When does $f(X) = g(X+1)-g(X)$ where $f, g \in \mathbb{C}(X)$? Is there an interesting or useful necessary and sufficient condition for a rational function $f \in \mathbb{C}(X)$ to be expressible as the difference $g(X+1)-g(X)$ of another rational function $g \in \mathbb{C}(X)$?

As an example, every polynomial $f$ is expressible in such a way since we have the Faulhaber's formula.
Alternately, we can observe that $(X+1)^n - X^n = \sum\limits_{k=0}^{n-1}{n\choose k}X^k$ and so given a polynomial $f(X) = a_nX^n + \ldots + a_1X + a_0$ we may solve the system of equations obtained by equating the $k$-th terms of $f$ with those of $\sum\limits_{k=0}^{n}\left(\sum\limits_{i=k+1}^{n+1}b_i{i \choose k}\right)X^k$. The associated matrix consists of the upper triangular matrix with entries $A_{ij} = {j \choose i-1}, 1 \leq i \leq j \leq n+1$, and its determinant is clearly nonzero since it is the product of the diagonal entries ${i \choose i-1} = i$, so the determinant is $(n+1)!$.
The solution for the $b_i$, $1 \leq i \leq n+1$ will give a polynomial $g(X) = b_{n+1}X^{n+1} + \ldots b_1X + b_0$ such that $f(X) = g(X+1)-g(X)$ ($b_0$ is arbitrary).
Another example is the rational function $f(X) = \frac{1}{X(X+1)}$ where $g(X) = -\frac{1}{X}$.

As a non-example, consider the function $f(X) = \frac{1}{X}$. Suppose there exist polynomials $p(X), q(X) \in \mathbb{C}[X]$ such that $\frac{p(X+1)}{q(X+1)}-\frac{p(X)}{q(X)} = \frac{1}{X}$. Assume moreover that $p$ and $q$ are coprime after dividing by any common factors.
Then we have $Xp(X+1)q(X)-Xp(X)q(X+1)=q(X+1)q(X)$ so that $q(X)$ divides $Xq(X+1)$ and $q(X+1)$ divides $Xq(X)$. This means we have polynomials $\alpha(X), \beta(X)$ such that $$Xq(X+1) = \alpha(X)q(X)$$ $$Xq(X) = \beta(X)q(X+1)$$
By degree considerations, we have $\deg \alpha = \deg \beta = 1$.
Then $X^2q(X) = \alpha(X)\beta(X)q(X)$ which means $X^2 = \alpha(X)\beta(X)$ and finally this forces $\alpha(X) = cX$ and $\beta(X) = \frac{1}{c}X$ for some $c \neq 0$. Substituting in the equations above we obtain $q(X+1) = cq(X)$ which is only possible if $q$ is constant because any zero would lead to infinitely many zeros.
But this is impossible because $\frac{p(X)}{q(X)}$ would then be a polynomial, and its forward difference cannot equal $\frac{1}{X}$.
I believe this proof can generalize to other cases where $f(X) = \frac{1}{g(X)}$ where $g$ is irreducible (that would just be the linear functions since $\mathbb{C}$ is algebraically closed).

