= Solution
Take <Poisson process> arrivals of rate $a$, independent holding times with the <exponential distribution> of parameter $\mu$, unit resource requirement per call, no waiting room, and integer capacity $C\ge1$. The occupancy is a <birth-death process> on $\{0,\ldots,C\}$ with birth rate $a$ below $C$ and death rate $k\mu$ at occupancy $k$. Writing $\nu=a/\mu$ for <offered traffic>, <detailed balance for a continuous-time Markov chain> gives
$$
\pi(k)a=\pi(k+1)(k+1)\mu,\qquad \pi(k)=\frac{\nu^k/k!}{Z_C(\nu)},\qquad Z_C(\nu)=\sum_{h=0}^C\frac{\nu^h}{h!}.
$$
An arrival is lost exactly in state $C$, so <Poisson arrivals see time averages> proves the <Erlang loss formula>
$$
\boxed{E(\nu,C)=\frac{\nu^C/C!}{\sum_{h=0}^C\nu^h/h!}.}
$$
This derivation assumes exponential holding times; the <insensitivity of loss networks> extends the same formula to independent general holding times with the same mean. For completeness, $E(\nu,0)=1$; for $C\ge1$, set $E(0,C)=0$.
A <loss network> with <fixed routing> has finitely many resources, capacities $C_j$, and call types indexed by nonempty subsets $r$ of resources. A call of type $r$ uses one unit of every resource in $r$ for its entire holding time and is accepted only if all are available. Let $A_{jr}=\mathbf1_{\{j\in r\}}$ be the <link-route incidence matrix> and $\nu_r$ the type's <offered traffic>, arrival rate times mean holding time. Different types have independent <Poisson processes> and independent holding times.
In the <reduced-load approximation>, availability at distinct resources is treated as independent, and screening at the resources other than $j$ is modeled as independent Poisson thinning. The load offered to $j$ after that screening is
$$
a_j(B)=\sum_{r:j\in r}\nu_r\prod_{i\in r\setminus\{j\}}(1-B_i).
$$
Applying the <Erlang loss formula> to this isolated-resource load gives the <Erlang fixed point approximation>, $B_j=E(a_j(B),C_j)$. These equations are an approximation because simultaneous admission induces dependence between resource occupancies, and the screened arrival process need not actually be Poisson.
Here is a proof of existence and uniqueness for positive capacities. Write $e_C(a)=E(a,C)$. Its truncated-Poisson mean is the <carried load of an Erlang loss resource>,
$$
m_C(a)=\frac{aZ_C'(a)}{Z_C(a)}=a[1-e_C(a)].
$$
For $a>0$, differentiating its stationary probabilities gives
$$
m_C'(a)=\frac{\operatorname{Var}_a(K)}a>0,\qquad e_C'(a)=\frac{e_C(a)}a[C-m_C(a)]>0.
$$
The variance is positive and $m_C(a)<C$ because every state has positive probability. Moreover $e_C$ increases continuously from zero to one, and $m_C$ increases from zero to $C$. Thus $e_C$ has a continuous inverse on $[0,1)$.
Use acceptance coordinates $y_j=-\log(1-B_j)\ge0$ and define
$$
u_C(y)=e^{-y}e_C^{-1}(1-e^{-y})=m_C\bigl(e_C^{-1}(1-e^{-y})\bigr),\qquad u_C(0)=0.
$$
Here $u_C$ is strictly increasing, continuous, and tends to $C$ as $y\to\infty$. The <convex potential for the Erlang fixed point> is
$$
F(y)=\sum_r\nu_r\exp\left(-\sum_{i\in r}y_i\right)+\sum_j\int_0^{y_j}u_{C_j}(z)\,dz.
$$
Each exponential term is convex, and each integral is a <strictly convex function> of its coordinate because its derivative is strictly increasing. Hence $F$ is strictly convex on the nonnegative orthant. It is also a <coercive function>: for sufficiently large $z$, $u_{C_j}(z)\ge C_j/2$, so each integral tends at least linearly to infinity. Therefore $F$ attains a unique minimum $y^*$.
Its partial derivative is
$$
\partial_jF(y)=u_{C_j}(y_j)-\sum_{r:j\in r}\nu_r\exp\left(-\sum_{i\in r}y_i\right)=u_{C_j}(y_j)-e^{-y_j}a_j(B).
$$
At an interior minimum this derivative is zero. Since $u_C(y)=e^{-y}e_C^{-1}(1-e^{-y})$, this is exactly $B_j=e_{C_j}(a_j(B))$. At a boundary coordinate $y_j=0$, the one-sided minimizing condition is $\partial_jF\ge0$, whereas the displayed derivative is $-a_j(B)\le0$. Thus $a_j(B)=0$ and $B_j=E(0,C_j)=0$ there as well. Conversely every solution of the <Erlang fixed point approximation> has these derivative conditions and hence minimizes the convex potential. No solution has $B_j=1$ at positive capacity, since every $a_j(B)$ is finite and $E(a,C)<1$ for finite $a$. The coordinates therefore give a bijection between solutions and minima, proving
$$
\boxed{\text{The fixed-routing Erlang equations have exactly one solution }B\in[0,1)^J\text{ when all }C_j>0.}
$$
If a capacity is zero, its blocking probability is forced to one. Routes using it contribute zero reduced load to every other resource. Delete those routes and the zero-capacity resources and apply the proof to the remaining network; unused positive-capacity resources have blocking probability zero. This establishes existence and uniqueness also with zero capacities, allowing $B_j=1$ precisely at those resources.
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