= Solution
Write $T=\tanh(ax)$ and $S=\operatorname{sech}^2(ax)$. Using
$$
T'=aS,
\qquad
S'=-2aST,
$$
two differentiations of
$$
\psi(x)=Ne^{ikx}(aT-ik)
$$
give
$$
-\frac{\hbar^2}{2m}\psi''
-\frac{\hbar^2a^2}{m}S\psi
=\frac{\hbar^2k^2}{2m}\psi.
$$
Thus it solves the time-independent <Schrödinger equation> in the stated <Pöschl-Teller potential>, with
$$
\boxed{E=\frac{\hbar^2k^2}{2m}}.
$$
As $x\to-\infty$ and $x\to+\infty$, respectively,
$$
\psi(x)\sim N(-a-ik)e^{ikx},
\qquad
\psi(x)\sim N(a-ik)e^{ikx}.
$$
There is no left-moving term, so this is a <reflectionless potential>:
$$
\boxed{P_{\rm ref}=0}.
$$
The incident and transmitted amplitudes have equal modulus because
$$
|-a-ik|^2=|a-ik|^2=a^2+k^2.
$$
Hence
$$
\boxed{P_{\rm tr}=1}.
$$
Now set $k=i\lambda$ with $\lambda>0$. Then
$$
\psi(x)=Ne^{-\lambda x}\bigl(a\tanh(ax)+\lambda\bigr),
\qquad
E=-\frac{\hbar^2\lambda^2}{2m}.
$$
At $+\infty$ this decays for every positive $\lambda$. At $-\infty$, however, the factor tends to $\lambda-a$ while $e^{-\lambda x}$ grows. Normalizability therefore requires $\lambda=a$. In that case
$$
\psi(x)=Na\,e^{-ax}(1+\tanh ax)
=Na\,\operatorname{sech}(ax),
$$
which decays at both ends. The bound-state energy is
$$
\boxed{E=-\frac{\hbar^2a^2}{2m}}.
$$
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