CHAP. XXV] WAKING'S PROBLEM. 719 



positive cubes. For a. and a' odd and relatively prime and for every 

 A' such that a<a f <a*l8, Saa'^A'^a' 3 , it is shown that there exist positive 

 integers m and ra' satisfying the earlier conditions and also am+a'm' = A f . 

 Hence every integral multiple QA of 6, for which 



6(a 3 + /3 )+48aa / ^6A^6(a 3 + Q: /3 )+6a' 3 , a < a' < a 2 /8, 



with a, a' odd and relatively prime, is a sum of at most twelve positive cubes. 

 Taking a = y 2, a' = 7, we see that the intervals obtained by varying 7 

 overlap if 7 exceeds a finite limit and is odd. Hence every multiple of 6 

 exceeding a certain finite limit is a sum of at most twelve positive cubes, 

 whence N 3 = 12+5 (at least five cubes being or 1). 



G. Oltramare 12 proved that any positive cube is the sum of 9 smaller 

 cubes ^0. Any number N is the sum a 2 +6 2 +c 2 +d 2 of four squares. Then 

 8x 3 +6xN is the sum s of the cubes of xa, xb, xc, xd. For N odd, 

 N = 2x+l, we have N 3 = ! 3 +s. For Ni = 2 k N, where N is odd, we multiply 

 the last formula by 2 3fc . 



G. B. Mathews 13 argued that there is a considerable probability that 

 all sufficiently large integers are expressible as sums of p+l pth powers, 

 at least for some positive integers p. According to Kempner 42 , this is not 

 true when p is 6 or any power of 2. 



E. Maillet 14 proved that if (t>(x)=ax 5 +aix*-{ \-a* equals a positive 



integer for every integer x I^AI, then every integer n exceeding a certain func- 

 tion of a, , a 6 is the sum of a limited number N of positive numbers 0(z) 

 and a limited number of units, where N is at most 6, 12, 96, 192 when <f> is 

 of degree 2, 3, 4, 5, respectively. For each function tf>(x), the number of 

 representations of n obtained increases indefinitely with n. 



E. Lemoine 15 stated that every integer equals p+s, where s is a cube or 

 a sum of distinct cubes, while p is one of the 24 numbers 6, 8 17, 27 33. 



L. Ripert 16 proved this statement. 



R. D. von Sterneck 17 gave a table showing the number of cubes needed for 

 the representation of all numbers ^40000. From 8042 on, six cubes suffice. 

 He stated incorrectly [Fleck 20 ] that 3& 3 is not the sum of three cubes un- 

 less they are equal. He conjectured incorrectly [Kempner 42 ] that always 

 about ten of any thousand consecutive numbers are sums of two cubes. 



G. Vacca, 18 after citing Euler's statement, noted that 2 n -v l is the 

 sum of v 1 numbers each 2" and 2 n 1 units. [Thus, for n = 2, 7 is the 

 sum of 4, 1, 1, 1, but not a sum of fewer than 4 squares; for n = 3, 23 is the 

 sum of 8, 8 and seven units, but not a sum of fewer than 9 positive cubes; 

 for n = 4, 79 is the sum of 16, 16, 16, 16 and 15 units, but not a sum of fewer 

 than 19 bi quadrates.] 



12 L'interme'diaire des math., 2, 1895, 30. 



13 Messenger of Math., 25, 1895-6, 69. 



"Jour, de Math., (5), 2, 1896, 363-380; Bull. Soc. Math. France, 23, 1895, 40-49. Cf. 

 papers 68, 72, 73, 117, 181-2 of Ch. I. 



16 Nouv. Ann. Math., (3), 17, 1898, 196. 

 Ibid., (3), 19, 1900, 335-6. 



17 Sitzungsber. Akad. Wiss. Wien (Math.), 112, Ha, 1903, 1627-66. 



18 L'interme'diaire des math., 11, 1904, 292-3. 



