How to show that $\dfrac{n^3 + 2n}{3}$ is an integer Show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer
My try:
Let P(n) be the statement that $n^3 + 2n$ is divisible by $3$.
Base step:
When $n = 0$ we have $0^3 + 0 = 0 = 3 \times 0$
So, the base case true.
Inductive hypothesis:
Assume that $P(k)$ is true.
which means $\dfrac{k^3 + 2k}{3}$ is divisible by $3$ and $\dfrac{k^3 + 2k}{3}=p$ for some integer $p$.
Now we need to show that $P(k+1)$ is true.
$(k+1)^3+2(k+1)$ and we will show that this divisible by $3$.
Proof:
$(k+1)^3+2(k1)=k^3+3k^2+3k+1+2k+2$
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$=3(p+k^2+k+1)$
As $p+k^2+k+1$ is an integer we have that $(k+1)^3+2(k+!)$ is divisible by $3$.
Is my above attempt correct?
Did I show that for each natural number $n, \dfrac{n^3 + 2n}{3}$ is an integer?
 A: Yes, you have proved the statement.
Note: Natural numbers are positive integers, so the base case is actually $1$. You made a few typing errors. Please define what $p$ is. (A simple statement like where $p=P(x)$ works)
A: Write $$n^3+2n$$ as $$n(n^2+2)$$
If $$n\equiv 0 \mod 3$$ then all is clear.
If $$n\equiv 1 \mod 3$$ then $$n^2+2\equiv 0\mod 3$$
If $$n\equiv 2\mod 3$$ then $$n^2+2\equiv 0\mod 3$$
and another idea
:
$$n^3-n+3n=n(n^2-1)+3n=(n-1)n(n+1)+3n$$
A: Hint
$$n^3+2n=n^3-n+3n=(n-1)n(n+1)+3n$$
Now use the fact that among any 3 consecutive integers you can find one divisible by $3$.
A: Or you could directly let $P'(n)$ be the proposition that "$\frac{n^3+2n}{3}$ is an integer".
Assume $P'(k)$ is true for some $k \in \mathbb N$, then consider $P'(k+1)$,
$$\begin{align*}
\frac{(k+1)^3+2(k+1)}{3} &= \frac{k^3+3k^2+3k+1 + 2k + 2}3\\
&= \frac{k^3+2k}3 + k^2+k+1\\
&\in \mathbb Z
\end{align*}$$
A: $n^3 +2n=n(n^2+2) $, now, check that all square numbers will leave a remainder $0$(means divisible by $3$) or, $1$ when we divide $n$ by $3$. Hence, when $3|n$ we have $3|n(n^2+2)$ and when $n=3k+1$ or $3k+2$, then $3|n^2+2\implies 3|n(n^2+2)$. 
