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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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$(n!)!=(n!-1)!\cdot n!$ - is it trivial or not? –  Nurdin Takenov Sep 20 '10 at 11:37
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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. –  muad Sep 20 '10 at 11:38
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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:$(m+1)!=(m-1)!\cdot n!$ –  Nurdin Takenov Sep 20 '10 at 12:11
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The OEIS sequence oeis.org/classic/A003135 is related to this question. –  tdnoe Sep 20 '10 at 18:21
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Kevin, are you thinking of Guy's book, Unsolved Problems In Number Theory? The reference is problem B23. –  Gerry Myerson Sep 21 '10 at 4:38
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1 Answer

up vote 9 down vote accepted

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

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In the 1975 Erdos paper mentioned in OEIS sequence A003135, this is considered a trivial solution. See pages 27-28 of that paper. –  tdnoe Sep 20 '10 at 19:54
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