# (c-a-b)2F1+a(1-z)2F1(a+1)-(c-b)2F1(b-1)=0

The following formula holds: $$(c-a-b){}_2F_1(a,b;c;z)+a(1-z){}_2F_1(a+1,b;c;z)-(c-b){}_2F_1(a,b-1;c;z)=0,$$ where ${}_2F_1$ denotes hypergeometric 2F1.