Proving the sum of two independent Cauchy Random Variables is Cauchy

We may exploit the Lagrange identity: $$ (a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2 \tag{1}$$ to state: $$ I_z=\int_{-\infty}^{+\infty}\frac{dx}{(1+x^2)(1+(z-x)^2)}=\int_{-\infty}^{+\infty}\frac{dx}{(1+x(z-x))^2+(z-2x)^2}\tag{2}$$ and by replacing $x$ with $x+\frac{z}{2}$ in the last integral, we get: $$ I_z = \int_{-\infty}^{+\infty}\frac{dx}{(1+\frac{z^2}{4}-x^2)^2+4x^2}\tag{3}$$ hence $I_z$ just depends on $\left(1+\frac{z^2}{4}\right)$. For the sake of brevity, let: $$ J(m)=\int_{0}^{+\infty}\frac{dx}{(x^2-m)^2+4x^2}\tag{4} $$ for any $m\geq 1$. With the change of variable $x-\frac{m}{x}=u$ we have: $$ J(m) = \int_{-\infty}^{+\infty}\frac{1-\frac{u}{\sqrt{4m+u^2}}}{8m+2m \,u^2}\,du = \frac{1}{2m}\int_{-\infty}^{+\infty}\frac{du}{4+u^2}=\frac{\pi}{4m}\tag{5}$$ from which it follows that:

$$ I_z = \frac{\pi}{2+\frac{z^2}{2}}=\frac{2\pi}{4+z^2}.\tag{6}$$

The interesting thing is that this proof is just a variation of the proof of the relation between the arithmetic-geometric mean (AGM) and the complete elliptic integral of the first kind ($K(k)$).

So after no satisfactory answer to this question, here I am posting the ultimate hint which I found after a long hard search and from which the problem becomes immediately obvious.

Decompose $\dfrac{1}{(1+x^2)(1+(z-x)^2)}=\dfrac{1}{z^2(z^2+4)}\big[\dfrac{2zx}{1+x^2}+\dfrac{z^2}{1+x^2}+\dfrac{2z^2-2zx}{1+(z-x)^2}+\dfrac{z^2}{1+(z-x)^2}\big]$

I post this keeping in mind that there must be an online record which I may also use for my personal computations at a later stage.