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Is anybody know a solution of this problem?

flag	asked Sep 18 2010 at 10:09 Alexey Ustinov 1,356●4●13

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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. – Hugo van der Sanden Sep 18 2010 at 10:17

Charles Matthews · Accepted Answer · 2010-09-18 10:17:39Z UTC

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Bertrand's postulate (http://en.wikipedia.org/wiki/Bertrand%27s_postulate).

answered Sep 18 2010 at 10:17

Charles Matthews
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	Not the first time ... I added the Wikipedia reference and thought twice, so was 18 seconds behind Robin. – Charles Matthews Sep 18 2010 at 10:18
	Sorry Charles! It's an old chestnut though :-) – Robin Chapman Sep 18 2010 at 10:19

Robin Chapman · Answer 2 · 2010-09-18 10:22:47Z UTC

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From Bertrand's postulate it follows swiftly that there are no solutions with $n>1$.

edited Sep 18 2010 at 10:22

answered Sep 18 2010 at 10:17

Robin Chapman
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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). – danseetea Sep 18 2010 at 10:29

Solve in positive integers $n!=m^2$

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