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I am self-studying number theory and went back to an earlier section and was not able to solve this. I have tried many things in order to write it as a linear combination equal to 1. I really just want a hint because I haven't been stuck this long so I may just be overlooking something simple. Thanks.

Derek Luna
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    Any common divisor divides the difference $(n+1)!+1 - (n!+1) = (n+1)!-n! = n!(n)$. But no nontrivial factor of $n!(n)$ divides $n!+1$. – Arturo Magidin Apr 06 '19 at 20:47
  • You might find this past MSE post fruitful -- https://math.stackexchange.com/questions/25688/prove-that-gcdn1-n11-1 – PrincessEev Apr 06 '19 at 20:47
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    This problem is kind of calling out for a proof by contradiction. That plus @ArturoMagidin 's comment is all you need. – Leonhard Euler Apr 06 '19 at 20:48
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    @stuartstevenson: Why do a proof by contradiction, when you can do a direct proof that any common divisor divides $1$? – Arturo Magidin Apr 06 '19 at 20:52
  • I got to the point of equating the difference to $n!(n)$ but never made the connection that none of the nontrivial factors would divide $n!+1$. Oops. Thank you all. – Derek Luna Apr 06 '19 at 20:53
  • @ArturoMagidin It's faster. – Leonhard Euler Apr 06 '19 at 21:11
  • @stuartstevenson: Not really;you have to add an extra hypothesis at the top (“assume not”), and then deduce a contradiction from the conclusion you reached without every using that assumption, adding “but this contradicts our assumption that we had a nontrivial factor”. Sounds like two extra steps to me. – Arturo Magidin Apr 06 '19 at 21:12

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Hint: use $1=n!+1-(n-1)!n,\,n=(n+1)(n!+1)-((n+1)!+1)$ to write $1$ in the desired form.

J.G.
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