Inspired by [this](https://math.stackexchange.com/questions/716182/consecutive-quadratic-residues) recent question, I wondered if a similar result is true for quadratic non-residues, namely, if **it is true that for every $k \in \mathbb{N}$ there exists a prime $p$ such that exists $k$ consecutive quadratic non-residues mod $p$?** or even a stronger result like **for every $k \in \mathbb{N}$ there exists $N$ such that for every prime $p > N$ we have $k$ consecutive quadratic non-residues mod $p$?**


Obviously, the idea used in the link could not be adapted to this case because it relies on the fact that the product of two quadratic residues is a quadratic residue, which is not true for quadratic non-residues. I searched the web and did not find anything relevant on this question. Does anybody have any ideas?


Thank you!


PS: If anyone could give me a reference to a book that approaches this kind of problems (consecutive quadratic or non-quadratic residues), I would be grateful.



Consecutive non-quadratic residues
The answer to both your questions is positive and indeed, every given pattern of quadratic residues and non-residues of fixed length appears among consecutive elements of ${\mathbb F}\_p$, for all $p$ large enough; moreover, it appears about the expected number of times. This is non-trivial, but fairly standard. 


Fix $\epsilon=(\epsilon\_1,\ldots,\epsilon\_k)\in\{-1,1\}^k$ ("the pattern")
and for a prime $p$, let $N\_\epsilon(p)$ denote the number of those
$x\in{\mathbb F}\_p$ with
 $$ \left(\frac{x+1}p\right)=\epsilon\_1,\ldots, \left(\frac{x+k}p\right)=\epsilon\_k, $$
where $\left(\frac xp\right)$ is the [Legendre symbol](http://en.wikipedia.org/wiki/Legendre_symbol). Clearly, we have
\begin{align\*}
 N\_\epsilon(p) &= 2^{-k}\sum\_{x=0}^{p-k-1} \prod\_{j=1}^k \left(1+\epsilon\_j\left(\frac{x+j}{p}\right) \right) \\
 &= 2^{-k}\sum\_{x=0}^{p-1} \prod\_{j=1}^k \left(1+\epsilon\_j\left(\frac{x+j}{p}\right) \right) - \theta \frac k2,\quad 0\le \theta\le 1,
\end{align\*}
as the contribution of each $x\in[p-k,p-1]$ to the whole sum is at most $2^{k-1}$.


Expanding the product, one can write $N\_\epsilon(p)$ as a sum of the main term
$2^{-k}\sum\_x1=2^{-k}p$ and $2^k-1$ remainder terms, each of the form $2^{-k}\sum\_x \left(\frac{Q(x)}p\right)$ with a non-square polynomial $Q(x)$ of degree at most $k$. Now, [Weil's bound](http://www.math.ias.edu/~avi/PUBLICATIONS/MYPAPERS/CKW11/CKW11.pdf) implies that each of these remainder terms does not exceed $(k-1)\sqrt p$ in absolute value; as a result,
 $$ N\_\epsilon(p)=2^{-k}p+\theta k(\sqrt p+1/2),\ |\theta|<1, $$
which is certainly positive for $p$ sufficiently large (like $p>\exp(ck)$ with a suitable constant $c$).


This argument readily extends to count, say, the number of those elements $x$ of a finite field such that for a given system of square-free, pairwise co-prime polynomials $P\_1,\ldots,P\_k\in{\mathbb Z}[X]$, the values $P\_1(x),\ldots,P\_k(x)$ follow a prescribed quadratic residue / non-residue pattern. Indeed, in a similar way one can handle the joint distribution of $P\_1(x),\ldots P\_k(x)$ in the cosets of any subgroup of the multiplicative group of a finite field, not just the subgroup of quadratic residues.


