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2008 a fourth power? Baseless! (Posted on 2010-10-09) Difficulty: 3 of 5
Can the base B number 2008 be a perfect fourth power, where B is a positive integer ≥ 9?

If so, give an example. If not, prove that the base B number 2008 can never be a perfect fourth power.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution solution | Comment 2 of 3 |
There is no such number.

2008 base B = 8 + 2*B^3 = k^4 for some k

8+2*B^3 = 2(B^3 + 4).

Now, since this side is even, then k^4 is also even, which means k itself is even and k^4 must be divisible by 16 (each k contributes at least a factor of 2)

So we'd have 2(B^3+4) = 16x for some x or
B^3 + 4 = 8x for some x.

If B is odd, the l.h.s. is odd and can't ever be a multiple of 8. But if B is even then B^3 is itself a multiple of 8 so B^3+4 is still not divisible by 8. Thus, whether B is odd or even, there is no solution.
  Posted by Paul on 2010-10-09 17:54:53
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