theorem Th29:
  m divides n*k & m gcd n=1 implies m divides k
proof
  assume that
A1: m divides n*k and
A2: (m gcd n)=1;
  consider m1,n1 such that
A3: m*m1+n*n1=1 by A2,Th28;
  k=(m*m1+n*n1)*k by A3
    .=m*(m1*k)+(n*k)*n1;
  hence thesis by A1,Th5;
end;
