= Solution
Let $L=\Delta+c$ with real $c$, and let
$$
N=\{z\in C^{2,\alpha}(\overline\Omega):Lz=0,\ z|_{\partial\Omega}=0\}.
$$
The <Fredholm alternative for an elliptic Dirichlet problem> says that $N$ is finite-dimensional. If $N=0$, all the prescribed data admit a unique solution. In general the exact <boundary compatibility in the self-adjoint Fredholm alternative> is
$$
\boxed{\int_\Omega fz\,dx+\int_{\partial\Omega}\psi\,\partial_\nu z\,dS=0
\quad\text{for every }z\in N.}
$$
These conditions are necessary and sufficient; when they hold, all solutions form $u_0+N$. Thus in the second branch some data are incompatible, while compatible data have nonunique solutions. The outward <normal derivative> fixes the displayed sign.
To prove the alternative, put $X=\{w\in C^{2,\alpha}(\overline\Omega):w|_{\partial\Omega}=0\}$ and $Y=C^{0,\alpha}(\overline\Omega)$. The standard zero-boundary <Poisson equation> theorem makes $\Delta:X\to Y$ an isomorphism. Multiplication by $c$, followed by $\Delta^{-1}$, is a <compact operator> $K:X\to X$: the embedding $C^{2,\alpha}\hookrightarrow C^{0,\alpha}$ is compact, and multiplication is bounded. Writing $u=\psi+w$ gives
$$
(I+K)w=\Delta^{-1}F,\qquad F=f-\Delta\psi-c\psi.
$$
The <Fredholm alternative for a compact operator> implies finite kernel and cokernel of equal dimension, and invertibility exactly when the kernel vanishes.
For the explicit compatibility conditions, use the <self-adjoint> Dirichlet realization of $L$ on $L^2$. It has compact resolvent: a sufficiently large positive shift of $-L$ is <coercive>, its inverse gains two derivatives, and the <Rellich-Kondrachov compactness theorem> makes that inverse compact. The <Fredholm solvability condition for a self-adjoint operator> is $F\perp N$. Standard boundary <elliptic regularity> upgrades the resulting <weak solution> to $C^{2,\alpha}$ for these data and boundary. Thus it gives the same kernel and solvability as the preceding Hölder-space formulation.
Finally <Green second identity>, with $z=0$ on the boundary and $Lz=0$, gives
$$
\int_\Omega zL\psi\,dx=-\int_{\partial\Omega}\psi\partial_\nu z\,dS.
$$
Consequently $\int zF=\int zf+\int\psi\partial_\nu z$, proving both necessity and sufficiency with the stated sign. If using complex data and real $c$, replace $z$ by $\overline z$ in the pairings.
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