I have the following systems of equations that I would like to solve for $A$ and $B$:
\begin{align} A&=\frac{1}{N}\sum_i^N\frac{(a_i-x+\tau A)}{B-\left(a_i-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}\\ 1&=\frac{1}{N}\sum_i^N\frac{-1}{B-\left(a_i-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}\\ \end{align}
Is there a way to solve this equation at least one of them?
I tried something in those lines: Starting from equation $(2)$ we have:
\begin{align} \sum_i^N\frac{1}{B-\left(a_i-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}&\equiv\sum_i\frac{1}{B-C_i}\\ &=\frac{\sum_i\left(\prod_{j\neq i}(B-C_i)\right)}{\prod_i(B-C_i)} \end{align} Then equation $(2)$ becomes: \begin{align} \implies -N&=\frac{\sum_i\left(\prod_{j\neq i}(B-C_i)\right)}{\prod_i(B-C_i)}\\ -N\prod_i^N (B-C_i)&=\sum_i\left(\prod_{j\neq i}(B-C_i)\right) \end{align}
But this does not tell me much. Is there an obvious method to solve $A$ or $B$?
Edit: $a_i$ represents the eigenvalues of a matrix. If half of the eigenvalues are $1$ and the other half $-1$ the equations then become:
\begin{align} A&=\frac{(1-x+\tau A)}{B-\left(1-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}+\frac{(-1-x+\tau A)}{B-\left(-1-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}\\ 1&=\frac{-1}{B-\left(1-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}+\frac{-1}{B-\left(-1-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}\\ \end{align}
Which are equivalent to solving polynomial equations.
I was wondering if the first equations could be simplified (if not solved) assuming that the sum converges.
An other way to look at it, is assuming that $a_i$ follows a distribution and $N\to \infty$. Then the equations become: \begin{align} A&=\int p(a)\frac{(a-x+\tau A)}{B-\left(a-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}\mathrm{d}a\\ 1&=\int p(a)\frac{-1}{B-\left(a-x+\tau A\right)^2-\frac{y^2}{(\tau-1)^2}}\mathrm{d}a\\ \end{align}
In that case is it possible to simplify or solve the equations?