reserve a,b,c,k,m,n for Nat;
reserve p for Prime;

theorem Th36:
  (2*n+1)^2 mod 8 = 1
  proof
    defpred P[Nat] means (2*$1+1)^2 mod 8 = 1;
A1: P[0] by NAT_D:24;
A2: P[k] implies P[k+1]
    proof
      assume
A3:   P[k];
      thus (2*(k+1)+1)^2 mod 8 = (4*k^2+4*k+1 + 8*(k+1)) mod 8
      .= (1 + (8*(k+1) mod 8)) mod 8 by A3,NAT_D:66
      .= (1 + 0) mod 8 by NAT_D:13
      .= 1 by NAT_D:24;
    end;
    P[k] from NAT_1:sch 2(A1,A2);
    hence thesis;
  end;
