Prove $n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0

Question

The statement I'm trying to prove is:

$n^3 + 7n + 3$ is divisible by 3 for all integers n ≥ 0

I eventually need to prove $(k + 1)^3 + 7(k + 1) + 3$ is divisible by 3.

I don't really understand how to deal with $k + 1$, so I'm a little lost.

I've know that the base case of P(0) is true, but I'm not sure about proving the inductive case.

Answer 1

The statement is not true. Take $n=1$ as a counter example.

Since it's not true, you won't manage to prove it (by induction or otherwise).

Answer 2

Take $k+1=j$ and take modulo $3$. So the question is if $j^3+j$ is a multiple of $3$, but $j^3+j=j(j^2+1)$, which is never a multiple of $3$ if $j$ isn't.

Answer 3

Set $n=k+1$ then your expression is equivalent modulo 3 to $$ n^3+7n+3 \equiv n+7n+3 \equiv 8n \equiv -n \pmod{3}. $$ In particular, this is a multiple of $3$ iff $3$ divides $k+1$.