Let $p \ge 2$ be a positive integer, and let $Q \in \mathcal P(\mathbb Z_p)$ be a probability distribution on $\mathbb Z_p$.

Question. What are necessary and sufficient conditions on $Q$ to ensure that the later admits a square-root w.r.t convolution, i.e such that there exists $D \in \mathcal P(\mathbb Z_p)$ verifying $D \star D = Q$ ?

I'm particularly, interested in the case where $Q$ is Zipf, i.e $Q(k) \propto (k+1)^{-\beta}$ for some $\beta \gt 1$.

An illustrative example

Consider the case where $p = 2$. Let $a := Q(0)$, $b:=Q(1)$, $x:=D(0)$, and $y := D(1)$. We are interested in the feasibility of the following system. \begin{align} a &= x^2 + y^2,\\ b &= xy + yx = 2xy,\\ 1 &= x + y,\\ x &\ge 0,\\ y &\ge 0. \end{align} Substituting $y = x - 1$ gives $2x(1-x) = b$, i.e $2x^2 - 2x + b = 0$, which evaluates to \begin{eqnarray} x_\pm = \frac{2 \pm \sqrt{4 - 8b}}{4} = \frac{1 \pm \sqrt{1 - 2b}}{2}. \end{eqnarray} For this to be real, we require $$ b \le 1/2. $$ With this condition, note that $x_+ + x_- = 1$ and $x_+, x_- \ge 0$. Take $x=x_+$ and $y=x_-$. Now, the first equation (the circle) reduces to the requirement

$$ a = x_+^2 + x_-^2 = 2\left(\frac{1}{4} + \frac{1-2b}{4}\right) = 1-b, $$ i.e $a+b=1$, which is satisfied since $Q$ is a distribution. We conclude that in the case $p=2$, the answer to our question is affirmative if $b \le 1/2$. It's easy to check that this condition is also necessary (i.e $D$ doesn't exist when $b \gt 1/2$).