I know that we are gonna need to use one of the identities that the $\gcd$ is equal to but I can't remember one that would be useful for this problem. Any help?
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2Bézout. $ $ $ $ – Did Feb 27 '13 at 19:29
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1Duplicate, full answer here. – Math Gems Feb 27 '13 at 19:36
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Can you think of an integer $m$ with the property that if you multiply $n!+1$ with $m$, you will get something that is relatively close to $(n+1)!+1$? That gives you a first step of Euclid. – Jyrki Lahtonen Feb 27 '13 at 19:36
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Opps, sorry y'all. Didn't see that one. Thank you for your help!!! – user63342 Feb 27 '13 at 19:50
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Just note that a common divisor $d$ of $(n+1)!+1$ and $n!+1$ divides also $$ (n+1) (n!+1) - ((n+1)!+1) = (n+1)! + n+1 - (n+1)! - 1 = n. $$ And then $\gcd(n!+1, n ) = 1$, as $n!+1$ divided by $n$ leaves remainder $1$.
Andreas Caranti
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