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

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