Htykuut

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 $n=2k+1$, $kin mathbb N$ 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?

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

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

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

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

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 $ninBbb N$ not $frac{n}{3}+frac{n^2}{3}+frac{n^3}{6}$

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

Insane overkill incoming: if a polynomial $p(x)$ with degree $d$ takes integer values at $xin{0,1,ldots,d}$, it takes integer values at any $xinmathbb{Z}$, 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 $ninmathbb{N}$, and these values are clearly non-negative.