= Solution
Put $W=1+\rho'^2+\rho^2\phi'^2$. The Euler–Lagrange equations are
$$
\frac d{dz}\left(\frac{n\rho'}{\sqrt W}\right)-n_\rho\sqrt W-\frac{n\rho\phi'^2}{\sqrt W}=0,
$$
$$
\frac d{dz}\left(\frac{n\rho^2\phi'}{\sqrt W}\right)-n_\phi\sqrt W=0.
$$
When $n=n(\rho)$, the helical ansatz $\rho=R$, $\phi=\phi_0+\omega z$ automatically satisfies the second equation. The first becomes
$$
n(R)R\omega^2+(1+R^2\omega^2)n'(R)=0,
$$
so $a=1$.
The endpoints have the same cylindrical angle exactly when $\omega L=2\pi k$ for some integer $k$. Thus a positive $L$ with $L=2\pi k/\omega(R)$ admits the two special solutions of opposite handedness, using $\omega$ and $-\omega$.
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