608 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxi 



c so that 1 -rr/3 >0. Since the third cube is negative, we apply 406 the 

 Porism. Here Bachet was again embarrassed; he confessed that he could 

 express the difference of two given cubes as a sum of two cubes only when 

 the greater of the given cubes exceeded the double of the smaller." 



Ludolph van Ceulen 407 (1540-1610), at the end of his Dutch work on the 

 circle, proposed 100 problems the 68th and 69th of which are to find three 

 and four numbers such that if each be subtracted from the cube of their 

 sum the remainders are cubes. For three numbers his solution, communi- 

 cated in letters, is the one published by van Schooten, 408 his successor at 

 the University of Leyden. After learning van Ceulen's method, N. Huberti 

 obtained answers for four numbers, quoted by van Schooten : 



867160/C, 787400/C, 13527640/D, 14087528/D 



(C = 4657463, D = 125751501) ; 



12172736/&, 11296152/&, 9112168/&, 4724776/A; (k = 64481201); 

 and the further answer for three numbers : 



15817815000/G, 9568152000/G, 8925120000/& (G = 86526834967) . 



Frans van Schooten 408 first found three cubes such that on subtracting 

 them from 4 3 , the sum of the remainders is a square. Let the roots of 

 the cubes be N l, 4 N, 2. Subtracting their cubes from 64, we get 

 G5-3N+3N*-N 3 , 48N~12N 2 +N 3 and 56. Equate their sum 



121+45N-9N 2 



to (11 +A0 2 ; hence AT = 23/10. Thus the above remainders equal 

 a = 61803/1000, 6 = 59087/1000, 56. 



Now let the three desired numbers be an 3 , bn 3 , 56n 3 and their sum 4w, 

 whence n = 20/1 33. Hence the answer is 



494424/D, 472696/I>, 448000/D (D = 2352637) . 



Their sum is 80/133, whose cube diminished by the three numbers gives 

 as remainders the cubes of 26/133, 34/133, 40/133. 



J. H. Rahn and J. Pell 409 treated Diophantus V, 18, 19, 20 at length. 

 Pell's solution differs little from van Schooten's, except in using (11+mAO 2 

 in place of (11 +A 7 ") 2 , and is given in Wallis' Algebra, Engl. ed., 1685, p. 219, 

 with van Schooten's answer. 



J. Kersey 410 employed, for Diophantus V, 19, a t = 53/144, a 2 = 27/144, 

 a 3 = 16/144, whence 3 - Sa'J = (247/144) 2 , s = 144/247 ; thus the desired num- 



406 To secure Diophantus' value (x 1) 2 = 2\, we must take z = 5/2 or 1/2, whence 



27c 3 = 364/5 or 8-26, so that c is irrational. Hence Fermat's process is not general, 

 although it leads to a solution by setting c = 5/3, whence x = 13/11, and the sides of 

 the cubes are 20/33, 72/33, -65/33, as noted by Heath, " Diophantus," ed. 2, 214. 



407 Van den Circkel, 1596, 1615. Latin transl. by W. Snellius, 1619. Cf. Bull. Bibl. Storia 



Sc. Mat. e Fis., 1, 1868, 141-156. 



408 Exercitationvm Math., 1657, Liber V, Sect. 13, 434-6. Reproduced by C. Hutton, The 



Diarian Miscellany, London, 1, 1775, 138-9. 



409 Rahn's Algebra, Zurich, 1659. An Introduction to Algebra, transl. out of the High-Dutch 



by T. Brancker, much altered and augmented by D. P[ell], London, 1668, 105-131. 

 Cf. Wallis' Algebra, Ch. 59. 



410 The Elements of Algebra, London, Book III, 1674, 111-4, 104-5. 



