Can both
n + 3 and
n^2 + 3 be perfect cubes if n is an integer ?
Looking at the one's place:
0 0 0
1 1 1
2 4 8
3 9 7
4 6 4
5 5 5
6 6 6
7 9 3
8 4 2
9 1 9
Since the ones place of n^2 + 3 can end in any digit, it comes down to the squares. Since squares can only end in a certain way, this means that n^2+3 must be the cube of something ending in 2, 3, 4, 7, 8, 9.
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Posted by Gamer
on 2003-08-12 22:34:19 |