 Solution

In the multiplicative group ${\mathbb{F}}_p^{\times }$, the squaring map is a group homomorphism with kernel $\{1,-1\}$, since $p$ is odd. Every nonzero square therefore has exactly two square roots. There are $(p-1)/2$ quadratic residues and the same number of nonresidues, and consequently ${\sum }_{a=1}^{p-1}\left(\frac{a}{p}\right)=0$.

For $n{m}_n\equiv 1(\mathrm{mod}p)$, direct multiplication gives $n^2(1+m_n)\equiv n^2+n=n(n+1)$. Multiplicativity of the Legendre symbol, including its value zero on multiples of $p$, impliesInversion permutes ${\mathbb{F}}_p^{\times }$, so $1+m_n$ runs through all residue classes except $1$. Thus