
theorem SFS:
  for a be Nat holds a is square & a is square-free iff a = 1
  proof
    let a be Nat;
    A1: a is square & a is square-free implies a = 1
    proof
      assume
      B1: a is square & a is square-free; then
      reconsider k = 2-root a as Element of NAT by INT_1:3;
      a < 2
      proof
        assume a >= 2; then
        consider p be Element of NAT such that
        C1: p is prime & p divides a by INT_2:31;
        not p|^2 divides k|^2 by B1,C1,MOEBIUS1:def 1;
        hence contradiction by NAT_3:5,WSIERP_1:14,C1;
      end;
      hence thesis by B1,MOEBIUS1:21,NAT_1:23;
    end;
    1*1 is square-free square by MOEBIUS1:22;
    hence thesis by A1;
  end;
