TITLE: Binomial coefficients equal to a prime squared
QUESTION [5 upvotes]: 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 ${n + k - 1 \choose k} = p^2$.

REPLY [3 votes]: 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 $e\ge 3$.