But I have not found a "nice" condition on $f$ which could be necessary and sufficient. The continuous analogue of the problem is easy, we know that $f$ has a rational antiderivative if and only if its partial fraction decomposition does not contain denominators of degree $1$.
 A: Every rational fraction $f\in \mathbb{C}(X)$ can be written uniquely as a sum 
$$f(X) = P(X) + R(X)$$ where $P$ is a polynomial and $R$ is the polar part.
Now the polar part is uniquely expressed as a finite sum
$$R(X)= \sum_{k\ge 1, \alpha \in \mathbb{C}}\frac{c_{k,\alpha}}{(X-\alpha)^k}$$
The map $g(X) \mapsto g(X+1) -g(X)$ takes polynomial (polar) part to polynomial (polar) part. So $f$ is in the image of this map if and if only if its polar part is. It is easy to see that the necessary and sufficient condition is
$$\sum_{n\in \mathbb{Z}} c_{k,\alpha +n}=0$$ for all $k\ge 1$ and $\alpha \in \mathbb{C}$. (finitely many linear conditions).
ADDED: Consider now the case $f\in \mathbb{Q}(X)$.  The decomposition in partial fraction of $f$ happens in a finite (Galois) extension $K$ of $\mathbb{Q}$. Assume that there exists $g\in \mathbb{C}(X)$ so that $f(X) = g(X+1) - g(X)$.
Since the linear conditions above are satisfied then we can find $g\in K(X)$ so that $f(X) = g(X+1)-g(X)$. Now average all the equalities $f(X) =g^{\sigma}(X+1)-g^{\sigma}(X)$ for all $\sigma \in Gal(K/\mathbb{Q})$. We get $\bar g\in \mathbb{Q}(X)$ so that $f(X) = \bar g(X+1) - \bar g(X)$. 
ADDED:  
The linear necessary and sufficient conditions follow from the following combinatorial lemma:   for a sequence $(a_n)_{n\in \mathbb{Z}}$ of finite support there exists a sequence of finite support  $b= (b_n)_{n\in \mathbb{Z}}$ so that $a= \Delta b$ if and only if $\sum a_n =0$. Moreover, in that case $b$ is uniquely defined. This is the discrete analogue of the result: a function with compact support on $\mathbb{R}$ has an antiderivative with compact support if and only if its total integral is $0$. In our case we want $b_{n+1}-b_n =a_n$ for all $n$. The unique solution with finite support is given by $b_n=\sum_{k<n}a_k$ for all $n$. 
ADDED: As @Steven Stadnicki: noticed, if a solution $g$ exists, it is unique up to a constant. Moreover, if $f$ has coefficients in a subfield $K$, so can $g$ be taken with coefficients in $K(X)$.
We give now some results in particular cases:


*

*Assume that $f(X) = \frac{P(X)}{Q(X)}$ where $Q$ has simple integral roots and 
$\deg Q \ge \deg P +2$. Then $f$ is a difference. Indeed, it is enough to check that the sum of residues of $f$ is $0$, and this follows from Cauchy's formula.

*Assume that $\deg f =-1$. Then 
$f$ is not a difference.  This can be seen in two ways: one, since the sum of residues is not $0$; another way is to notice that $\deg f = \deg g -1$, but $g$  is at most of degree $-1$ ( basically we look at the rate of decrease at $\infty$.
ADDED:
If the equation $f(X)= h(X+2)-h(X)$ has a solution $h$, then the equation $f(X)=g(X+1)-g(X)$ has a solution $g(X)= h(X+1)+h(X)$. 
A: If $f(z)$ has a singularity at $0$ (WLOG) then we must have either $g(z)$ having a singularity at $0$ or at $1$ (or both.)
If $0$ is a singularity of $g$ then since $f(-1)=g(0)-g(-1)$, either $f$ has a singularity at $-1$ or $g$ does (or both.) If $g$ does, then we can keep moving down and find that either $g$ has a singularity at $-k$ or $f$ has singularities at $0$ and $-k.$ So for some $k> 0$, we must have $f(z)$ with singularities at $0,-k$ and $g$ has a singularity at $-1,\dots,-(k-1).$ We have to be able to stop at some point because $f$ must have finitely many singularities.
If $1$ is a singularity of $g$, then either $2$ is a singularity of $g$ or $1$ is a singularity of $f$. We then move up or down, and find that $0$ and $k$ must be a singularity of $f$ for some integer $k>0.$
This means that if $f$ has a singularity at $z_0$ then there must be an integer $n\neq 0$ such that $f$ has a singularity at $z_0+n.$
An Algorithm
Here is an algorithm that doesn't require you to find the exact roots.
The general strategy: Each time through the loop, we will reduce the problem to another $f$ with fewer singularities. 
Write $f(x)=\frac{p(x)}{q(x)}$ with $\deg p<\deg q,$ and $p,q$ are relatively prime. (Any polynomial part can be handled separately.) Compute $q_n(x)=\gcd(q(x),q(x+n))$ for $n=1,2,\dots$ until either:


