Is $(n+\ell)^{-1}\binom{kn}{n}$ an integer for only $(\ell,k)=(1,2)$? 
Find all pairs $(\ell,k)$ of natural numbers, such that the number $\dfrac1{n+\ell}\dbinom{kn}{n}$ is an integer for all natural $n$.

Is $(\ell,k)=(1,2)$ the only solution?
 A: Suppose there is a solution $(k,\ell)$ with $\ell>1$. Then setting $n=1$ shows that $k$ is a multiple of $\ell+1$. Suppose $p$ is a prime dividing $\ell$; then $k\equiv 1\mod{p}$ as well. Let $n = p$:
$$
  \dbinom{kn}{n} 
    = \frac{kp(kp-1)\cdots(kp-(p-1))}{p!}
     = \frac{k(kp-1)\cdots(kp-(p-1))}{(p-1)!},
$$
which is not divisible by $p$ since $k$ is not. But the denominator is
$$n+\ell = p+\ell \equiv 0\mod{p}.$$
So the only solutions have $\ell=1$ (and $k$ even).
Now suppose that $\dbinom{kn}{n}$ is a multiple of $n+1$ for all natural numbers $n$. If $k>2$, choose $p\,\mid\,k-1$ with $p$ odd, so that $k\equiv 1\mod{p}$. Then set $n=p-1$. By a computation similar to the one above
\begin{align*}
  \dbinom{k(p-1)}{p-1}
    &= \frac{k(p-1)(k(p-1)-1)(\cdots)(k(p-1)-(p-2))}{(p-1)!} \\
    &= \frac{k(p-1)(kp-k-1)\cdots(kp-k-(p-2))}{(p-1)!}.
\end{align*}
But none of 
$$k,\ kp-k-1,\ kp-k-2, \dotsc,\ kp-k-(p-2) \equiv 1,\ -2,\ -3, \dotsc,\ -(p-1) \mod{p}$$ 
is divisible by $p$ since $k\equiv 1\mod{p}$, so the expression is not divisible by $p$. But $n+1 = p-1+1=p$ and the quotient is not an integer.
So $\ell=1$, $k=2$ is the only solution.
