= Solution
A <Papperitz symbol> records the three regular singular points of a second-order <Fuchsian differential equation> and the two local exponents at each. Exponents at infinity refer to solutions behaving as $z^{-\gamma}$, that is, to powers of the local coordinate $1/z$. Their sum satisfies the Fuchs relation: the six exponents add to one. The symbol represents the two-dimensional solution space, not one normalized solution by itself.
The specified exponents give the <Gauss hypergeometric equation>
$$
z(1-z)u''+[c-(a+b+1)z]u'-ab\,u=0.
$$
Set $d=c-a-b$ and let $v=F(c-a,c-b;c;z)$. Its equation is
$$
z(1-z)v''+[c-(2c-a-b+1)z]v'-(c-a)(c-b)v=0.
$$
For $u=(1-z)^dv$, differentiation gives $u'=(1-z)^d[v'-dv/(1-z)]$ and $u''=(1-z)^d[v''-2dv'/(1-z)+d(d-1)v/(1-z)^2]$. Substitution into the first equation leaves $(1-z)^d$ times exactly the second equation: its $v'$ coefficient is $c-(a+b+1+2d)z$, and its $v$ coefficient simplifies to $-(c-a)(c-b)$. Hence $u$ solves the same equation as $F(a,b;c;z)$.
The symbol gives the same check geometrically: multiplication by $(1-z)^d$ shifts the exponents at one from $\{0,-d\}$ to $\{d,0\}$ and those at infinity from $\{c-a,c-b\}$ to $\{b,a\}$; it does not change the exponents at zero. Choose the branch with $(1-0)^d=1$. Then $u$ is analytic at zero and $u(0)=1$. For $c\notin\{0,-1,-2,\ldots\}$ the analytic recurrence uniquely determines this normalized solution, proving
$$
\boxed{F(a,b;c;z)=(1-z)^{c-a-b}F(c-a,c-b;c;z).}
$$
The identity continues wherever both sides are defined with compatible branches. At exceptional parameters where the stated normalization does not uniquely define an analytic function, a parameter-continuation convention is needed; the generic proof does not silently assume uniqueness there. This is <Euler's hypergeometric transformation>.
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