= Solution
A <bound state> has $E<0$ and $\kappa=\sqrt{-2mE}/\hbar>0$, since outside the <delta potentials> a square-integrable solution must decay exponentially. The <wavefunction> is continuous at each delta: a jump would produce a delta derivative in $\psi''$ with no matching term in the equation. Integrating the <Time-independent Schrödinger equation> across either $x_0=\pm a$ gives
$$
\boxed{\psi(x_0^+)=\psi(x_0^-),\quad\psi'(x_0^+)-\psi'(x_0^-)=-\frac{2m\lambda}{\hbar^2}\psi(x_0)}.
$$
The symmetry of the <symmetric double-delta potential> permits even and odd solutions. Up to a nonzero normalization factor, the even candidate is $\cosh(\kappa x)$ for $|x|\le a$, and $\cosh(\kappa a)e^{-\kappa(|x|-a)}$ outside. Its derivative jump at $a$ gives
$$
\kappa(1+\tanh\kappa a)=\frac{2m\lambda}{\hbar^2}.
$$
The odd candidate is $\sinh(\kappa x)$ inside and $\operatorname{sgn}(x)\sinh(\kappa a)e^{-\kappa(|x|-a)}$ outside. Its jump gives
$$
\kappa(1+\coth\kappa a)=\frac{2m\lambda}{\hbar^2}.
$$
Parity ensures the jump at $-a$ as well. Put $t=\kappa a$ and $L=2m\lambda a/\hbar^2$. The continuous strictly increasing function $t(1+\tanh t)$ ranges from $0$ to infinity, so there is exactly one even <bound state> for every $L>0$. The continuous strictly increasing function $t(1+\coth t)$ ranges from $1$ to infinity, so the <odd bound state of a symmetric double-delta potential> exists if and only if $L>1$. Strict increase follows from the allowed monotonicity facts plus the strictly increasing term $t$. Therefore
$$
\boxed{\text{even: every }\lambda>0;\qquad\text{odd: }\lambda>\frac{\hbar^2}{2ma}},\qquad E=-\frac{\hbar^2\kappa^2}{2m}.
$$
At equality the odd limiting solution has $\kappa=0$ and is not square integrable, so it is not a <bound state>. No nonnegative energy can supply another full-line square-integrable state, since its exterior solutions are oscillatory or affine.
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