 Solution

SetThis Möbius change sends $z=1,-1,\infty$ to $u=0,1,\infty$, respectively, without changing the corresponding exponents. The hypergeometric P-symbol has exponentsMatching it with $(0,1/2)$ at both finite singularities and $(-n,n)$ at infinity givesThe exponent-zero solution normalized to one at $u=0$ is therefore

The second standard local solution at $u=0$ isAfter absorbing the constant $2^{-1/2}$ into its normalization, this becomesTheir exponents at $z=1$ are $0$ and $1/2$, so they are linearly independent.