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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.

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

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    $\begingroup$ 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. $\endgroup$ – Carl Mummert Mar 9 at 14:15
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    $\begingroup$ Hint of an easy way to generate infinitely many solutions: set $x=1$. $\endgroup$ – Minus One-Twelfth Mar 9 at 14:19
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    $\begingroup$ 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. $\endgroup$ – JMoravitz Mar 9 at 14:26
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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

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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 $x!y!=z!$

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