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How to find the exact number of possible ways to represent a factorial as a product of three consecutive natural numbers?

$$m \cdot (m+1) \cdot (m+2) = n!$$

where $m, n \in \mathbb{N}$

I found these cases:

$1 \cdot 2 \cdot 3 = 3!\\2 \cdot 3 \cdot 4 = 4!\\4 \cdot 5 \cdot 6 = 5!\\8 \cdot 9 \cdot 10 = 6!$

But I need to find the exact number of all possible cases using some formula (or maybe prove that there are only four cases).

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    You might consider primes less than $m$ and whether they can all be factors of $m$ or $m+1$ or $m+2$ – Henry Apr 19 '22 at 08:24
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    Maybe, someone can prove that there is no further solution. What you should do first is to determine the (unique) positive solution of $x(x+1)(x+2)=n!$ for $n=1$ to , say , $n=100$ and check whether it is an integer. This will give already a huge lower bound for $m$ for a solution you still did not find. Another approach is to consider the largest prime factor of $m(m+1)(m+2)$. ALL smaller prime numbers must then divide $m(m+1)(m+2)$ which will very rarely be the case. – Peter Apr 19 '22 at 08:55
  • PARI/GP allows to check whether a polynomial is irreducible over $\mathbb Q[x]$. This reveals that there is no further solution for $n\le 8\ 000$ which is convincing enough for me , although it of course does not prove yet that there is no further solution. – Peter Apr 19 '22 at 09:30
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    @Peter, do you think it is possible to draw an analogy with Brocard's problem? –  Apr 19 '22 at 09:47
  • @EuropeX This problem at least shows how difficult such equations are ususally to solve. If the problems are equivalent, we would be screwed, but I do not think this is the case. – Peter Apr 19 '22 at 09:51
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    According to this it seems to be an open problem (I don't have the original source though). This question was already asked on MSE here and there it is connected with the abc conjecture. – Fabius Wiesner Apr 19 '22 at 12:24
  • @BillyJoe, my question, although similar to the one you indicated, is different, because I'm not asking you to solve the equation, but at least calculate the number of cases that satisfy the equation. –  Apr 19 '22 at 14:32

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