SOLUTION: Find all natural n such that n^3 + 8 is prime.

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Question 1033739: Find all natural n such that
n^3 + 8 is prime.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
n%5E3+%2B+8+=+n%5E3+%2B+2%5E3+=+%28n%2B2%29%28n%5E2-2n%2B4%29.
Since n%5E3+%2B+8+ is supposed to be prime, then either
(i) n+2 = 1 and n%5E2-2n%2B4+=+n%5E3+%2B+8
OR
(ii) n%2B2+=+n%5E3+%2B+8 and n%5E2-2n%2B4+=+1.
The first case (i), the 2nd equation +n%5E3-+n%5E2+%2B2n%2B+4+=+0 is equivalent to %28n%2B1%29%28n%5E2+-+2n%2B4%29, which when equated to zero will only give the real solution n = -1. This also satisfies the first equation n+2=1. But n = -1 is not a natural number, so we get no solutions from this case.
For the second case, we get n%5E3+-+n+%2B+6+=%28n%2B2%29%28n%5E2-2n%2B3%29=+0 and n%5E2-2n%2B3=+0. The first equation has only real solution n = -2, while the second equation has only complex solutions, and clearly n = -2 does not satisfy n%5E2-2n%2B3=+0. Thus, there are no solutions arising from the second case.


Therefore we conclude that there are no natural numbers n such that n%5E3+%2B+8+ is prime.