theorem
  a>1 & b>0 & a gcd b = 1 implies not a divides (a+b)|^n
  proof
    assume
    A0: a>1 & b>0 & a gcd b =1;
    then a gcd b|^n = 1 by WSIERP_1:12;
    then not a divides a gcd b|^n by WSIERP_1:15,A0;
    then not a divides b|^n by INT_2:def 2;
    hence thesis by Th11;
  end;
