The identity in my [recent answer](https://mathoverflow.net/q/334919) can be stated in a particularly neat form:
$${}\_2F\_0\left({-n, n+1\atop{}};\frac{x}{2}\right) ~\cdot~ {}\_2F\_0\left({-n, n+1\atop{}};-\frac{x}{2}\right) ~=~ {}\_3F\_0\left({\frac{1}{2}, -n, n+1\atop{}};x^2\right).$$
Is this by any chance a partial case of something more general and/or has a straightforward proof?



A hypergeometric identity related to Bessel functions
In ["Higher Trancendental Functions", Vol. 1](https://authors.library.caltech.edu/43491/1/Volume%201.pdf), by A. Erdelyi (ed.), on page 86, equation (4) says 
$$
\_2F\_0(α,β;z)\ \_2F\_0(α,β;-z) = \, \_4F\_1(α,β,\frac{1}{2}(α+β),\frac{1}{2}(α+β+1);α+β;4z^2),
$$
from which your formula can be derived by setting $\alpha=−n$, $\beta=n+1$.


Unfortunately Erdelyi, et al., do not give a proof only references, which I reproduce here:


[W.N. Bailey (1928), *Proc. London Math. Soc.* (2) 28, 242-254](https://doi.org/10.1112/plms/s2-28.1.242)


[Preece, C.T. (1924), *Proc. London Math. Soc.* (2) 22, 370-380](https://doi.org/10.1112/plms/s2-22.1.370),


but I cannot say more about these papers and their content.


