= Solution
For a <bound state>, $0<E<U_0$: the <Hamiltonian operator> is nonnegative, and an energy at or above the exterior potential does not give a nonzero <normalizable wavefunction>. The <Time-independent Schrödinger equation> and the hard wall at zero therefore give
$$
\psi(x)=A\sin(kx)\quad(0<x<a),\qquad \psi(x)=B e^{-\kappa(x-a)}\quad(x>a),
\qquad k=\frac{\sqrt{2mE}}{\hbar},\quad \kappa=\frac{\sqrt{2m(U_0-E)}}{\hbar}.
$$
Continuity of the <wavefunction> and its derivative at the finite step gives $k\cot(ka)=-\kappa$. Put $z=ka$ and $z_0=a\sqrt{2mU_0}/\hbar$. The matching equation becomes
$$
z\cot z=-\sqrt{z_0^2-z^2},\qquad 0<z<z_0.
$$
It requires $\cot z<0$, so solutions lie in the intervals $((j+1/2)\pi,(j+1)\pi)$ for $j=0,1,\ldots$. Squaring, with that sign restriction retained, gives $z_0=z/|\sin z|$. On each such interval this function increases strictly from $(j+1/2)\pi$ to infinity: its derivative is $(|\sin z|-z\,\operatorname{sgn}(\sin z)\cos z)/\sin^2z>0$, because the second term in the numerator is positive there. Each interval therefore contributes exactly one <bound state> if and only if $z_0>(j+1/2)\pi$. These are the <bound-state thresholds for a square well with one hard wall>.
\b[Exactly one normalizable bound state exists if and only if]
$$
\boxed{\frac{\pi^2\hbar^2}{8ma^2}<U_0\leq\frac{9\pi^2\hbar^2}{8ma^2}.}
$$
\b[The printed strict upper inequality misses the threshold case.] At the upper endpoint, the prospective second state has $E=U_0$ and $\kappa=0$. Its exterior solution is constant or linear, hence is not square-integrable unless zero; the zero exterior solution and matching would force the entire solution to vanish. Thus there is still exactly one <bound state> at that endpoint. The strict upper inequality is valid if threshold values are additionally excluded, but that exclusion is not part of the stated assumptions.
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