A261296 Smaller of pairs (m, n), such that the difference of their squares is a cube and the difference of their cubes is a square.

6, 384, 4374, 5687, 24576, 17576, 27783, 64350, 93750, 354375, 279936, 113750, 363968, 166972, 370656, 705894, 263736, 1572864, 1124864, 1778112, 3187744, 4225760, 4118400, 3795000, 3188646, 4145823, 4697550, 1111158, 730575, 6000000, 8171316, 2413071, 8573750

Offset

1,1

Comments

The numbers come in pairs: (6,10), (384, 640) etc. The larger numbers of the pairs can be found in A261328. The sequence has infinite subsequences: Once a pair is in the sequence all its zenzicubic multiples (i.e., by a 6th power) are also in this sequence. Primitive solutions are (6,10), (5687, 8954), (27883, 55566), (64350, 70434), ....

Assumes m, n > 0 as otherwise (k^6, 0) will be a solution. Sequence sorted by increasing order of largest number in pair (A261328). - Chai Wah Wu, Aug 17 2015

References

H. E. Dudeney, 536 Puzzles & Curious Problems, Charles Scribner's Sons, New York, 1967, pp 56, 268, #177

Links

Chai Wah Wu, Table of n, a(n) for n = 1..302

Gianlino, in reply to Smci, Solution method for "integers with the difference between their cubes is a square, and v.v.", Yahoo! answers, 2011

Example

10^3 - 6^3 = 784 = 28^2, 10^2 - 6^2 = 64 = 4^3.
8954^3 - 5687^3 = 730719^2, 8954^2 - 5687^2 = 363^3.

Prog

(Python)
def cube(z, p):
....iscube=False
....y=int(pow(z, 1/p)+0.01)
....if y**p==z:
........iscube=True
....return iscube
for n in range (1, 10**5):
....for m in range(n+1, 10**5):
........a=(m-n)*(m**2+m*n+n**2)
........b=(m-n)*(m+n)
........if cube(a, 2)==True and cube(b, 3)==True:
............print (n, m)

Crossrefs

Cf. A000290, A000578, A001014, A261328.

Sequence in context: A270558 A245398 A078207 * A060871 A370845 A193133

Adjacent sequences: A261293 A261294 A261295 * A261297 A261298 A261299

Keyword

nonn

Author

Pieter Post, Aug 14 2015

Extensions

Added a(6) and more terms from Chai Wah Wu, Aug 17 2015

Status

approved