CHAP. XXI] SUM OF CUBES OF NUMBERS A SQUARE. 585 



E. B. Escott 261 proved that, for 2^n^5, 



has only the following integral solutions: (li) and 



3 2 +4- = 5 2 , (-2) 3 +(-l) 3 +0 :i +r ! +-- 



L. Matthiessen 262 noted that if fractional values of x, v are allowed in 

 (7), we may set r = l. Write u = 2x+n 2, v = pu-\-n/2. The usual form 

 of (7) becomes a quadratic in u : 



Evident solutions are obtained by equating to zero the first or third coeffi- 

 cient. In the second case, integers x are found only for n = 2, n = 4. 



F. Hromadko 263 noted that x = 3 is the only positive integral solution of 

 x 3 +(x+l) 3 +(x+2) 3 = (x+3Y [Lebesgue 249 ]. 



E. Grigorief 264 obtained the special solutions 



15 2 +52 3 +89 3 H ----- f-348 3 = 495 3 , 76 3 +477 3 +878 3 H ----- |-2883 3 = 3016 3 , 

 435 3 +506 3 +577 3 +648 3 +719 3 +790 3 = 1155 3 . 



" L. N. Machaut " 265 treated (2) by setting xfr = u and obtaining a cubic 

 for u with a real positive root (u = 3) only for n = 3, leading to (li). 



J. N. Vischers 266 proved Lebesgue's 249 first result when ?i = 3. 



L. Aubry 267 proved that 3, 4, 5 are the only three consecutive integers 

 the sum of whose cubes is a cube. 



SUM OF CUBES OF NUMBERS IN ARITHMETICAL PROGRESSION A SQUARE. 



To find five integers in A. P. the sum of whose cubes is a square (or 

 sum of squares is a cube), J. Stevenson 268 used nx 2x, nxx, nx, nx+x, 

 nx+2x, the sum of whose cubes 5n 3 3 +30nx 3 will equal m 2 x- by choice of x 

 (or sum of squares 5n 2 a; 2 +10x 2 = m 3 x 3 by choice of x). Several solved both 

 questions simultaneously by using x 2 , 2x 2 , 3x 2 , 4x 2 , 5x 2 , whose sum of cubes 

 is (15x 3 ) 2 and sum of squares is 55z 4 = a 3 3 , if z = a 3 /55; take a = 55. 



Several 269 made the sum n 2 (2n 2 1) of the cubes of the first n odd integers 

 a square by using Euler's 83 solutions (Ch. XII) n = 1, 5, 29, of 2n 2 1 = D . 



A. Genocchi 253 discussed the rational solutions of 



(1) z 3 +(z+r) 3 +(z+2r) 3 H ---- + (x+nr-r}* = y\ 



In view of (3) of Lebesgue 249 the problem is nsa - 8?/ 2 . Set 2y = ?is. Solving 

 ff = 2nst 2 for s, we see that nH 4 -(n 2 -l}r 2 = D = (nt'-rp} 2 . Hence 



*-lt 2 or 



261 L'intermediaire des math., 5, 1898, 254-6; 7, 1900, 141. 



262 Zeitsch. Math. Naturw. Unterricht, 33, 1902, 372-5. 

 2 " Ibid., 34, 1903, 258. 



264 L'intermediaire des math., 9, 1902, 319. 



265 Ibid., 15, 1908, 163-4. 



266 Wiskundig Tijdschrift, 5, 1908, 65. 

 M7 Sphinx-Oedipe, 6, 1911, 142-3. 



268 The Gentleman's Diary, or Math. Repository, London, 1814, 36-7, Quest. 1010. 

 Ladies' Diary, 1832, 36, Quest. 1529. 



