Prove that $\frac{n}{3}+\frac{n^2}{3}+\frac{n^3}{6}$ is natural number

Question

Prove that for every $n \in \mathbb N$, $\frac{n}{3}+\frac{n^2}{3}+\frac{n^3}{6}$ is natural number.

I try in this form $\frac{2n+2n^2+n^3}{6}=\frac{n(n^2+2n+2)}{6}$ so I must show that $2| n(n^2+2n+2)$ and $3|n(n^2+2n+2)$ then I try that $n=2k$ or $n=2k+1$, $k\in \mathbb N$ to prove that if $n=2k$ it is trivial if $n=2k+1$ then $(2k+1)((2k+2)^2+1)$ then it is odd number multiply odd number so I did not prove that 2 divide that number, the same is for 3 do you have some idea?

Answer 1

By below $\,6\mid nf(n)\!=\! n(n^2\!+3n+2)\ $ by $\ 6\mid f(1)\!=6,\,$ and $\,3\mid f(-1)\!=\!0$

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Theorem $\ \forall n\!:\ 6\mid nf(n)\! \iff\! 6\mid f(1),\ 3\mid f(-1)\ $ for a polynomial $f(x)$ with integer coef's

Proof $ $ It is true $\iff nf(n)\,$ has roots $\,n\equiv 0,1\pmod{\! 2}$ and $\,n\equiv 0,\pm1 \pmod{\!3}$

But $\,0\,$ is always a root, and $\,1\,$ is a root $\!\iff\!\!\! \underbrace{6\mid f(1)}_{\large 2,3\ \mid\ f(1)\ \ \ \ }\!$ and $\,-1$ is a root $\iff\!3\mid f(-1)$

Answer 2

$m=n/3+n^2/3+n^3/6=n(n^2+2n+2)/6=n((n+1)^2+1)/6$. If $n$ is odd then so is $(n+1)^2+1.$ So, for $n$ odd $m$ is not an integer.

Answer 3

It is a typo because $$\frac{n}{3}+\frac{n^2}{2}+\frac{n^3}{6}={n(n+1)(n+2)\over 6}$$ is natural number for all $n\in\Bbb N$ not $\frac{n}{3}+\frac{n^2}{3}+\frac{n^3}{6}$

Answer 4

Insane overkill incoming: if a polynomial $p(x)$ with degree $d$ takes integer values at $x\in\{0,1,\ldots,d\}$, it takes integer values at any $x\in\mathbb{Z}$, as a consequence of the properties of the forward difference operator $\delta:p(x)\mapsto p(x+1)-p(x)$. Since $$ \frac{n}{3}+\frac{n^2}{\color{red}{2}}+\frac{n^3}{6} $$ takes integer values at $\{0,1,2,3\}$, it takes integer values at any $n\in\mathbb{N}$, and these values are clearly non-negative.