 Solution

Since $n$ is nonzero modulo $p$, it has a unique multiplicative inverse $m$ in the nonzero residues. The inverse cannot be $-1$ because that would imply $n=-1$, excluded by the range. Thus $1\le m\le p-2$ as claimed. The congruence $mn=1$ givesThe Legendre symbol of the nonzero square $n^2$ is one. Using multiplicativity of the Legendre symbol therefore gives