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Are there infinitely many nontrivial integer solutions $(a,b,c)$ of

$$\Gamma(a)\Gamma(b)=\Gamma(c)\hspace{10mm}?$$

This seems like it may have been asked before but I didn't find an earlier question (yet). I checked for about $c < 200$ and $b-a < 10$ and found only $(a,b,c)=(4,6,7),(7,8,11).$

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daniel
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    If $a, b, c$ must be integers then are you just asking about factorials? – badjohn Apr 20 '17 at 08:20
  • @badjohn it came up in context of gamma but yes. – daniel Apr 20 '17 at 08:22
  • So, we can assume that $a < b < c$. Consider the prime factorisation of both sides. If there is a prime $p$ between $b$ and $c$ then it will be factor of the RHS but not the LHS. This will place severe limits on possible solutions. – badjohn Apr 20 '17 at 08:26
  • @badjohn: So must be $\pi(b) = \pi(c)$? – daniel Apr 20 '17 at 08:30
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    If you are only asking about factorials, this may be what you're looking for : https://math.stackexchange.com/questions/112670/on-the-factorial-equations-a-b-c-and-abc-d – Arnaud D. Apr 20 '17 at 08:35
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    Or this: https://math.stackexchange.com/questions/150192/solutions-of-pq-r?noredirect=1&lq=1 – wythagoras Apr 20 '17 at 08:38
  • I would expect so. There is the well known example $10!=7!×6!$ but this satisfies that restriction. – badjohn Apr 20 '17 at 08:43
  • @daniel I don't really care about the reputation, but I think it is good to leave this one around. Other people might also be looking for the gamma variant of this factorial equation. – wythagoras Apr 20 '17 at 08:46

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Note that $(d+1, d! , d!+1)$ is a solution for all $d$:

$$\Gamma(d+1)\Gamma(d!) = d!(d!-1)! = (d!)! = \Gamma(d!+1)$$

However, this might be considered as a trivial solution. But since you included $(4,6,7)$, which is this solution with $d=3$, I thought it was worth pointing out.

I think the really trivial solutions are $(1,d,d)$ and $(2,d,d)$.

A bit more advanced stuff: Under the abc conjecture, there are only finitely many other solutions than those I mentioned above. Probably, the only one is $(7,8,11)$, which corresponds with $6!7!=10!$. See also: Solutions of $p!q! = r!$

wythagoras
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