"Binomiable" numbers Is there a nice criterion to determine whether a given natural $m$ can be written as a binomial number $\binom{n}{k}$ with $1 &lt; k &lt; n-1$?

I've been thinking on this problem with a friend and all we concluded is that primes aren't "binomiables".


  CLAIM 1. Primes aren't "binomiables".
  
  Proof. Let $p$ a prime and suppose $p = \binom{n}{k}$.
  
  
  If $n$ is prime, then $n$ must be equals $p$, because $n\mid \binom{n}{k}$. But, in this case, $n\mid\mid \binom{n}{k}$, so $k=1$. (This would work to prime powers too.)
  If $n$ is composed, note that $\binom{n}{k} = p$ implies, a fortiori, that $p\mid \binom{n}{k}$ and consequently $p\leqslant n$. But under our conditions, $\binom{n}{k} &gt; n$, contradiction. $\square$
  


Also, I've already saw in Proofs from THE BOOK the problem "Binomial coefficients are (almost) never powers", and although it is highly related, it doesn't provide much insight towards this problem (at least as far as I understood).



EDIT: Prime powers aren't binomiables too!

Let's state three results first:


Lemma. Let $p$ a prime. If $p\mid n$ and $0&lt;k&lt;p$, then $p\mid \binom{n}{k}$.

Proof. Let $n= pm$ and assume $0&lt;k&lt;p$. Consider the Vandermonde's identity:

$$\binom{pm}{k} = \sum_{j=0}^k \binom{p}{j}\binom{p(m-1)}{k-j}.$$

It follows that $\binom{pm}{k}\equiv \binom{p(m-1)}{k} \mod{p}$ (the
term $j=0$ in the sum). Repeating it $m$ times, we conclude
$\binom{pm}{k}\equiv \binom{p}{k} \mod{p}$, and from $p\mid
    \binom{p}{k}$ follows our statement. $\square$
Bertrand's postulate. If $n\geqslant 2k$, then the binomial $\binom{n}{k}$ has a prime factor $q &gt; k$.
Theorem. [Erdös] The binomial $\binom{n}{k}$ is never a perfect power for $4\leqslant k\leqslant n-4$.

(This is the paragraph I mentioned in Proofs from THE BOOK.)


So there is only two cases to analyse, $k=2$ and $k=3$. In both cases we can assume $n \geqslant 6$.


  CLAIM 2. Prime powers aren't "binomiables".
  
  Proof. Let $p$ a prime, $m&gt;1$, and suppose $p^m = \binom{n}{k}$.
  
  
  If $k=2$, then $p^m = \frac{n(n-1)}{2}$. By Bertrand's postulate, $p\neq 2$. Now observe that $\gcd(n, n-1) = 1$, so
  $$ p^m \mid n\quad \text{xor}\quad p^m\mid n-1. $$
  Then:
  $$2 =\frac{n(n-1)}{p^m} \geqslant n-1 \implies n \leqslant 3,$$
  that contradicts $n\geqslant 6$.
  If $k=3$, then $p^m = \frac{n(n-1)(n-2)}{6}$. Bertrand's postulate allows us to take $p\neq 2,3$. Hence, similarly to the previous case, we can deduce:
  $$p^m \mid n \quad \text{xor}\quad p^m\mid n-1 \quad \text{xor} \quad p^m\mid n-2.$$
  But then:
  $$ 6 = \frac{n(n-1)(n-2)}{p^m} \geqslant (n-1)(n-2) \implies n \leqslant 4,$$
  a contradiction. $\square$
  




EDIT 2: I've just realized that I didn't use the Lemma at all. Anyway, I think I'll let it here, it may be useful someday hahaha
Here is a partial answer for a special case: if $\sqrt{8n+1}$ is an odd integer, then $n$ is a binomial coefficient of the form $n \choose 2$.  Such a formula is possible because $n \choose 2$ is quadratic in $n$ and therefore has a straightforward solution.
I know they are composite.

To show this, suppose not, that is, $\binom mn=p$, a prime. Then $m\ge p$. Since we are excluding the trivial cases, $m\ge p+1$ and then

$$\binom mn\ge\binom{p+1}2=p\cdot\frac{p+1}2\ge\frac32p&gt;p$$

a contradiction.

But not every composite number is a non-trivial binomial coefficient. For example, $9$ is not.