Question regarding the Cauchy functional equation

Is it true that, if a real function $f$ satisfies $f(x+y) = f(x) + f(y)$ and vanishes at some $k \neq 0$, then $f(x) = 0$? Over the rationals(or, allowing certain conditions like continuity or monotonicity), this is clear since it is well known that the only solutions to this equation are functions of the form $f(x) = cx$. The reason I'm asking is to see whether or not there's "weird" solutions other than the trivial one.

Some observations are that $f(x) = f(x+k) = -f(k-x)$. $f$ is periodic with $k$.

It is easy to see that at $x=\frac{k}{2}$ the function also vanishes, and so, iterating this process, the function vanishes at and has a "period" of $\frac{k}{2^n}$ for all $n$. If the period can be made arbitrarily small, I want to say that implies the function is constant, but of course I don't know how to preclude pathological functions.

Answer

The values of $f$ can be assigned arbitrarily on the elements of a Hamel basis for the reals over the rationals, and then extended to all of $\mathbb{R}$ by $\mathbb{Q}$-linearity. So (assuming the Axiom of Choice) there are indeed weird solutions.