reserve a,b,k,m,n,s for Nat;
reserve c,c1,c2,c3 for Complex;
reserve i,j,z for Integer;
reserve p for Prime;
reserve x for object;

theorem Th26:
  for n being Nat holds
  divisors(3|^(2*n+1),4,3) = { 3|^m where m is Nat: m is odd & m <= 2*n+1 }
  proof
    let n be Nat;
    set A = divisors(3|^(2*n+1),4,3);
    set B = {3|^m where m is Nat: m is odd & m <= 2*n+1};
    thus A c= B
    proof
      let x be object;
      assume x in A;
      then consider k such that
A1:   x = k and
A2:   k mod 4 = 3 and
A3:   k divides 3|^(2*n+1);
      consider t being Element of NAT such that
A4:   k = 3|^t and
A5:   t <= 2*n+1 by A3,PEPIN:34,41;
      t is odd by A2,A4,Th23;
      hence thesis by A1,A4,A5;
    end;
    let x be object;
    assume x in B;
    then consider m such that
A6: x = 3|^m and
A7: m is odd and
A8: m <= 2*n+1;
A9: 3|^m mod 4 = 3 by A7,Th24;
    3|^m divides 3|^(2*n+1) by A8,NEWTON:89;
    hence thesis by A6,A9;
  end;
