CHAP, xxi] CUBE OF SUM PLUS ANY ONE A CUBE. 607 



TO FIND n NUMBERS THE CUBE OF WHOSE SUM INCREASED (OR DIMINISHED) 



BY ANY ONE OF THEM GIVES A CUBE. 



Diophantus, V, 18 [19], required three numbers x { such that, if s de- 

 notes their sum, s 3 +x t - [s 3 #J are cubes. Set x f = (a* l)s 3 [_Xi= (1 a'])s 3 ]. 

 Since 2Zi = s, we have (2a< 3)s 2 = l [=--!]. For the first problem, 



i 3 = D, take ai = m+l, a 2 = 2 jn, a 3 = 2; then 



if w = 2/15; thus s = 5/18. For the second problem, 3 2a?=D, a']<l, 

 Diophantus took the square to be 2J, whence 2al = J = 162/216. Hence 

 we have to express 162 as the sum of three cubes. Now 162 = 125+6427. 

 By the theorem in the " Porisms," the difference of two cubes is always a 

 sum of two cubes. Having thus the three cubes [not given by Diophantus] 

 and 2s 2 = 1, whence s = 2/3, we obtain the numbers x t . Cf. Bachet. 404 



Diophantus, V, 20, required three numbers x* of sum s, such that Xi s 3 

 are cubes. Set x i =(a*+l')s 3 . Then Sa?+3 is to be a square 1/s 2 . Let 

 ai = m, a 2 = 3-m, a 3 = l. Then 9ra 2 -27ra+31 = D = (3w-7) 2 , say, whence 

 m = 6/5, s = 5/l7. 



C. G. Bachet 404 believed that Diophantus had found by accident the 

 square 1\ which 3 exceeds by a number expressible as a sum of three cubes 

 < 1, and stated that he could not solve the problem if 2 \ be replaced by 2f . 

 He completed the computation omitted by Diophantus. [By Vieta's 38 

 formula (1)], 64-37 is the sum of the cubes of 40/91 and 303/91. Thus 

 162/216 is the sum of the cubes 125/216, 20 3 /(31 3 -27), 101 3 /(91 3 -8). Sub- 

 tracting them from unity and multiplying the remainders by s 3 = (2/3) 3 , 

 we obtain the answers 91/27 2 , etc., which Bachet expressed as fractions with 

 a common denominator, but with the common factor 27 in all terms. 415 

 The reduced denominator is 549353259 = 91 3 -27 2 . 



A. Girard 39a noted that we may employ Bachet's value 2f since 

 3 2j = 162/9 3 is a sum of three cubes. Or we may employ 2^-f which 3 

 exceeds by the sum 440/1000 of the cubes 216/1000, 216/1000 and 8/1000; 

 the resulting solution of Diophantus V, 19 is 49/256, 49/256, 62/256 [since 



Fermat 405 would not admit that Diophantus was led to 2| by chance 

 and remarked that it is not difficult to rediscover his method. " Take 

 x 1 as the side of the required square between 2 and 3. Then 3 (x I) 2 

 is to be the sum of three cubes. Take as sides of two of the cubes linear 

 functions of x such that, if the sum of their cubes be subtracted from 

 2+2# z 2 , the result contains only two terms in x of consecutive degrees. 

 This can be done in an infinitude of ways. Take lx/3 and 1+rc as the 

 sides of two of the cubes ; then the result mentioned is 



-13 26 



^^_^^___ /y* /yO 



3 27 



Equating this to c 3 z 3 , we have x = 117/(27c 3 26). We are to choose 



404 Diophanti Alex. Arith., 1621, 324. 



405 Oeuvres, III, 25S-9. 



