Can both
n + 3 and
n^2 + 3 be perfect cubes if n is an integer ?
If n was even, then n² must end in 0, 4 or 6. This means that n²+3 must end in 3, 7 or 9, and this means b³ (the perfect cube) ends with the same numbers, so b must end with 3, 7, 9
a, (the one whose cube is n + 3), must end in 1, 3, 5, 7, 9.
Ends with:
n a b
0 7 7
2 5 3
4 3 9
6 9 9
8 1 3
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Posted by Gamer
on 2003-08-19 18:54:00 |
Even idea?
