How find all postive integer number such $(n+k)\nmid \binom{2n}{n}$

Question:

Find the all integer $k$,such there are exist infinitely many $n$ such $$(n+k)\nmid \binom{2n}{n}$$

This is china 2014 (CMO problem 4),it's have been end exam three hours ago.

I think we can use Chinese Remainder(or Luca's) Theorem? It's nice problem.

It is said by some students,$k\neq 1,k\in Z$ is answer.

Solution 1:

If $k\le 0$ then for any odd prime $p>-2k$ let $n=p-k$ then $2n<3p$ so $p^2\|(2n)!$ and $(n+k) \not\mid \binom{2n}{n}$. Infinitely many choices of $p$ lead to infinitely many such $n$.

If $k>1$ then by Bertrand's postulate there is a prime $p>2$ with $k<p<2k$. Then for any $t>1$ let $n=p^t+(p-k)$, then $2n=2p^t+2(p-k)$, and since $0<2(p-k)<p$ it follows from Lucas's theorem that $\binom{2n}n \equiv 2\binom{2p-2k}{p-k}\not\equiv 0\pmod{p}$ and hence that $(n+k)\not\mid\binom{2n}n$. Infinitely many choices of $t$ lead to infinitely many such $n$.

In the remaining case $k=1$ we always have $(n+1)\mid\binom{2n}n$. Since $\gcd(2n+1,n+1)=1$ $$ \frac{2n+1}{n+1}\binom{2n}{n} = \binom{2n+1}{n} \in \mathbb{Z} \implies (n+1)\left\vert \binom{2n}n\right. $$