Let $k$ a field. It is known (Lüroth's theorem) that every subfield of $k(x)$ is of the form $k(\phi(x))$.
We expect that if we take two random non-constant fractions $R(x)$, and $S(x)$ that they generate the field $k(x)$ ( random -- a suggestive term). One can check that in concrete examples using Gröbner bases (elimination). It is definitely so if the degrees of $R(x)$ and $S(x)$ are relatively prime ( the degree of a fraction is the largest of the degrees of the numerator and denominator, equals the cardinality of a generic fibre of the map $x\mapsto R(x)$. Note that the degree of the composition is the product of the degrees).
From now one, $k$ is of characteristic $0$.
I am pretty convinced that given a non-constant fraction $R(x)$, the fraction $R(x)$, $R(x+1)$ generate the whole field of rational fractions.
- Why do I care about that: say we have a sequence $a_n \colon n\mapsto R(n)$ where $R(x) \in \mathbb{C}(x)$. The above implies for instance that the sequence $a_n$ satisfies a rational recurrence of order $2$
$$a_{n+2} = T(a_n, a_{n+1})$$
for all $n\ge 0$. Note that if $R$ is a polynomial of degree $m$ then the sequence $a_n = P(n)$ satisfies a linear recurrence of order $m+1$. But ( perhaps surprising), it will satisfy a rational recurrence of order $2$.
- How the proof might work. Assume the contrary, that is both $R(x)$ and $R(x+1)$ are functions of a fraction $\phi(x)$ of degree $> 1$
$$R(x) = U(\phi(x)) \\ R(x+1) = V(\phi(x))$$
We get $U(\phi(x+1)) = V(\phi(x))$. Let's try to get a contradiction.
Hand-wave approach: the function on RHS is constant on the fibres of $\phi(x))$, while the one on LHS is constant on the translates of the fibres of $\phi$ by $1$. If $\deg \phi(x) > 1$, we "get" a contradiction
We can prove the impossibility of some particular cases of the above. For instance, we cannot have
$$\phi(x+1) = V(\phi(x)$$
if $\deg \phi> 1$. Indeed, on RHS we need $\deg V = 1$, but then $V$ is a Mobius transform, so conjugate either to a dilation or a translation. Dilation gets eliminated, and translation means in the end $\phi$ of degree $1$.
Note also that the fact about $R(x)$ is certainly true if $R(x)$ is a polynomial. Indeed, $R(x+1) - R(x)$ has degree relatively prime to $\deg R(x)$.
This method works to show the statement in explicit cases: if the degrees of $R(x)$ and $R(x+1) - R(x)$ ( or other combinations) are relatively prime, the statement follows.