Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 4 / 7E / b / ii

The hypergeometric exponent differences give $1-c=m$, $c-a-b=-m$, and exponents $a,b$ at infinity. Hence one may takeso a solution is ${}_2{F}_1(n,1-n;1-m;(1-z)/2)$, with the prefactor from part (i) when returning to the original dependent variable.