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?
Alice, Bob, and Carol used bases from 1 to 12, giving three solutions:
1 6 8 giving 1 + 216 + 512 = 729 = 9 cubed
3 4 5 giving 27 + 64 + 125 = 216 = 6 cubed
6 8 10 giving 216 + 512 + 1000 = 1728 = 12 cubed
The first and third share bases 6 and 8 so are Alice and Bob (in either order), and the second (which shares no base with either) must be Carol's list (6,8,10).
Testing Diane just for two-digit bases greater than 12, gives
9 12 15 for 729 + 1728 + 3375 = 5832 = 18 cubed QED
Cubes
