Binomial coefficients equal to a prime squared I am looking for some reading on when binomial coefficients are equal to $p^2$ for $p$ a prime. In general I imagine this is rare, as there are simply too many factors. Concretely, I am looking for pairs $(n, k)$ such that ${n + k - 1 \choose k} = p^2$. 
 A: For the equality
$$
{m\choose k}=p^2
$$
to hold for a prime $p$, we must obviously have $2p\le m$. Otherwise $m!$ won't be divisible by $p^2$ so neither will the binomial coefficient. 
I claim that this implies $k=1$.
Without loss of generality we can assume that $k\le m/2$. If $2<k\le m/2$, then
$$
{m\choose k}\ge {m\choose 3}=\frac{m(m-1)(m-2)}6,
$$
which is larger than $m^2/4\ge p^2$ whenever $m\ge6$. This means that we must have $k=1$ or $k=2$ or $m<6$. The last possibility won't concern us - a brute force check tells in few seconds that there are no counterexamples.
So we need to take a look the case $k=2$. But
$$
{m\choose2}=\frac{m(m-1)}2.
$$
Here $m$ and $m-1$ have no common factors, so from $m(m-1)=2p^2$ we get that either $m-1=2, m=p^2$ or $m-1=p, m=2p$ and both are impossible. The claim follows from this.
Obviously you can arrange ${m\choose 1}=m$ to be anything you want.
A: No nontrivial pairs exist.  According to the abstract of this paper, $\binom{n}{s}$ has at least as many distinct prime factors as $n$.  If we desire $\binom{n}{s}=p^2$, it follows that $n$ must be a prime power.  If $n=q^e$ for some prime $q$, then $q|\binom{n}{s}$ so that $q=p$, and $n=p^e$ with $e\ge 2$.
We have the solution $e=2$ with $s=1$ or $s=p^2-1$.  If $e\ge 3$, it follows that:
$$\binom{p^e}{s}\ge p^e>p^2$$
if we exclude $s=0,p^e$.  It follows that there are no solutions for $e\ge 3$.
