To show that Fermat's Last Theorem applies only to sums of two terms, Fred asked his friends, Alice, Bob, Carol and Diane, to list three perfect cubes that added up to another perfect cube. Each came up with his or her own list, different from the others'.

All except Diane limited their lists of three cubes to the first 12 cubes. Alice and Bob had two of the same numbers in their lists, but Carol's list had no numbers in common with either of those lists.

Diane, not limiting herself to the first 12 cubes, did use in her list two of the sums from among the three sums of cubes found by Alice, Bob and Carol.

What were Carol's and Diane's lists of cubes?

Carol's list of cubes:
27, 64, 125

27 + 64 + 125 = 216
3³ + 4³ + 5³ = 6³

Diane's list of cubes:
729, 1728, 3375

[Diane's is the same as Carol's, but multiplied by a factor of 3³]729 + 1728 + 3375 = 5832
9³ + 12³ + 15³ = 18³


Alice's and Bob's list of cubes:
1, 216, 512 and 216, 512, 1000

1 + 216 + 512 = 729
1³ + 6³ + 8³ = 9³

216 + 512 + 1000 = 1728
6³ + 8³ + 10³ = 12³
[The second list, provided by either Alice or Bob, is also the same as Carol's list, but multiplied by a factor of 2³.]

Posted by Dej Mar on 2008-05-05 12:10:41