p}_{ij}^{11}}+\frac{1}{{\hat p}_{ij}^{01}}-\frac{1}{{\hat p}_{ij}^{10}}-\frac{1}{{\hat p}_{ij}^{00}}}
{\frac{1}{{\hat p}_{ij}^{11}}+\frac{1}{{\hat p}_{ij}^{01}}+\frac{1}{{\hat p}_{ij}^{10}}+\frac{1}{{\hat p}_{ij}^{00}}},$$ so that with little assistance of Maple, we may finally reach the expression given in [@NgBe] $$\begin{aligned}
[\hat \chi^{-1}]_{ij} = \Bigl[\frac{1-d_i}{1-m_i^2} &+\sum_{k\in \partial i}\frac{1-m_k^2}{(1-m_i^2)(1-m_k^2)
-\hat \chi_{ik}^2}\Bigr]\delta_{ij}\nonumber\\[0.2cm]
&-\frac{\hat \chi_{ij}}{(1-m_i^2)(1-m_j^2)-\hat \chi_{ij}^2}\ \delta_{j\in \partial i},\label{eq:invchis}\end{aligned}$$ equivalent to the original one derived in [@WeTe], albeit written in a different form, more suitable to discuss the inverse Ising problem. This expression is quite paradoxical since the inverse of the $[\chi]_{ij}$ matrix, which coefficients appear on the right hand side of this equation, should coincide with the left hand side, given as input of the inverse Ising problem. The existence of an exact solution can therefore be checked directly as a self-consistency property of the input data $\hat\chi_{ij}$: for a given pair $(i,j)$ either:

-   $[\hat\chi^{-1}]_{ij}\ne 0$, then this self-consistency relation (\[eq:invchis\]) has to hold and $J_{ij}$ is given by (\[eq:Jij\]) using $\hat m_{ij}= \hat m_i\hat m_j+\hat \chi_{ij}$.

-   $[\hat\chi^{-1}]_{ij} =  0$ then $J_{ij}=0$ but $\hat\chi_{ij}$, which can be non-vanishing, is obtained by inverting $[\hat\chi^{-1}]$ defined by (\[eq:invchis\]).

Finally, complete consistency of the solution is checked on the diagonal elements in (\[eq:invchis\]). If full consistency is not verified, this equation can nevertheless be used to find approximate 