A hypergeometric identity related to Bessel functions The identity in my recent answer 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: The coefficient of $x^n$ in $_2F_0(\alpha,\beta;z) {}_2F_0(\alpha,\beta;-z)$, multiplied by $n!$,
is 
$$\begin{aligned}
\sum_{k=0}^n (-1)^{n-k}\binom nk &(\alpha)_k(\beta)_k (\alpha)_{n-k}(\beta)_{n-k}\hfill\\
&=(-1)^n (\alpha)_n(\beta)_n\,
  {}_3F_2\left({-n,\atop}\!{\alpha,\atop 1-\alpha -n,}\,
  {\beta\atop1-\beta-n}\biggm | 1\right).
\end{aligned}
$$ 
The right side can be evaluated by Dixon's theorem to give the identity cited by Johannes Trost.
A: In "Higher Trancendental Functions", Vol. 1, 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
Preece, C.T. (1924), Proc. London Math. Soc. (2) 22, 370-380,
but I cannot say more about these papers and their content.
A: $a(x) = {}_2F_0(-n,n+2; x/2)$ satisfies the differential equation $$ \left( -{n}^{2}-n \right) a \left( x \right) + \left( 2\,x-2
 \right) {\frac {\rm d}{{\rm d}x}}a \left( x \right) +{x}^{2}{\frac {
{\rm d}^{2}}{{\rm d}{x}^{2}}}a \left( x \right) 
=0$$
$b(x) = a(-x)$ satisfies 
$$ \left( -{n}^{2}-n \right) b \left( x \right) + \left( 2\,x+2
 \right) {\frac {\rm d}{{\rm d}x}}b \left( x \right) +{x}^{2}{\frac {
{\rm d}^{2}}{{\rm d}{x}^{2}}}b \left( x \right)=0$$
$c(x) = {}_3F_0(1/2,-n,n+1; x^2)$ satisfies
$$ 
  \left( -4\,{n}^{2}x-4\,nx \right) c \left( x \right) +
 \left( -4\,{n}^{2}{x}^{2}-4\,n{x}^{2}+6\,{x}^{2}-4 \right) {\frac 
{\rm d}{{\rm d}x}}c \left( x \right) +6\,{x}^{3}{\frac {{\rm d}^{2}}{
{\rm d}{x}^{2}}}c \left( x \right) +{x}^{4}{\frac {{\rm d}^{3}}{
{\rm d}{x}^{3}}}c \left( x \right)=0$$
Substitute $c(x) = a(x) b(x)$, use the differential equations for $a(x)$ and $b(x)$, and this simplifies to $0=0$.  That is, $c(x)$ and $a(x)b(x)$ satisfy the same
third order linear differential equation. There is only a one-dimensional family of analytic solutions of this equation ($0$ being an irregular singular point), and the fact that $a(0) b(0) = c(0) = 1$ completes the proof.   
