 Solution

Write $\chi (x)=(x/p)$. Choose a quadratic nonresidue $a$. Multiplication by $a$ permutes ${\mathbb{F}}_p$, while multiplicativity of the Legendre symbol gives $\chi (ax)=-\chi (x)$. Thereforeso the complete Legendre-symbol sum is zero.

Since $4x(x+1)=(2x+1{)}^2-1$ and $\chi (4)=1$, the bijection $x\mapsto 2x+1$ reduces the next sum to ${\sum }_y\chi (y^2-1)$. Count pairs $(y,t)\in {\mathbb{F}}_p^2$ satisfying $t^2=y^2-1$. On one hand their number isOn the other hand $(y-t)(y+t)=1$, and every $u\in {\mathbb{F}}_p^{\times }$ gives exactly one pair by $y-t=u$, $y+t=u^{-1}$. There are therefore $p-1$ pairs, proving the quadratic Legendre-symbol correlation

For $1\le z\le p-2$, the indicator that both $z$ and $z+1$ are quadratic residues is $(1+\chi (z))(1+\chi (z+1))/4$. Summing and using the two identities above gives the number of consecutive nonzero quadratic residues modulo an odd prime