CHAP. XXI] SUM OF CUBES OF NUMBERS A SQUARE. 587 



equation, for the case a = l, whence the sum of x, x+r, , x+(n l)r 

 is 6 2 . In the most interesting case = 1, r = 2, the eliminant becomes 

 a 2 _ (4_|_ j) n 2 = _ 1 for 6 = nZ. It has an infinitude of solutions (a, n) = ( 2 , 1), 

 (4 6 +3Z 2 , 4 4 +l), etc. Taking t= 1, we have z = 1 and the following result: 

 While the sum of n consecutive odd numbers 1,3, is always a square n 2 , 

 the sum of the cubes of the same n odd numbers will be the square of an 

 when a 2 -2n 2 = -1. Examples are (a, n) = (1, 1), (7. 5), (41, 29), (239, 169). 



E. Lucas 275a stated that the sum of the cubes of five consecutive integers 

 is a square only when the middle number is 2, 3, 98 or 120. The sum of 

 two consecutive cubes is a square only for the cubes 1 and 8. 



G. R Perkins' 276 solution of (1) differs only in notation from Genocchi's 253 . 



E. Lucas 277 asked when the sum of 7 consecutive cubes is a square. 



Several 278 found that the sum of the cubes of the first n odd integers is 

 a square if 2n 2 1 = D, w = l, 5, 29, 



M. A. Gruber 279 attempted to show that a sum of cubes of n consecutive 

 integers is a square only for 1 3 +2 3 H ----- \-n*= (1-1 ----- |-n) 2 . 



A. Cunningham 280 desired a sum of successive odd cubes equal to a 

 square. The sum S r of the successive odd cubes 1, 3 3 , , (2r I) 3 is 

 r 2 (2r 2 1) and is a square if r = 5. Next, 



is a square z 2 if, upon setting x = 2r 2 , y = 2p 2 , 



Solutions are found by making special assumptions. 



W. A. Whitworth 281 expressed V2 as a continued fraction, took a conver- 

 gent N/D, with D odd, and got 



1 3 +3 3 H ----- K2D-1) 3 = N 2 D 2 . 

 Cunningham 282 asked for a sum of successive cubes 



S m , n = (n+l) 3 +(n+2Y+ -+m 3 

 equal to the product of a square by q. Since 



S m ,o=l 3 +2 3 +---+m 3 = 7t, T m = m(m+l), S m , n = S m , -S n< , 

 we set T m = T n and see that S m , n +q is a square if ( 2 -l)/# is a square. 

 For each such , we test T m = T n by a table of triangular numbers (de 

 Joncourt's, 1772) and find suitable pairs m, n. Solutions are found for 

 5 = 2, .--,11. 



M. A. Gruber 283 noted that n = l and n = 5 are the only cases in which 



! 3 +3 3 +5N ----- t-(2rc-l) 3 = D, (2n-l) 3 = D. 



2750 Recherches BUT Panalyse indeterminde, Moulins, 1873, 92. Extract from Bull. Soc. 

 d'Emulation du Departement de 1'Allier, 12, 1873, 532. 



276 The Analyst, Des Moines, 1, 1874, 40. 



277 Nouv. Corresp. Math., 2, 1876, 95. 



278 Math. Quest. Educ. Times, 53, 1890, 55. Cf. Brocard 92 of Ch. XXIII. 



279 Amer. Math. Monthly, 2, 1895, 197-8. 



280 Math. Quest. Educ. Times, 72, 1900, 45-46 (error); 73, 1900, 132-3. 



281 Ibid., 72, 1900, 46. 

 Ibid., 75, 1901, 87-88. 



283 Amer. Math. Monthly, 7, 1900, 176. 



