Prove that $n^3+2$ is not divisible by $9$ for any integer $n$ How to prove that $n^3+2$ is not divisible by $9$?
 A: Hint: It's clear that if $3|n$ then $n^3+2$ is not divisible by $9$. You just now have to consider the remaining two cases: $n=3k+1$ and $n=3k-1$. The $\pm1$ makes the binomial expansion very straightforward.
A: Consider the following $9$ cases:


*

*$n\equiv0\pmod9 \implies n^3\equiv         0\pmod9 \implies n^3+2\equiv        2\pmod9$

*$n\equiv1\pmod9 \implies n^3\equiv         1\pmod9 \implies n^3+2\equiv        3\pmod9$

*$n\equiv2\pmod9 \implies n^3\equiv         8\pmod9 \implies n^3+2\equiv10\equiv1\pmod9$

*$n\equiv3\pmod9 \implies n^3\equiv 27\equiv0\pmod9 \implies n^3+2\equiv        2\pmod9$

*$n\equiv4\pmod9 \implies n^3\equiv 64\equiv1\pmod9 \implies n^3+2\equiv        3\pmod9$

*$n\equiv5\pmod9 \implies n^3\equiv125\equiv8\pmod9 \implies n^3+2\equiv10\equiv1\pmod9$

*$n\equiv6\pmod9 \implies n^3\equiv216\equiv0\pmod9 \implies n^3+2\equiv        2\pmod9$

*$n\equiv7\pmod9 \implies n^3\equiv343\equiv1\pmod9 \implies n^3+2\equiv        3\pmod9$

*$n\equiv8\pmod9 \implies n^3\equiv512\equiv8\pmod9 \implies n^3+2\equiv10\equiv1\pmod9$

A: Suppose, $\exists n\in \mathbb{N}$ such that $n^3+2\equiv 0 \pmod{9}\implies n^{6}\equiv 4\pmod{9}$, which is not true since\begin{array}{rl}
n^6\equiv 1\pmod 9 & \mbox{if $n$ and $9$ are relativley prime,  by Euler's theorem, since $\phi(9)=6$.}\\ 
n^6\equiv 0\pmod 9 & \mbox{otherwise} 
\end{array}  
A: The cubes modulo $9$ are $0$, $1$ and $8$; now
$$
0+2\equiv2,\quad 1+2\equiv3,\quad 8+2\equiv1\pmod{3}
$$
Computations are easy, because $(n+3)^3=n^3+9n^2+27n+27\equiv n^3\pmod{3}$, so we just need to do $0^3$, $1^3$ and $2^3$.
A: Write $n=3k+j$ and 
$$(3k+j)^3+2=9(3k^3+3k^2+k)+j^3+2$$
and use induction.
