DA: The original question plus (ax+b)/(cx+d)brings up the question: Which rational functions over GF(n) (n = prime power) are permutations of GF(n) ? WDS: One of my quibbles with RCS probably due to miscommunication. My point is, the cycle in the multiplicative group of GF(p^e) is generated by x := g*x where g is a plain element of GF(p^e). Re DA's question about rational functions: It seems more natural to me if we ask, not about permuting GF(q), but rather about permuting GF(Q) U {infinity}, which has q+1 elements. That is a different question, which DA may or may not prefer. My question has a nice partial answer which is that A/(x+B) with A nonzero causes a permutation. Not so nice in DA's version. And plainly composing those and RCS's two polynomial generators in all possible ways yields all permutations of GF(Q) U {infinity}. The point of RCS's proof was, every permutation of GF(q) is realizable by a polynomial of degree<q. Further, that polynomial is unique, since we can do lagrange interpolation to find it. Can every perm of GF(q) U {infinity} can be realized by poly1(x)/poly2(x) with degree1+degree2<=q because of rational interpolation? For DA's original problem without infinity, The denominator cannot be linear(x) since it will have a zero. It perhaps could be quadratic though, e.g: denominator = x^2 - A where A is nonsquare in GF(q). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)