Divisibility among discriminants Let $f(x)$ be an algebraic function over the field $\mathcal F$ of algebraic numbers over $\mathbb{Q}$.  Suppose that $r \in \mathcal F$.  Does the discriminant of $f(r)$ divide the discriminant of $f(f(r))$? 
Example: take $r$ to be any algebraic number and  $$f(x) = (x + \sqrt{4 + x^2} )/2 = 
[x,x,x, \ldots]$$  Writing $D$ for discriminant, in this example and many others, something much stronger may be true: $(D(f(r)))^2 | D(f(f(r))).$  Are there known conditions (on $f$ and $r$) for this stronger divisibility? 
Thanks, GNiklasch.  Yes, I mean "discriminant" to be the field discriminant, as given by Mathematica's "NumberFieldDiscriminant".  Let's assume that $\mathbb{Q}(f(f( r)))$ contains $\mathbb{Q}(f( r))$ as a subfield -- maybe later consider other possibilities.    
 A: There are a number of papers that deal with fields generated by points in the inverse image of iterates. So in your setting, let $c=f(f(r))=f^2(r)$, then $f(r)$ is in $f^{-1}(c)$, the first inverse image, and $r\in f^{-2}(c)$. So some of these papers could be relevant. For example, the paper "Discriminants of Chebyshev Radical Extensions," Thomas Alden Gassert, http://arxiv.org/abs/1304.6055 discusses discriminant of iterates. Actually, 
Gassert's thesis "Prime Decomposition in Iterated Towers and Discriminant Formulae" (UMass 2014) probably has lots of useful formulas and references. You could also write directly to him, he's currently a post-doc at the University of Colorado at Boulder.
A: A partial answer:
I am not quite sure what "the discriminant" of an algebraic number $\alpha$ is supposed to mean (that of the number field $\mathbb{Q}(\alpha)$?), but for most $f$, there will be algebraic values of $r$ resulting in trivial counterexamples.
E.g., take $f(x) = x^2-2$ and $r$ a root of $f(f(x))-1$.  Then $r$ is an algebraic integer of degree $4$ in an obvious number field of discriminant 2304, and $f( r)$ is a square root of $3$ in the quadratic subfield of discriminant $12$, and $f(f( r))$ is the algebraic integer $1$, to which I would assign the discriminant $1$ under any reasonable definition.
To get somewhere, you'll probably want to assume that the field $L = \mathbb{Q}(f(f( r)))$ contains $K = \mathbb{Q}(f( r))$ as a subfield, which (as the example shows) isn't necessarily the case but which readily entails divisibility of the field discriminants $D_K\mid|D_L$. When $K$ happens to be a proper subfield of $L$, a proper power of $D_K$ will divide $D_L$. Of course the latter case can't occur when $f$ happens to be a polynomial with rational coefficients and would thus map $K$ into itself.
