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

Question

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.

@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
@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, Aug 14 '17 at 19:32
@Famke that is the behavior we should suspect (I'm actually not so sure of this?), so maybe no solution for sufficiently large $n$? — Andres Mejia, Aug 14 '17 at 20:24
Possible duplicate of Is there any perfect squares that are also binomial coefficients? — José Carlos Santos, Aug 15 '17 at 00:11
@JoséCarlosSantos have you read my edits? That question is linked, I have further questions. — Andres Mejia, Aug 15 '17 at 00:52
If ${n\choose 2}=p^j$ then $n(n-1)=2p^j$. Since the numbers $n$ and $n-1$ are relatively prime, one of then is divisible by $p^j$. Then the other is $1$ or $2$, thus $n\le 3$. So $n=3=p^j$. — Alex Ravsky, Aug 15 '17 at 05:35

score 3 · Accepted Answer · answered Aug 15 '17 at 16:00

3

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.

answered Aug 15 '17 at 16:00

Mike Bennett

3,494

Could you give some hint as to what Gyory's results were? In your first sentence of the second paragraph, you mean to say that this is a negative result (can't be written as a prime power)? – Andres Mejia Aug 15 '17 at 16:57
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 01:24
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
Oh, I see. I'm sorry, I'm not sure why your main point was so opaque to me. Thank you – Andres Mejia Aug 16 '17 at 17:49

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

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