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

Question

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.

Answer 1

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

Answer 2

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 $x!y!=z!$