View source for 2F0(a,b;;z)2F0(a,b;;-z)=4F1(a,b,a/2+b/2,a/2+b/2+1/2;a+b;4z^2)

==Theorem==
The following formula holds:
$${}_2F_0(a,b;;z){}_2F_0(a,b;;-z)={}_4F_1 \left(a, b, \dfrac{a}{2} + \dfrac{b}{2}, \dfrac{a}{2} + \dfrac{b}{2} + \dfrac{1}{2}; a+b; 4z^2 \right),$$
where ${}_2F_0$ denotes [[hypergeometric 2F0]] and ${}_4F_1$ denotes [[hypergeometric 4F1]].
==Proof==

==References==
* {{BookReference|Higher Transcendental Functions Volume I|1953|Harry Bateman|prev=0F1(;r;z)0F1(;r;-z)=0F3(r,r/2,r/2+1/2;-z^2/4)|next=findme}}: $4.2 (4)$ 

[[Category:Theorem]]
[[Category:Unproven]]