Is a continuous random variable always dependent on a linear combination which includes itself?

Question

For example, where $Z$ may or may not be independent of $X$, is $X$ ever independent of $X+Z$? Intuitively, it seems absurd to suggest that a random variable may be independent of a function of itself. Indeed, I can think of cases which would provide zero covariance, however I can't think of a way to prove or disprove the claim that $X$ can be independent of $X+Z$, aside from citing the obviously degenerate case of $Z=-X$. Is there an approach which excludes this case and any other cases like it which boil down to just subtracting away the X variable?

Answer 1

@Arthur suggests $Z = Y - X$, where $Y$ is independent of $X$, and OP asks "well, sure, but are there other examples?"

No, there are not. Here's why.

Suppose we have some "other" answer, i.e., suppose that $X$ is independent of $X + Z$.

Then we can let $Y = X + Z$, and we have a $Y$ that is independent of $X$, and the $Z$ we found is then $Z = Y - X$.

In short: no, there is no other possible answer, but this "simple" answer really covers quite a lot.