Existence of Consecutive Quadratic residues For any prime $p\gt 5$,prove that there are consecutive quadratic residues of $p$ and consecutive non-residues as well(excluding $0$).I know that there are equal number of quadratic residues and non-residues(if we exclude $0$), so if there are two consecutive quadratic residues, then certainly there are two consecutive non-residues,therefore, effectively i am seeking proof only for existence of consecutive quadratic residues. Thanks in advance.

The number of $k\in[0,p-1]$ such that $k$ and $k+1$ are both quadratic residues is equal to:
$$ \frac{1}{4}\sum_{k=0}^{p-1}\left(1+\left(\frac{k}{p}\right)\right)\left(1+\left(\frac{k+1}{p}\right)\right)+\frac{3+\left(\frac{-1}{p}\right)}{4}, $$
where the extra term is relative to the only $k=-1$ and $k=0$, in order to compensate the fact that the Legendre symbol $\left(\frac{0}{p}\right)$ is $0$, although $0$ is a quadratic residue. Since:
$$ \sum_{k=0}^{p-1}\left(\frac{k}{p}\right)=\sum_{k=0}^{p-1}\left(\frac{k+1}{p}\right)=0, $$
the number of consecutive quadratic residues is equal to 
$$ \frac{p+3+\left(\frac{-1}{p}\right)}{4}+\frac{1}{4}\sum_{k=0}^{p-1}\left(\frac{k(k+1)}{p}\right). $$
By the multiplicativity of the Legendre symbol, for $k\neq 0$ we have $\left(\frac{k}{p}\right)=\left(\frac{k^{-1}}{p}\right)$, so:
$$ \sum_{k=1}^{p-1}\left(\frac{k(k+1)}{p}\right) = \sum_{k=1}^{p-1}\left(\frac{1+k^{-1}}{p}\right)=\sum_{k=2}^{p}\left(\frac{k}{p}\right)=-1,$$
and we have $\frac{p+3}{4}$ consecutive quadratic residues if $p\equiv 1\pmod{4}$ and $\frac{p+1}{4}$ consecutive quadratic residues if $p\equiv -1\pmod{4}$.
