p-1=nk p=nk+1 p|n^6-1=(n-1)(n+1)(n^2-n+1)(n^2+n+1) We will now check the cases: 1. p|n-1 is impossible because from n|p-1 we have p≥n+1>n-1 so that gives us n-1=0 what is impossible because n > 1 2. p|n+1 gives us p=n+1 so p-n=(n+1)-n=1 what is a perfect square 3. p|n^2-n+1 Using p=nk+1 we have nk+1|n^2-n+1-(nk+1)=n^2-n-nk=n (n-k-1) It's obvious that (n, nk+1)=1 so nk+1|n-k-1. We have nk+1>n>n-k-1 so n-k-1=0, k=n-1 p=nk+1 p=n(n-1)+1=n^2-n+1 p-n=(n^2-n+1)-n=(n-1)^2 4. p|n^2+n+1 nk+1|n^2+n+1-(nk+1)=n^2+n-k=n (n-k+1) so nk+1|n-k+1 nk+1>n>n-(k-1) so n-(k-1)=0, k=n+1 p=n(n+1)+1=n^2+n+1 p+n=(n^2+n+1)+n=(n+1)^2