How do I show the [Clausen identity](https://en.wikipedia.org/wiki/Clausen%27s_formula)
$$
{}\_2F\_1\left(a, b; a+b+\frac{1}{2}; z\right)^2 = {}\_3F\_2\left(2a, 2b, a+b; a+b+\frac{1}{2}, 2a+2b, z\right)?
$$
I saw this on [MathWorld](https://mathworld.wolfram.com/ClausensProductIdentity.html) 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!



Proving Clausen hypergeometric identity
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.


