Question 1033739
{{{n^3 + 8 = n^3 + 2^3 = (n+2)(n^2-2n+4)}}}.

Since {{{n^3 + 8 }}} is supposed to be prime, then either

(i) n+2 = 1 and {{{n^2-2n+4 = n^3 + 8}}}

OR

(ii) {{{n+2 = n^3 + 8}}} and {{{n^2-2n+4 = 1}}}.

The first case (i), the 2nd equation {{{ n^3- n^2 +2n+ 4 = 0}}} is equivalent to {{{(n+1)(n^2 - 2n+4)}}}, 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^3 - n + 6 =(n+2)(n^2-2n+3)= 0}}} and {{{n^2-2n+3= 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^2-2n+3= 0}}}.  Thus, there are no solutions arising from the second case.



Therefore we conclude that there are no natural numbers n such that {{{n^3 + 8 }}} is prime.