How to demonstrate that an elliptical circulation distribution can be induced by an elliptical chord distribution? I intend to demonstrate that a wing with an elliptical chord distribution, without torsion ($\alpha(z)=\alpha=\text{const}$) and constant section ($\beta(z)=\beta=\text{const}$, and $a_{2\text{D}}(z)=a_{2\text{D}}=\text{const}$), can induce an elliptical circulation distribution.
Well, for an elliptical chord distribution, the chord is defined by:
$$c(z)=c_C\sqrt{1-\left(\frac{z}{s}\right)^2}$$
where $c_C$ is a constant.
On the other hand, the circulation distribution is defined by:
$$\Gamma(z)=\frac{1}{2}a_{2\text{D}}V_\infty c(z)\left[\alpha+\beta+\alpha_i(z)\right]$$
So, by substituting $c(z)$ on $\Gamma(z)$, we get:
$$\Gamma(z)=\frac{1}{2}a_{2\text{D}}V_\infty \left[c_C\sqrt{1-\left(\frac{z}{s}\right)^2}\right]\left[\alpha+\beta+\alpha_i(z)\right]$$
which can be written as:
$$\Gamma(z)=\Gamma_C(z)\sqrt{1-\left(\frac{z}{s}\right)^2}$$
where $\Gamma_C(z)=\frac{1}{2}a_{2\text{D}}V_\infty c_C\left[\alpha+\beta+\alpha_i(z)\right]$
So, to demonstrate that an elliptical chord distribution can induce an elliptical circulation distribution, we would also need to demonstrate  that $\alpha_i(z)=\alpha_i=\text{const}$, to make $\Gamma_C$ a constant.
I don't know how to do this last step. How would you do it?
Note that:
$s:=$ semispan of the wing,
$z:=$ transversal coordinate of the wing, it goes from $-s$ to $s$,
$\alpha:=$ geometric angle of attack,
$\alpha_i:=$ induced angle of attack,
$\beta:=$ symmetric of the null lift (geometric) angle of attack,
$a_{2\text{D}}:=$ airfoil lift coefficient slope,
$V_\infty:=$ freestream airspeed.
First, let's transform the coordinates to polar coordinates for simplicity by introducing a variable $\theta$ such that:
$$z=-s\cos{\theta}$$
In this coordinate, the chord distribution is then:
$$c(\theta)=c_C\sin{\theta}$$
Let's write the unknown lift distribution of the bound vortex as a Fourier sine series, again in the $\theta$ coordinate:
$$\Gamma(\theta) = \sum_{n=1}^{\infty}A_n\sin{n\theta}$$
Through Biot-Savart, the induced AOA can be computed from the trailing vortices, such that:
$$\alpha_i(z) = \frac{1}{4\pi V_\infty}\int_{-s}^{s}{\frac{(d\Gamma/d\zeta)}{z-\zeta}d\zeta} = \frac{1}{4\pi V_\infty}\int_{0}^{\pi}{\frac{\sum_{n=1}^{\infty}{nA_n\cos{n\theta_0}}}{\cos{\theta_0}-\cos{\theta}}d\theta_0} = \frac{1}{4V_\infty}\sum_{n=1}^{\infty}{nA_n\frac{\sin{n\theta}}{\sin{\theta}}}$$
Now collecting everything together and put into your original equation, we have:
$$\Gamma(\theta) = \frac{1}{2}a_{2D}V_{\infty}c_C\sin{\theta} \left[ \alpha - \beta - \frac{1}{4V_\infty}\sum_{n=1}^{\infty}{nA_n\frac{\sin{n\theta}}{\sin{\theta}}} \right] = \sum_{n=1}^{\infty}A_n\sin{n\theta}$$
The key, at this point, is to realize that each $\frac{\sin{n\theta}}{\sin{\theta}}$ is linearly independent to the others, so in order for the other terms to satisfy the above relation for all $\theta$, they must be simultaneously zero. That is,
$$A_n = \begin{cases}
      A_1 &amp; n=1\\
      0 &amp; n&gt;1\\
    \end{cases}$$
So now the circulation and chord distributions have the same form, which are elliptical.