document.write( "Question 1033739: Find all natural n such that
\n" ); document.write( "n^3 + 8 is prime. \n" ); document.write( "

Algebra.Com's Answer #648401 by robertb(5830) $\"\"$ $\"About$

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$\"n%5E3+%2B+8+=+n%5E3+%2B+2%5E3+=+%28n%2B2%29%28n%5E2-2n%2B4%29\"$ .\r
\n" ); document.write( "\n" ); document.write( "Since $\"n%5E3+%2B+8+\"$ is supposed to be prime, then either\r
\n" ); document.write( "\n" ); document.write( "(i) n+2 = 1 and $\"n%5E2-2n%2B4+=+n%5E3+%2B+8\"$ \r
\n" ); document.write( "\n" ); document.write( "OR\r
\n" ); document.write( "\n" ); document.write( "(ii) $\"n%2B2+=+n%5E3+%2B+8\"$ and $\"n%5E2-2n%2B4+=+1\"$ .\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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.\r
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\n" ); document.write( "\n" ); document.write( "Therefore we conclude that there are no natural numbers n such that $\"n%5E3+%2B+8+\"$ is prime.
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