reserve X for set;
reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve r,s for Real;
reserve p,p1,p2,p3 for Prime;
reserve z for Complex;

theorem Th22:
  2|^n-1 is prime & 2|^n+1 is prime implies n = 2
  proof
    assume that
A1: 2|^n-1 is prime and
A2: 2|^n+1 is prime;
    per cases;
    suppose
A3:   n is odd;
      3 divides 2+1;
      then 3 divides 2|^n + 1|^n by A3,NEWTON03:25;
      then 2|^n+1 = 3 by A2;
      hence thesis by A1;
    end;
    suppose n is even;
      then consider k such that
A4:   n = 2*k;
      now
        assume n <= 1;
        then n = 0 or n = 1 by NAT_1:25;
        then 1-1 is prime by A1,NEWTON:4;
        hence contradiction;
      end;
      then
A5:   n is prime by A1,GR_CY_3:28;
      2 divides 2*k;
      hence thesis by A4,A5;
    end;
  end;
