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



SUM OF CUBES OF NUMBERS IN ARITHMETICAL PROGRESSION A CUBE. 



L. Euler 245 treated the problem to find three consecutive numbers 

 x 1, x, x+1, the sum 3x 3 -\-Qx of whose cubes is a cube. Since a; = 4 gives 

 a solution, set x = 4 J [-y. Then 6 3 +150?/+362/ 2 +32/ 3 is to be the cube of a 

 number, say 6-\-fy. The coefficients of y are equal if 108/=150, and then 

 1871i/=-7452, x = 32/1871. Or we may take 3x*+6x = 27xW, whence 

 2 (18,z 3 2) =4, and 18z 3 2 is to be a square. Since this is the case for 

 2 = 1, set z = l+y; the cubic in v is the square of 4+27^/4 if v=- 15/32. 



J. R. Young 246 required that the sum of the cubes of a a/x, a, a+afx 

 be a cube. Hence, as by Euler, 3+6/z 2 is to be a cube. To make x 2 = 2n z , 

 take x = 2nq, whence n = 2q*. Then 3n 3 +3 = 24# 6 +3 is to be a cube, which 

 is true if q = l. 



C. Pagliani 247 treated the problem to find 1000 consecutive numbers 

 the sum of whose cubes is a cube. The sum of the cubes ofz+1, - -,x-}-m 

 is s = m(?/+l)(?/ 2 +2i/+m 2 )/8 for y = 2x+m. Let m = Sn 3 . Then s will be 

 the cube of n?+4n 2 if = Q or 



Writing v for 2n, we see that this is equivalent to saying that 



if 6x=(v~ I) 2 3(^+1). Then x is integral if v is not divisible by 3. 

 The cases v = 2, 4, 10 give 



(1) 3 3 +4 3 +5 3 = 6 3 , 6 3 +7 3 H ----- h69 3 = 180 3 , 1134 3 + -+2133 3 = 16830 3 . 



W. Lenhart 248 treated the problem of m consecutive cubes whose sum 

 is a cube. First, let m = 2n. The sum of the cubes of s+1, , s+n, s, 

 s 1, , s n+1 is a = (2s+l)(ns 2 +ns+ft 3 ). Set ?z = 4nf and divide a 

 by (2tti) 3 ; we get 



if 3s = 8nl 4ni 1. To make s an integer >1, take n\ prime to 3. For 

 ni = l, the roots of the 8 cubes are 2, 1, 3, 0, 4, 1, 5, 2, leading to (li). 

 For ni = 2 or 5 we get (1 2 ), (Is). Again, we can equate a to the cube of 

 n+s(2n 2 +l)/(3n) by choice of s in terms of n. Second, let m = 2n+l. 

 Then 



2 = a + (s n) 3 = ms 3 + -sm(m? 1 ) . 



Since S is a cube for s = l/2, set s = l/2+ and take m = m\. Thus 



if < = (mt-2mi-2)/6, whence s = (m?-l) 2 /6. Again, let S = p 3 ?^ 3 s 3 . Then, 



245 Algebra, 2, 1770, art. 249; French transl., 2, 1774, p. 365. Opera Omnia, (1), I, 497-8. 



246 Algebra, 1816; Amer. ed., 1832, 332. 



247 Annales de math, (ed., Gergonne), 20, 1829-30, 382-4. 



248 Math. Miscellany, New York, 2, 1839, 127-132; French transl., Sphinx-Oedipc, 8, 1913, 



81-4. 



