Solution
= Solution
Each nonzero <quadratic residue> is attained twice as $x^2$ for $1\le x\le p-1$. Therefore twice the sum of those <residues> is $\sum_{x=1}^{p-1}x^2$, which is zero modulo $p$ by (ii), since $p-1>2$. As two is invertible modulo $p$, the sum of the <residues> is zero. The sum of all nonzero <residues> is $p(p-1)/2\equiv0$, so subtracting gives the same result for the nonresidues. Hence $\boxed{\text{both indicated sums are }0\pmod p}$.