You can take q-deformed number derivatives of field elements. Work in GF(p^n). Define [a] as (q^a - 1)/(q - 1). Let x be a generator of the field. D( x^a ) = ((qx)^a - x^a) / (qx - x) = (q^a - 1)/(q-1) x^(a-1) = [a] x^(a-1) If x^a + x^b = x^c, then [a] x^(a-1) + [b] x^(b-1) (qx)^a + (qx)^b - (x^a + x^b) = ----------------------------- qx - x So if q is such that (qx)^a + (qx)^b = (qx)^c, then we have (qx)^c - x^c = ------------ qx - x = [c] x^(c-1) and the derivative is linear. Setting q = x^(p^k - 1) will satisfy the equation, where 1 < k < n. Of course, you can do "differential equations" this way, too. For example, take GF(2^3) mod x^3+x+1; then x is a generator and D(x^2+x+1) = x^2+x+1. -- Mike Stay staym@clear.net.nz http://www.cs.auckland.ac.nz/~msta039