g 3 , so that/ 3 g 3 is a cube 2u. 



J. A. Euler 9 noted that, if p 3 +g 3 +r 3 = is possible, x = p 2 q, y = q 2 r, 

 z=r 2 p satisfy xly+y/z+z[x = Q or x 2 z+y 2 x+z 2 y = 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. Euler 10 noted that p*+q 3 = r 3 implies AB(A+B) = 1 for A = p 2 /(gr), 

 B = q 2 l(pr). Set A = aB. Then 5 3 a(o!+l) = l, whereas (+!) is not a 

 cube. 



N. Fuss I 11 noted that a 3 = fr 3 +c 3 implies that 

 a 6 46 3 c 3 = (V c 3 ) 2 . Conversely, a 6 4d 3 = D im- 

 plies a 3 = p 2 +p<? 3 (since the square root of A 2 dB 2 

 is of the form p 2 dq 2 ), whence p = r 3 , p+<? 3 = cube. 

 J. Glenie 12 constructed on a given right line 

 BC as base a triangle ABC such that A 3 +AC 3 

 = C 3 . Through the mid point G of BC draw a 

 perpendicular GH to it and take 



, 



2 A/31 



Draw the circle HBC; let it cut the parallel FA to 

 BC at ^4. Without proof he stated that ABC is the required triangle. 

 To make AB 3 +AC 3 = 2BC 3 or 3BC 3 (Probs. 2, 3), take 



or 

 He treated the corresponding three problems on the difference of cubes. 



9 L. Euler's Opera postuma, I, 1862, 230-1 (about 1767). 



10 Ibid., 236-7 (about 1769). 



11 Ibid., 242 (about 1778). 



"The Antecedental Calculus . . . and the Constructions of Some Problems, London, 1793, 

 16 pp., p. 13. 



