Q:

Proving Clausen hypergeometric identity

How do I show the Clausen identity
$$
{}_2F_1\left(a, b; a+b+\frac{1}{2}; z\right)^2 = {}_3F_2\left(2a, 2b, a+b; a+b+\frac{1}{2}, 2a+2b, z\right)? 
$$
I saw this on MathWorld but am unsure how to progress.  I tried the Cauchy product but that was no success.  I am thinking maybe Apell functions due to the order on the left hand side, but am not entirely sure.  A hint would be very much appreciated.  Thank you! 

A:

Let
$$y_1(z):={}_2F_1\left(a, b; a+b+\frac{1}{2}; z\right)^2$$
and
$$ y_2(z):= {}_3F_2\left(2a, 2b, a+b; a+b+\frac{1}{2}, 2a+2b, z\right) $$
You can verify that $y_1$ and $y_2$ both satisfy the following differential equation:
$$ z^2(z-1)y''' -3z \left (a+b+\frac{1}{2} - (a+b+1)z \right) y'' + ((2(a^2+b^2+4ab)+3(a+b)+1)z-(a+b)(2a+2b+1))y'+4ab(a+b)y=0 $$
Then, you just need to find some point $z_0$ for which:
$$ y_1(z_0)=y_2(z_0); \, y_1'(z_0)=y_2'(z_0) $$
It is well known that there exists a unique solution $y$ to a liner ODE which satisfies $y(z_0)=x_1$, $y'(z_0)=x_2$ for some $x_1$, $x_2$. Consequently, $y_1$ and $y_2$ must be the same, and Clausen's identity follows.