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> Do there exist such three non-constant pairwise independent random variables $X, Y, Z$ such that $X + Y + Z = 0$?
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I managed only to prove the following two facts:



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> If such $X, Y, Z$ exist, they are not independent.
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*Proof:*


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



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



Do there exist three pairwise independent random variables, such that their sum is zero?
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.


