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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 ${n + k - 1 \choose k} = p^2$.

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You are right in the sense that usually there are too many factors. This guided me to my answer. – Jyrki Lahtonen Jun 12 '13 at 21:19
up vote 9 down vote accepted

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.

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many thanks for this response. A neat proof! However I wonder if you could elaborate on why it is necessary that $2p \leq m$, and not the more trivial bound that $p^2 \leq m$? – Moderat Jun 13 '13 at 14:19
Never mind, I now see that you would need $2p \cdot (2p - 1) \cdots (p)(p-1)\cdots$. Thanks again – Moderat Jun 13 '13 at 14:38

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 $e\ge 3$.

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I accepted Jyrki's answer, but I want to thank you for the referral to the paper, this is useful for the more general aspects of my project – Moderat Jun 13 '13 at 14:21

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