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
  n is even & n > 4 implies 2|^n-1 is having_at_least_three_different_divisors
  proof
    assume n is even;
    then consider k such that
A1: n = 2*k;
    assume that
A2: n > 4 and
A3: not 2|^n-1 is having_at_least_three_different_divisors;
    reconsider k as non zero Nat by A1,A2;
A4: now
      assume 2|^k-1 <= 1;
      then 2|^k-1 = 0 or 2|^k-1 = 1 by NAT_1:25;
      then 2|^k = 2|^1;
      then k = 1 by PEPIN:30;
      hence contradiction by A1,A2;
    end;
    2|^(2*k) = 2|^k|^2 by NEWTON:9
    .= (2|^k)^2 by WSIERP_1:1;
    then
A5: 2|^n-1 = (2|^k+1)*(2|^k-1) by A1;
    then
A6: 2|^k-1 divides 2|^n-1;
A7: 2|^k+1 divides 2|^n-1 by A5;
A8: 2|^k-1 < 2|^k+1 by XREAL_1:8;
    then
A9: 2|^k+1 > 1 by A4,XXREAL_0:2;
    now
      assume not 2|^k+1 is prime or not 2|^k-1 is prime;
      then per cases;
      suppose not 2|^k-1 is prime;
        then consider a such that
A10:    a divides 2|^k-1 and
A11:    a <> 1 and
A12:    a <> 2|^k-1 by A4;
A13:    now
          assume a <= 1;
          then a = 0 or a = 1 by NAT_1:25;
          hence contradiction by A10,A11;
        end;
A14:    a,2|^k-1,2|^k+1 are_mutually_distinct by A8,A10,A12,NAT_D:7;
        a divides 2|^n-1 by A6,A10,INT_2:9;
        hence contradiction by A3,A4,A6,A7,A9,A13,A14;
      end;
      suppose not 2|^k+1 is prime;
        then consider a such that
A15:    a divides 2|^k+1 and
A16:    a <> 1 and
A17:    a <> 2|^k+1 by A4,A8,XXREAL_0:2;
A18:    now
          assume a <= 1;
          then a = 0 or a = 1 by NAT_1:25;
          hence contradiction by A15,A16;
        end;
        now
          assume a = 2|^k-1;
          then 2|^k-1 divides (2|^k+1)-(2|^k-1) by A15,INT_5:1;
          hence contradiction by A4,NUMBER07:9;
        end;
        then
A19:    a,2|^k-1,2|^k+1 are_mutually_distinct by A17;
        a divides 2|^n-1 by A7,A15,INT_2:9;
        hence contradiction by A3,A4,A6,A7,A9,A18,A19;
      end;
    end;
    then k = 2 by Th22;
    hence contradiction by A1,A2;
  end;
