Equality with factorial: m(m+1)(m+2)=n!

EUCLA

Find all couples $(m,n)$ of integers such that $m(m + 1)(m + 2) = n!$

**Posted:** Tue Aug 11, 2009 4:12 am

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La guida al guado del fiume, Teeteto, diceva: sarà l'acqua stessa a mostrare quanto è profonda.
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Bugi

Small cases: $m=1,n=3;m=2,n=4;m=4,n=5$ .

It's kind of easy to prove that $m(m+1)(m+2)<(m+1)!$ for $m>4$ .

So, $m(m+1)(m+2)\ge m!$ . That means $(m+1)(m+2)\ge (m-1)!$ . Again easily we can prove that this is possible only for $m\le 5$ , so it must be $m=5$ . But, that case doesn't give solutions, so all solutions are the small cases above.

**Posted:** Tue Aug 11, 2009 12:11 pm

Rofler · Yang-Mills Theory Offline Joined: 10 Mar 2007 Posts: 572

How does $m(m + 1)(m + 2) < (m + 1)!$ imply $m(m + 1)(m + 2)\ge m!$ ?

**Posted:** Tue Aug 11, 2009 4:54 pm

fermat285 · New Member Offline Joined: 03 Apr 2009 Posts: 4

Bugi//How can you explain m=8, n=6?

**Posted:** Sun Aug 30, 2009 11:28 pm

jestrada · New Member Offline Joined: 26 Sep 2009 Posts: 3

what about trying to prove that if m>8, m must be infinitely large?
for example if m>8, then n>6, so n is at least 7 and 5040 divides m(m+1)(m+2) ==> m≥54, because we would have that 16 divides m(m+1)(m+2), 35 divides m(m+1)(m+2) and 9divides m(m+1)(m+2) and the smallest m that fulfills this is 54.
==> 27 divides n!, which implies n≥9, and so on...

**Posted:** Fri Oct 16, 2009 2:26 pm