# Square of the hypergeometric function

Given the hypergeometric function $\,_2F_1$, Pochhammer symbol $(m)_n$, and $0<a< 1$, anybody knows how to prove that,

$\,_2F_1(a,1-a;1;z) = \sqrt{\sum_{n=0}^\infty \frac{(a)_n (1-a)_n (\tfrac{1}{2})_n}{n!^3} \,\big(4z(1-z)\big)^n}$

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The identity is usually written as $$\, _2F_1(1-a,a\,;\,1\,;\,z){}^2 = \sum_{n=0}^\infty \frac{(a)_n (1-a)_n (\tfrac{1}{2})_n}{n!^3} \,\big(4z(1-z)\big)^n = \, _3F_2\left(1-a,a,\tfrac{1}{2} \,;\,1,1\,;\, 4 (1-z) z\right)$$
This is a special case of a more general (two-parameter) identity, combined with this quadratic transformation: $${}_2F_1\left(1-a,a\,;\,1\,;\,z\right) = {}_2F_1\left(\frac{1-a}{2},\frac{a}{2}\,;\,1\,;\,4 (1-z)z\right)$$
Then one also needs to compare main terms at $z=1$ and at $z=\infty$ to make sure that not only do the rhs and lhs belong to the same fundamental solution, but are actually the same globally. This may not be a very illuminating proof, but a proof nonetheless.