use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I prove that this isn't applicable for any number less than $1$? Also, where do I go after this? Thank you!
 A: In the induction step, we assume that it holds for $n = k$.
That means that $\exists a \in \mathbb{Z} : k^3 + 5k = 3a$.
Then, for $(k+1)$, we get
$$
\begin{split}
(k+1)^3 + 5(k+1)
 &= k^3 + 3k^2 + 8k + 6 \\
 &= (k^3 + 5k) + 3k^2 + 3k + 6 \\
 &= 3a + 3k^2 + 3k + 6 \\
 &= 3(a + k^2 + k + 2).
\end{split}
$$
So, if $k^3 + 5k$ is a multiple of $3$, then $(k+1)^3 + 5(k+1)$ must also be a multiple of $3$.
A: Just note that
$$
f(n+1)-f(n)=(n+1)^3 + 5(n+1)-(n^3 + 5n)=3 (n^2 + n + 2)
$$
I suspect this is the intended solution.
Actually, no induction is needed:
$$
n^3 + 5n = 6 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}
$$
and so is always a multiple of $6$.
A: You could be better to take the difference of successive terms as follows:
$$(n+1)^3+5(n+1)-n^3-5n= 3n^2+3n+1+5=3(n^2+n+2)$$ which is clearly a multiple of three. It follows that if $n^3+5n$ is divisible by $3$, so is $(n+1)^3+5(n+1)$, which is what you need for your inductive step.
Notice how taking the difference here immediately eliminates the highest order term in the each of the polynomial expressions - so $n^3$ disappears as does the $5n$ term. This is typical for such polynomial cases, and can leave you working with simpler expressions.
A: HINT
$$
\begin{split}
k^3 + 3k^2+8k+6
 &= 3(k^2+3k+2) + k^3-k \\
 &= 3(k^2+3k+2) + k(k^2-1) \\
 &= 3(k^2+3k+2) + k(k-1)(k+1)
\end{split}
$$
A: You correctly expanded $(k+1)^3+5(k+1)$ into $k^3+3k^2+8k+6$. What you now need to do is to show that $k^3+3k^2+8k+6$ is divisible by $3$, using the inductive hypothesis that $k^3+5k$ is divisible by $3$. To do this, simply write
$$k^3+3k^2+8k+6=(k^3+5k)+3k^2+3k+6$$
and note that each of the four terms on the right hand side is divisible by $3$.
As for your question,  how do I prove that this isn't applicable for any number less than $1$?, you don't need to prove anything about numbers less than $1$ (and in this case the assertion does still hold: $n^3+5n$ is divisible by $3$ for all integers $n$, positive or negative). Induction only needs to go in one direction.
