Solving the quadratic system $v_i = p_i + v_i \sum_{j \neq i} p_j v_{-j}$ Consider the system of equations for $v_i$
$$
v_i = p_i + v_i \sum_{j \neq i} p_j v_{-j} \tag{1}
$$
where $i = \pm 1, \pm2 \dots \pm m$ and $\sum_{i = -m}^{m} p_i = 1$. Apparently, the solution is given by
$$
v_i = \frac{1}{2p_{-i}}\left(-x + \sqrt{x^2 + 4 p_i p_{-i}}\right), \tag{2.a}
$$
where $x$ is a solution of
$$
0 = 1 + (m - 1) x - \sum_{i = 1}^{m}\sqrt{x^2 + 4 p_i p_{-i}}. \tag{2.b}
$$
I can't understand where Eq. (2) comes from. The source is Eq. (1) and the following paragraph in Ref. [1], but it's in Russian.  
[1]  E. B. Dynkin, M. B. Malyutov, “Random walk on groups with a finite number of generators”, Dokl. Akad. Nauk SSSR, 137:5 (1961), 1042–1045 
 A: This was a while ago, but I guess I'll post the answer for future internet travelers. Observe that
$$
\sum_{j \neq i} p_{j} v_{-j} = \left(\sum_{j} p_{j} v_{-j}\right) - p_{i} v_{-i} \tag{3}
$$
Note that the quantity in brackets is independent of $i$. Therefore, let $y = \sum_{j} p_{j} v_{-j}$. Substituting $(3)$ into $(1)$ along with the definition of $y$ gives
$$
v_i + p_i v_{i} v_{-i} = p_i + v_i y \tag{4}
$$
Interchanging $i$ for $-i$ gives
$$
v_{-i} + p_{-i} v_{i} v_{-i} = p_{-i} + v_{-i} y \tag{4}
$$
Now observe that $(4)$ and $(5)$ form a quadratic system for $v_i$ and $v_{-i}$. Solving and taking the positive root gives
$$
v_{i} = \frac{1}{2 p_{-i}}\left[-(1-y) + \sqrt{(1-y)^2 + 4 p_i p_{-i}} \right]
$$
Now, let $x = 1 - y$. We have
\begin{align}
x &= 1 - \sum_{j} p_{j} v_{-j} \\ 
&= 1 - \sum_{j = -m}^{m} \frac{1}{2}\left[-x + \sqrt{x^2 + 4 p_j p_{-j}} \right] \\ 
&= 1 - \sum_{j = 1}^{m}\left[-x + \sqrt{x^2 + 4 p_j p_{-j}} \right] \\
&= 1 + m x - \sum_{j = 1}^{m}\sqrt{x^2 + 4 p_j p_{-j}} \\
0 &= 1 + (m - 1)x - \sum_{j = 1}^{m}\sqrt{x^2 + 4 p_j p_{-j}}
\end{align}
as required.
