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Is anybody know a solution of this problem?

(Sorry, correct question is here.)

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    $\begingroup$ m=1, n=0 or 1. For larger n, there is a prime between n/2 and n, which guarantees an unsquared prime factor in the factorial. $\endgroup$ – Hugo van der Sanden Sep 18 '10 at 10:17
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Bertrand's postulate (http://en.wikipedia.org/wiki/Bertrand%27s_postulate).

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  • $\begingroup$ Not the first time ... I added the Wikipedia reference and thought twice, so was 18 seconds behind Robin. $\endgroup$ – Charles Matthews Sep 18 '10 at 10:18
  • $\begingroup$ Sorry Charles! It's an old chestnut though :-) $\endgroup$ – Robin Chapman Sep 18 '10 at 10:19
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From Bertrand's postulate it follows swiftly that there are no solutions with $n>1$.

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    $\begingroup$ Just to be more detailed for those who may need it: If $p$ be is the largest prime less than $n$ then $p$ divides $n!$, hence $p$ divides $m^2$ so $p^2$ divides $n!$. So $n>=2p$ which contradicts Bertrand's postulate (and the only solutions are the trivial ones). $\endgroup$ – danseetea Sep 18 '10 at 10:29

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