= Solution
For $g\geq0$, the <Hamiltonian operator> is nonnegative: its <quadratic form> is $\hbar^2\int|\psi'|^2dx/(2m)+g(|\psi(a)|^2+|\psi(-a)|^2)\geq0$. Thus it has no negative-<energy> <bound states>. A nonnegative-<energy> exterior <wavefunction> cannot be square integrable unless it vanishes identically, so there are no other <bound states> either.
For $g<0$, set $\nu=-mg/\hbar^2>0$. The even equation is the zero of $f(\kappa)=\nu(1+e^{-2a\kappa})-\kappa$. Since $f(0)=2\nu>0$, $f(\kappa)\to-\infty$ and $f'(\kappa)=-2a\nu e^{-2a\kappa}-1<0$, the <intermediate value theorem> and strict <monotonicity> give exactly one positive <equation root>.
The odd equation is the zero of $h(\kappa)=\nu(1-e^{-2a\kappa})-\kappa$. Here $h(0)=0$, $h'(0)=2a\nu-1$, $h''(\kappa)=-4a^2\nu e^{-2a\kappa}<0$, and $h(\kappa)\to-\infty$. If $2a\nu\leq1$, strict <concavity> makes $h(\kappa)<0$ for every $\kappa>0$. If $2a\nu>1$, $h$ initially increases, has exactly one maximum, and then decreases through zero exactly once. This proves the <bound-state count of an attractive symmetric double-delta potential>:
$$
\boxed{N(g)=\begin{cases}
0,&g\geq0,\\
1,&-\dfrac{\hbar^2}{2ma}\leq g<0,\\
2,&g<-\dfrac{\hbar^2}{2ma}.
\end{cases}}
$$
At the threshold $g=-\hbar^2/(2ma)$, the odd zero-<energy> solution is proportional to $x$ inside $(-a,a)$ and to $a\operatorname{sgn}x$ outside. It is bounded but not a <square-integrable function>, so it is a <zero-energy scattering resonance> rather than a second <bound state>. Each negative-<energy> <bound state> is either even or odd by <nondegeneracy of one-dimensional bound states>; the two branches above exhaust the count.
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