# are there known cases where $\binom{n}{k}$ is a perfect prime power?

I was wondering about cases where $\binom{n}{k}=p^j$ with $p$ a prime (nontrivially, so that $n-k>1$ and $n \neq p^j$.) I had the terrible idea of checking binomial expansions $$(x+y)^n \equiv x^n+y^n.$$

Upon some further googling, I learned that $\binom{50}{3}=140^2$, which means that this can occur for composite numbers.

Are there any ways that one can use primeness to argue that this will not occur? Or, are there any known counterexamples in which $\binom{n}{k}$ is a prime power?

Edit:

Thanks to the comments, this question answers the question for cases of $j \geq 2$ and $4 \leq k \leq n-4$.

The main answer mentions that there are examples where $k=j=2$, are any among these solutions with $p$ a prime?

this question gives a negative answer for $p=2$ with $n-k \geq 2$.

I suppose there is only one question left then: can the theorem be strengthened to contain the extreme cases with the assumption that $p>2$ is a prime.

• See this question. For $p=2$ see this question. – Dietrich Burde Aug 14 '17 at 18:44
• @DietrichBurde thank you very much! Both links were really interesting. I thankfully have a copy of proofs from the book, so I was able to locate the proof (although I haven't gone through it carefully.) – Andres Mejia Aug 14 '17 at 19:10
• You are welcome! – Dietrich Burde Aug 14 '17 at 19:11
• @DietrichBurde I hope I'm not missing anything, but there is still something to say here. Namely, for cases of $p>2$, the additional assumption of primeness could in principle strengthen the result. Are there results known in this way? – Andres Mejia Aug 14 '17 at 19:16
• Let $K:=\max \{ k, n-k \}$. Then the interval $[K,n]$, should not contaion any prime number. – Davood Khajehpour Aug 14 '17 at 19:32

Erdos proved (in 1951) the stronger result that $\binom{n}{k}$ is never a perfect power if $k > 3$ (assuming that $n \geq 2k$ as we may). His argument is completely elementary -- see https://www.renyi.hu/~p_erdos/1951-05.pdf.
For $k=2$ and $k=3$, the case of possible prime powers is easily treated by divisibility arguments. The more general situation of arbitrary perfect powers for these values of $k$ was handled by Gyory in 1997 (in Acta Arithmetica), by a rather more complicated proof.
• The equation $\binom{n}{k}=y^j$ has no solutions (assuming that $n \geq 2k$) if either $k \geq 4$, or if $k \in \{ 2, 3 \}$ and $j \geq 3$. If $k=3$ and $j=2$, the only solution is $n=50$, $y=140$. If $k=j=2$, there are infinitely many solutions. – Mike Bennett Aug 16 '17 at 1:24
• These rexults were already known (see the linked question in the OP.) The question is if you require that $y$ is prime, the result can be strengthened. – Andres Mejia Aug 16 '17 at 16:34
• Well, as I said, then remaining case where $k=j=2$ is essentially trivial to treat via divisibility arguments for prime powers, so what question are you asking? – Mike Bennett Aug 16 '17 at 17:47