# Show that $x! y! = z!$ has infinitely many solutions. (Hint: For example, $5! 119! = 120!$.) [closed]

Show that

$$x! ·y! = z!$$

has infinitely many solutions. (Hint: For example, $$5! 119! = 120!$$)

I am stuck on this problem. Within this section we are learning Congruence. So I know it involves something with mods.

## closed as off-topic by Carl Mummert, B. Goddard, John Omielan, Lee David Chung Lin, Delta-uMar 9 at 17:17

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• Welcome to Math.SE. When you asked the question, a link was given to How to ask a good question. In its present form, this question is unsuitable for the site, because it lacks context. You can edit the post to improve it - please add the source and motivation of the problem as well as your current thoughts about solving it. Also, please ensure the question is not only stated in the title. – Carl Mummert Mar 9 at 14:15
• Hint of an easy way to generate infinitely many solutions: set $x=1$. – Minus One-Twelfth Mar 9 at 14:19
• The hint should spell it out quite clearly what's going on. As an additional example, 4!23!=24! Consider the relationship between 4!, 23 and 24. "I know it has something to do with congruence and mods" I wouldn't say so... I think of it more directly. This has more to do with equality than with congruence. Note, the final proof will likely require induction. – JMoravitz Mar 9 at 14:26

Let $$a=p!$$ for some $$p\in\mathbb N$$.

Observe now that

$$a·(a-1)!=a!\iff p!·(a-1)!=a!$$

which has infinitely many solutions

The hint leads you to recognize that $$2!1!=2!, 3!5!=6!, 4!23!=24!, 5!119!=120!, 6!719!=720!$$ etc...

In general $$n!(n!-1)!=(n!)!$$

This is readily apparent that it is true. Arguably, an induction proof isn't even necessary as it can be shown directly. Remember that $$k\times (k-1)! = k!$$ for all natural $$k$$. Now, apply this for the case when $$k$$ happens to be $$n!$$.

Finally, recognize the connection to your originally phrased problem and conclude that since the above identity works for all natural $$n$$, there are infinitely many (non-trivial) solutions to $$x!y!=z!$$