*

*$q_n(x)\neq 1,$ or 

*$n$ is greater than the diameter of the set of (complex) zeros of $q(x).$


If 2. is reached, and $f\neq 0,$ then there is no $g$. (We can at least compute an upper bound for the diameter, so we only have to check finitely many $n.$)
Otherwise, we've found a $q_n(x)\neq 1.$ Write $q(x)=a(x)b(x)$ where $q_n(x)\mid a(x)\mid q_n^k(x)$ for some $k$ and and $\gcd(a(x),b(x))=1.$ Basically, let $a(x)=\gcd(q(x),q_n^k(x))$ for some large $k,$ so that $a(x)$ has all repetitions of the roots $z_0$ of $q$ such that $z_0+n$ is also a root.
We know that $b(x)$ is non-constant, because any root with a maximal real part is not a root of $q_n(x).$
Then we can write:
$$f(x)=\frac{p(x)}{q(x)}=\frac{c(x)}{a(x)} + \frac{d(x)}{b(x)},$$ using standard partial fraction approaches.
Let $$\begin{align}f_1(x)&=\frac{c(x-n)}{a(x-n)}+\frac{d(x)}{b(x)}.
\end{align}$$
Note that if $h(x)=\frac{c(x)}{a(x)},$ then $h(x)-h(x-n)=\Delta(h(x-1)+h(x-2)+\cdots h(x-n))$, so $f_1$ and $f$ differ by an element in the range of $\Delta.$ 
So $f_1$ is in the range of $\Delta$ if and only if $f$ is.
If $f_1=0$, we are done.
Otherwise we apply the process to $f_1.$
We know that $f_1$ has a smaller set of singularities - we have removed the singularities $z_0$ such that $z_0+n$ is also a singularity, and potentially added in more repetitions of the singularities at $z_0+n.$ What this means is that this process must eventually stop.

For example, if $$f(x)=\frac{3x^3+16x^2+28x+12}{(x+3)(x+2)^2x^2}$$ then $q_1(x)=x+3,$ and you get:
$$f(x)=\frac{-1}{x+3}+\frac{x^3+4x^2+8x+4}{(x+2)^2x^2}$$
Then you get $$f_1(x)=\frac{-1}{x+2}+\frac{x^3+4x^2+8x+4}{(x+2)^2x^2}=\frac{2x^2+8x+4}{(x+2)^2x^2}.$$
For this $f_1$, we get $q_2(x)=(x+2)^2$ and we solve:
$$f_1(x)=\frac{-(x+3)}{(x+2)^2} + \frac{x+1}{x^2}.$$
Then we get:
$$f_2(x)=\frac{-(x+1)}{x^2}+\frac{x+1}{x^2}=0$$
So our resulting expression is:
$$\begin{align}f(x)&=\left(\frac{-1}{x+3}-\frac{-1}{x+2}\right)+\left(\frac{-(x+3)}{(x+2)^2}+\frac{x+1}{x^2}\right)\\
&=-\Delta\left(\frac{1}{x+2} + \frac{x+2}{(x+1)^2}+\frac{x+1}{x^2}\right)
\end{align}$$
In this, I am obviously using the knowledge of the roots to do the work, you need not. All of the steps above can be computed without the roots. You can compute the $q_n,$ then the $a,b$, then $c,d$ all using standard operations.
For example, if the singularities of $f$ were at $0,1,2\pm i,3\pm i,5\pm i$, then your first $a(x)$ would be found with $a(x)$ having roots $0$ and $2\pm i.$ $f_1$ would have remaining singularities a subset of $\{1,3\pm i,5\pm i\}.$ The next step would find $n=2$ and remove singularities at $3\pm i.$ Then we'd find out if $f_2=0.$ If it isn't, the next attempt to find an $n$ will fail.

This algorithm also works in any $k(x)$ where $k\subseteq \mathbb C$ is a field, because all the algorithmic steps work inside the field. 
In particular, if $f(x)\in k(x)$ has a $g(x)\in\mathbb C(x)$ then it has a solution in $k(x).$
