TITLE: Do there exist three pairwise independent random variables, such that their sum is zero?
QUESTION [6 upvotes]: Do there exist such three non-constant pairwise independent random variables $X, Y, Z$ such that $X + Y + Z = 0$?

I managed only to prove the following two facts:

If such $X, Y, Z$ exist, they are not independent.

Proof:
If they are, then $X$ and $-X = Y + Z$ are also independent, which is impossible.

If such $X, Y, Z$ exist, then at least two of them do not have finite second moment.

Proof:
$\DeclareMathOperator\Var{Var}$Suppose, they all have finite second moments. Then $\Var(X) + \Var(Y) + \Var(Z) = 0$, which implies that all $X$, $Y$ and $Z$ are constants. Now suppose that without the loss of generality $Y$ and $Z$ have finite second moment. Then $\Var(X) = \Var(-Y-Z) = \Var(Y) + \Var(Z) \leq \infty$ and we return to the previous case.
However, those facts are clearly insufficient to solve this problem.

REPLY [10 votes]: Replace $Z$ by $-Z$, so that $Z=X+Y$. Let $f_X$ and $f_Y$ the characteristic functions of $X$ and $Y$, so that $f_X(s)=Ee^{isX}$ for real $s$. Suppose the pairwise independence. 
Then for all real $s$ and $u$
$$f_X(u)f_Y(u)f_X(s)=f_Z(u)f_X(s)=Ee^{iuZ+isX} \\
=Ee^{i(u+s)X+iuY}=f_X(u+s)f_Y(u). \tag{1}
$$
Therefore and because $f_Y$ is continuous with $f_Y(0)=1\ne0$, we have 
$$f_X(u+s)=f_X(u)f_X(s) \tag{2}
$$
for all real $u$ close enough to $0$ and all real $s$. 
Note that (2) (together with the conditions that $f_X$ is continuous with $f_X(0)=1\ne0$) implies that $f_X$ is nowhere $0$. Similarly, $f_Y$ is nowhere $0$. So, (2) actually holds for all real $u$ and $s$. So, $f_X(s)=e^{isa}$ for some real $a$ and all real $s$. 
So, $X$ is a constant almost surely. Similarly, $Y$ is a constant almost surely.