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So $(x-1) = np$ or $(x+1) = np$; $\Rightarrow x = np\pm1$  But because we are given limits that are between 1 and n, $x = np\pm1$ then $p=1$ and it must be $-1$, so you get $x=n-1$.  \item  $(n-1)! \equiv 1modn$  Let n=5:  $4! \equiv -1\ \textrm{mod}\ n$