= Solution
Let $\phi_1,\phi_2$ be two solutions with the same <Neumann boundary condition>, and set $u=\phi_1-\phi_2$. Then
$$
\nabla^2u-m^2u=0\quad\hbox{in }V,
\qquad
\frac{\partial u}{\partial n}=0\quad\hbox{on }S.
$$
Multiplying by $u$ and applying <Green's first identity> gives
$$
\int_V\left(|\nabla u|^2+m^2u^2\right)dV
=\int_Su\frac{\partial u}{\partial n}\,dS
=0.
$$
The integrand is nonnegative. If $m>0$, both terms can vanish only when $u=0$, so the solution of the <modified Helmholtz equation> is unique.
If $m=0$, the identity only forces $\nabla u=0$. On a connected region, $u$ may be any constant. Thus solutions of the <Laplace equation> with prescribed normal derivative are unique only up to an additive constant, so
$$
\boxed{\text{uniqueness fails when }m=0}.
$$
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