The sum of cubes n cubed plus eight is never prime

Statement

Prove that, for every integer $n > 1$, the number $n^3 + 8$ is not prime.

Solution

Write $8 = 2^3$ and use the sum of cubes factorization $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ with $x = n$, $y = 2$: $n^3 + 8 = (n + 2)\,(n^2 - 2n + 4).$ Check that both factors exceed $1$ when $n > 1$:

• $n + 2 > 3 > 1$;
• $n^2 - 2n + 4 = (n-1)^2 + 3 \ge 3 > 1$.

So $n^3 + 8$ is a product of two integers both $> 1$, hence composite (not prime). $\blacksquare$