Associated Legendre Integer Giving Full Asimuthal Range and Physical Interpretation of $l(l+1)$ In separating the Laplacian in spherical coordinates, one arrives at the associated Legendre equation (ALE).
Write the ALE as 
$$\frac{1}{\sin \theta} \frac{\partial }{\partial \theta}(\sin \theta \frac{\partial \phi }{\partial \theta}) + (k_{\theta}^2 - \frac{k_{\phi}^2}{\sin ^2 \theta})\phi = 0,$$ 
involving two separation constants.
The physical reason for defining $k_{\phi}^2 = m$ (an integer) seems to be to allow for the full azimuthal range, which apparently follows since $e^{i k_{\phi} \theta}$ is a solution to the separation equation for $\phi$ and because of this it must be an integer.
I don't really understand this point though, would someone mind explaining this? 
More importantly, what is the physical reason for defining $k_{\theta}^2 = l(l+1)$?
The issue is that I have a different way of arriving at the separation equations for the Laplacian, and judging from Jackson's derivation it falls out of slavishly going through the computations of separating each variable one at a time. It doesn't look necessary to me from my perspective, thus I'd just like a reason to randomly define $k_{\theta}^2 = l(l+1)$, and since an integral solution can be found for the associated-Legendre equation that doesn't require $l(l+1)$ for truncation of polynomial solutions I think there should be a purely physical, non-mathematical, reason for the $l(l+1)$, preferably both a classical and a quantum physical reason - thank you.
 A: Why $k^2_\phi = m^2$ ?
Assuming that our potential takes the form $V(r,\theta,\phi)=R(r)\Phi(\phi)\Theta(\theta)$, then we have:
$$\Phi''(\phi)=-k^2_\phi \Phi(\phi).$$
This equation has exponential solutions if $k^2_\phi<0$. In this case, it is impossible to match the continuity condition $V(\phi=0)=V(\phi=2\pi)$. Therefore, we  find that the only possible solutions are those with $k_\phi =m\in \mathbb{Z}$. This is the only case where we have $2\pi$-periodic functions.

Why $k^2_\theta=l(l+1)$ ? 
As far as I know, there is no deep physical reason for this choice. The important point is that any positive number can be written as $l(l+1)$ and with this definition the Legendre equation with solution $P_l^m(\cos\theta)$ can be easily identified. This gives rise to the spherical harmonics when you also consider the function depending in $\phi$.
The good question to ask is "why this parameter should be positive?". The answer is that if it wasn't, then the solution would not be regular at the poles $\theta=0,\pi$, which is completely nonphysical. Moreover, a similar reasoning explains why we must have $l>m^2$.
A: Since you're solving the Laplace equation in spherical coordinates, $V(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)$, $0\leq\theta\leq \pi$, $0\leq \phi\leq 2\pi$ and $\Phi''(\phi) = -m^2 \Phi(\phi)$, the coordinate values $\phi=2\pi n$, $n\in \mathbb{Z}$ represent the same point, therefore we must have $V(r,\theta,\phi=0) = V(r,\theta,\phi=2\pi n)$ $\forall n\in \mathbb{Z}$, which can only be true for $m \in \mathbb{Z}$ ($n$ is an integer and so should be $nm$, therefore $m$ must be an integer, too). 
The radial part is
$$\frac{1}{R}\frac{d}{dr}\left( r^2 \frac{ dR}{dr} \right) = k.$$ This becomes $r^2 R''(r)+ 2r R(r)-kR=0$, the Euler equation with solutions of the form $R\propto r^a$. If we substitute this, we obtain an equation quadratic in $a$, which yields $a = (-1 \pm \sqrt{ 1 + 4k})/2$. We define the one solution as $l\equiv (-1 + \sqrt{1 + 4k})/2$, therefore the other is $-1-l$, $k = l(l+1)$, hence the diff. eq. has solution $R(r) = Ar^l + \frac{B}{r^{1+l}}$, and the Laplace becomes: $$\frac{1}{\sin \theta} \frac{d}{d\theta} \left( \sin \theta \frac{d\Theta}{d\theta} \right)+ \left[ l(l+1) - \frac{m^2}{\sin^2\theta} \right] \Theta =0.$$ We substitute $x \equiv \cos \theta$, hence $$\frac{d}{dx} \left[ ( 1-x^2) \frac{d \Theta}{d \theta} \right] + \left[ l(l+1) - \frac{m^2}{1- x^2} \right] \Theta = 0,$$ the associated Legendre equation. For $m=0$ the solution that does not diverge at $x=\pm 1$ is the Legendre polynomials, $P_l(x)$ for $l=0,1,2,\ldots$; for other values of $m$ we use the Rodrigues-like formula $$P^m_l(x) = (-1)^m (1-x^2)^{m/2} \frac{d^m}{dx^m} P_l(x),$$ from which we see that $l=0,1,2,\ldots$ for every value of $m$. For $0\leq m \leq l$, the solution of the angular part comprises the spherical harmonics, $Y_{lm}(\theta,\phi)\propto e^{im\phi} P_{l}^m(\cos\theta)$, with $P_l^m$ the associated Legendre functions.
