CHAP, xxv] SUM OF FOUR POSITIVE CUBES. 729 



E. Lucas 70 remarked that Lebesgue 69 appears not to have guessed 

 what seems to have led Euler to obtain formula (4), viz., the problem to 

 express a number as a sum of two cubes. Any positive rational number N 

 is expressible in an infinitude of ways as a product or quotient of two sums 

 of two positive rational cubes. To prove the former (which corresponds 

 to Euler's theorem), employ the identities 



Divide their product (member by member) by (L 2 +3M 2 ) 3 . Hence 2 3 3 2 L 2 J M' 

 is expressed as a product of two sums of two cubes. Take L = Bb 3 , 

 M = 2 A - 3 3"- 2 Aa 3 . we get a decomposition of N = 2*3"AB 2 , and we can 

 choose a 3 /6 3 to make all the cubes positive. As a corollary, E. Fauquem- 

 bergue 70a proved that the quadruple and square of 4p 6 +27<f are sums of 

 two cubes, a problem proposed by Lucas. 



G. Oltramare 71 noted that every integer is a sum of five integral cubes. 



R. Norrie 72 gave the identity 



He expressed (p. 58), 5, 17 and 41 as sums of five integral cubes, not all 

 positive. Other solutions had been given by A. Cunningham. 73 



70 Nouv. Ann, Math., (2), 19, 1880, 89-91; Bull. Soc. Math. France, 8, 1879-80, 180-2. No 



reference is made to Euler's writings. The author of this History has found no formula 

 like (2) or (4) in Euler's papers or books. Nor did Libri 65 or Lebesgue 69 imply that 

 such a formula is due to Euler. The fact that Lebesgue spoke of (3) as the trans- 

 formation of Euler may have led Lucas to infer too hastily that also (2) is due to Euler. 

 700 Nouv. Ann. Math., (2), 19, 1880, 430. 



71 L'intermediaire des math., 1, 1894, 25. Cf. 165-6, 244; 2, 1895, 325. 



72 University of St. Andrews 500th Anniversary Mem. Vol., 1911, 68. 



73 Math. Quest. Educ. Times, (2), 4, 1903, 49. 



