Solution
= 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 $1\leq m\leq p-2$ as claimed. The congruence $mn=1$ gives
$$
n(n+1)=n^2(1+m)\pmod p.
$$
The <Legendre symbol> of the nonzero square $n^2$ is one. Using <multiplicativity of the Legendre symbol> therefore gives
$$
\boxed{\left(\frac{n(n+1)}p\right)=\left(\frac{1+m}p\right).}
$$