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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?

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    $\begingroup$ 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? $\endgroup$ – Will Jagy Apr 4 '13 at 1:17
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    $\begingroup$ 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. $\endgroup$ – coffeemath Apr 4 '13 at 3:57
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    $\begingroup$ I thought this one can be done completely because it's on my assignment sheet... Although it has been listed as optional. $\endgroup$ – Haikal Yeo Apr 4 '13 at 7:17
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    $\begingroup$ @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. $\endgroup$ – Will Jagy Apr 4 '13 at 18:31
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    $\begingroup$ mathoverflow.net/questions/39210/… is the same question, asked in 2010, the prevailing opinion then seemed to be that it is open. $\endgroup$ – Alex J Best May 29 '13 at 12:01
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It is very unlikely that this problem can be resolved with current techniques. See BROCARD and section D25 in Unsolved Problems in Number Theory by Richard K. Guy.

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 $m=x! \; , \; n = x! -1.$

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    $\begingroup$ Nicely Done Will. + $\endgroup$ – mrs Apr 7 '13 at 12:25

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