Show that $\displaystyle (n!+1, (n+1)!+1)=1$

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- Feb 27th 2010, 03:40 PMeykeA divisibility problem 2
Show that $\displaystyle (n!+1, (n+1)!+1)=1$

- Feb 27th 2010, 04:32 PMNonCommAlg
if $\displaystyle d \mid n! + 1,$ then $\displaystyle d \mid (n+1)! + n+1.$ so if we also have $\displaystyle d \mid (n+1)! + 1,$ then $\displaystyle d \mid n$ and hence $\displaystyle d \mid n!,$ which completes the proof because we're given that $\displaystyle d \mid n!+1.$