Consider the following Sturm–Liouville (SL) eigenvalue problem with $x\in(-\infty,0)$ $$(py')'+qy=-\lambda^2wy,$$ in which $p(x)=x^2$, $w(x)=1$, and $q(x)=-[(x/2+a)^2+a]$ with parameter $a>0$. It has a regular singularity $x=0$. We basically hope for something like homogeneous Dirichlet b.c.

It is solved by making the substitution $y(x)=e^{x/2}x^{-\frac{1}{2}+\sqrt{(a+\frac{1}{2})^2-\lambda^2}}u(x)$, leading directly to a standard *confluent hypergeometric equation* $$xu''(x)+(\gamma-x)u'(x)-\alpha u(x)=0,$$ in which $\alpha=\sqrt{(a+\frac{1}{2})^2-\lambda^2}-a+\frac{1}{2}$ and $\gamma=1+2\sqrt{(a+\frac{1}{2})^2-\lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.  
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $\infty$, the 1st kind is reduced to a polynomial when $-\alpha$ is a non-negative integer and eigenvalue $\lambda^2$ is attained. 

However, seen from this condition for $\alpha$, obviously **we only have a bounded and finite sequence of eigenvalues, which seems different from the infinite eigenspectrum that Sturm–Liouville theory claims.** 

What is wrong here?