I accidentally noticed that:

$$(7)_4=7 \cdot 8 \cdot 9 \cdot 10=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7=7!$$

Here $(n)_k$ is the Pochhammer symbol.

I wonder, are there any other non-trivial integer solutions $(n,k)$?


Among the ones I consider trivial we have $(0,0),(1,0),(1,1),(2,1)$. Somehow, I'm sure that I will get a lot of comments with these four solutions.

This is the implicit plot of the equivalent equation:

$$\Gamma (n+k)=n \Gamma^2 (n)$$

enter image description here

  • $\begingroup$ Nice question... $\endgroup$ – barak manos Sep 1 '16 at 8:41
  • $\begingroup$ I wish you the best of luck on this one. $\endgroup$ – Axoren Sep 1 '16 at 8:44
  • $\begingroup$ @Axoren, is the problem unsolved? If it is, I'm sure there are reasons... $\endgroup$ – Yuriy S Sep 1 '16 at 8:46
  • $\begingroup$ I think that the answer (no) lies in the fact that there is no number larger than $10$ which can be "split" into two parts containing the same prime factors, i.e., there will always be a prime factor which is present in one of the parts but not in the other. In your example, $10$ is "split" into $[1-7]$ and $[7-10]$, both parts containing the same prime factors (and even with the same multiplicity for each factor). $\endgroup$ – barak manos Sep 1 '16 at 8:47
  • $\begingroup$ You can easily prove that, if $n>1$, then $n$ must be prime. $\endgroup$ – Batominovski Sep 1 '16 at 8:56

If $n+k\geq 14$, then there are at two primes in the interval $\left(\frac{n+k-1}{2},n+k-1\right]$. Therefore, $(n+k-1)!$ cannot equal $n\cdot \big((n-1)!\big)^2$. This leaves the cases $n+k\leq 13$ to be dealt with.

  • $\begingroup$ I don't see the justification for the "therefore". If $k < n - 1$ then $n-1 \in \left(\frac{n+k-1}{2},n+k-1\right]$, so what's the obstacle? $\endgroup$ – Peter Taylor Sep 1 '16 at 10:38
  • $\begingroup$ Well, the primes in this interval must be factors of $n$, but then their product is greater than $n+k-1$. $\endgroup$ – Batominovski Sep 1 '16 at 10:44
  • $\begingroup$ Why must they be factors of $n$? Why can't they be $n-1$ and $n-3$? $\endgroup$ – Peter Taylor Sep 1 '16 at 10:47
  • 1
    $\begingroup$ Because each of them occurs with multiplicity $1$ in $(n+k-1)!$. $\endgroup$ – Batominovski Sep 1 '16 at 10:49

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