<s>[INST] 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?
 [/INST] $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 $3|n^2+2\implies 3|n(n^2+2)$. 
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