# Find the positive integer solutions of $m!=n(n+1)$

Find the positive integer solutions of $m!=n(n+1)$

I basically have $(m,n)=(2,1)$ or $(3,2)$ and I think these are the only solutions.

I don't have a complete proof but here's what I know so far. By Bertrand's Postulate, I can find prime $p$ in the "second half" of $m!$. If $m>4$, then $p$ is odd.

$(n,n+1)=1$

Suppose $n$ be even. Then $n+1$ be odd. Also, $n=2^kq$, where $k$ is the maximum number of times $2$ can divide $m!$.

What else can be done?

• The set of solutions to $m! = n(n+2)$ is not known, although it is likely that $m=7$ is the largest. Why do you think this one can be done completely? – Will Jagy Apr 4 '13 at 1:17
• I checked on maple and no solutions other than $m=2,3$ for $m \le 2000.$ I used that $a=n(n+1)$ implies $4a+1$ is a square, and maple has a fast test "issqr(n)" to see if $n$ is a square. – coffeemath Apr 4 '13 at 3:57
• I thought this one can be done completely because it's on my assignment sheet... Although it has been listed as optional. – Haikal Yeo Apr 4 '13 at 7:17
• @BabyDragon, the version $1 + m! = k^2$ is open, write it as $m! = (k-1)(k+1),$ then $m! = n (n+2).$ Maybe this version can be done. – Will Jagy Apr 4 '13 at 18:31
• mathoverflow.net/questions/39210/… is the same question, asked in 2010, the prevailing opinion then seemed to be that it is open. – Alex J Best May 29 '13 at 12:01

Berend and Osgood showed in 1992 that the set of solutions to $P(x) = n!$ has an integer solution has density zero if $P(x)$ is a fixed polynomial of degree at least two with integer coefficients. Among conditional results, F. Luca showed that the ABC conjecture implies that the number of solutions to $P(x) = n!$ is finite. Luca can be downloaded from the wikipedia page.
In case there has been a misunderstanding in stating the problem, there is an infinite family of solutions to $$m! = n! \; (n+1)!,$$ in that $$(x!)! = (x! -1)! \; x!,$$ so $m=x! \; , \; n = x! -1.$