Find all such positive integers $x(>1), y(>1)$ and $z$ so that $x! \cdot y! =z!$ holds.

Question

Upon inspection, I found out that: $3!\cdot5! =6! $ and $6!\cdot7! =10!.$ (couldn't dare to search for more such triplets)

How to check whether there exist more such triplets or not? Please show me a proper way to handle such problems.

This question is relevant, especially answer by Amir Hossein. — Adam Bailey, Jul 12 '20 at 21:43

Henry · Answer 1 · 2020-07-11T22:47:57.737

2

Your $3!\cdot 5! =6!$ is an example of the case for general $n$:

$$n! \cdot (n!-1)! = (n!)!$$

and when $n=4$ this gives $4! \cdot 23! = 24!$

edited Jul 11 '20 at 22:47

answered Jul 11 '20 at 22:43

Henry

1

This suggests that there are infinite number of such triplets. The form $n!\cdot(n! -1)! =(n!)!$ is generated on the basis of $3! ×5! =6!.$ But this form doesn't imply the existence of $6!×7! =10!$. So can't we find a single general form so that $x! \cdot y! =z! $ holds? Thank you for your answer. – Dhrubajyoti Bhattacharjee Jul 11 '20 at 23:49

score 1 · Answer 2 · answered Jul 11 '20 at 22:46

1

For example, you can observe that

$$(n!-1)!.(n!)=(n!)!$$

If $ n=3$, it gives

$$5!.3!=6!$$ for $ n=4 $, we get

$$23!.4!=24!$$ and so on.

answered Jul 11 '20 at 22:46

hamam_Abdallah

2 Answers2