TITLE: integer as sum of three binomials
QUESTION [9 upvotes]: Prove that for any nonnegative integer $n$ $\exists x,y,z \in \mathbb{N}$ and $0\leq x<y<z$ so
$$n=\binom{x}{1}+\binom{y}{2}+\binom{z}{3}$$
Please give me a hint, I don't have any idea.

REPLY [3 votes]: Consider $z_m={m\choose 3}$ for all $m$. Difference $z_{m+1}-z_{m}=\frac{m(m-1)}{2}$. We choose such $m$, that $z_m$ is maximal, non exceeding $n$.
The most difference $n_y=n-z_m$ we would have to cover, is $\frac{m(m-1)}{2}-1$.
Now we choose $y_k={k\choose 2}$ to be maximal, non exceeding $n_y$. Note that maximal $k$ could be $m-1$, since $\frac{m(m-1)}{2}-1$ is the maximal value for $n_y$ (i.e. it can't be $\ge\frac{m(m-1)}{2}$). Analogically, $y_{k+1}-y_k-1=k-1$ is the maximal value to cover with $x$.
So, we select $y=k$ and $z=m$.