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Does n!m!=t! have infinitely many solutions in positive interger besides trivial ones? (n=0 m=1 etc)

Can't work this one out. thanks.

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    $\begingroup$ $(n!)!=(n!-1)!\cdot n!$ - is it trivial or not? $\endgroup$ Sep 20, 2010 at 11:37
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    $\begingroup$ I'm not sure if this is considered trivial or not but $n! = (n!)!/(n!-1)!$. Let $n < m$, since $n! = t!/m!$ the range $(m,t]$ must avoid primes because if there was a prime in there $n!$ wouldn't contain it. $\endgroup$
    – muad
    Sep 20, 2010 at 11:38
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    $\begingroup$ Also - there is related question - mathoverflow.net/questions/39210/… Because, if there are any $(m,n)$, such that $n!=m(m+1)$, then we have:$(m+1)!=(m-1)!\cdot n!$ $\endgroup$ Sep 20, 2010 at 12:11
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    $\begingroup$ The OEIS sequence oeis.org/classic/A003135 is related to this question. $\endgroup$
    – tdnoe
    Sep 20, 2010 at 18:21
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    $\begingroup$ Kevin, are you thinking of Guy's book, Unsolved Problems In Number Theory? The reference is problem B23. $\endgroup$ Sep 21, 2010 at 4:38

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$(n!)!=(n!-1)!\cdot n!$ - is it trivial or not?

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  • $\begingroup$ In the 1975 Erdos paper mentioned in OEIS sequence A003135, this is considered a trivial solution. See pages 27-28 of that paper. $\endgroup$
    – tdnoe
    Sep 20, 2010 at 19:54

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