Hi, since about three weeks i am learning Haskell now. One of my first excercises is to decompose an Integer into its primefactors. I already posted discussion on the solution to the problem 35 in "99 excercises". My simple algorithm uses a datatype DivIter of 4 named fields together with the core iteration divstep :: DivIter -> DivIter divstep x | divisor x > bound x = x | ximod > 0 = x { divisor = (divisor x) +2 } | otherwise = x {dividend=xidiv, bound=intsqrt(xidiv), result = result x ++ [divisor x] } where (xidiv, ximod) = divMod (dividend x) (divisor x) (dividend x is already odd, when this is called). The problem to solve for really large Integers is how to call divstep iterated without not accumulating billions of stack frames. Here is one possibility: divisions = do y <- get if divisor y <= bound y then do put ( divstep y ) divisions else return y (this for a version of divstep without the first guard) called from res = execState divisions (DivIter { dividend = o1, divisor = 3, bound = intsqrt(o1), result = o2 }) ( where o1 "the odd part" of the number to decompose, o2 a list of its "contained" twos). This computes the primefactors of 2^120+1 in 17 minutes after all. But i cannot help feeling that this is an abuse of the State monad. The MonadFix has a function fix (a -> a) -> a and i added the first guard in divstep above for making this a fixpoint problem. For me the signature looks, as if using fix doesn't afford to create explicitely a variable of a MonadFix instance and a simple twoliner for divisions could do the job. What i do not understand at all from the documentation of fix is: "fix f is the least fixed point of the function f, i.e. the least defined x such that f x = x." What does "least" mean here ? There is nothing said about x being a variable of an instance of Ord. And why fix has not the type a -> (a -> a) -> a, means: How can i provide a starting point of the iteration x ==> f x ==> f (f x) ==> ...?