For any prime p>5 proving the existence of consecutive quadratic residues and consecutive quadratic non residues This question was asked in an assignment which I am solving and I couldn't solve it.

Prove that for any prime $p>5$ there exist integers $1\leq a,b \leq  p-1$ for which $\binom{a}{p}=\binom{a+1}{p}=1$ and $\binom{b}{p}=\binom{b+1}{p} =-1$.

Quadratic residues and non-residues are equal in number , if I prove that any 2 residues are consecutive then non-residues will be automatically proved consecutive.
But I am unable to prove and 2 residues to be consecutive. For theory, I am studying David M Burton's book.
Can you please help?

Here is a rather interesting solution involving a bit of combinatorical tricks:

We wil start with some basic theory:
Lemma $(1)$:

For any odd prime $p$, there are exactly $\frac{p-1}{2}$ quadratic residues (greater than $0$) and exactly $\frac{p-1}{2}$ non-quadratic residues.

Lemma $(2)$:

For any integer $n$, $n^2$ is a quadratic residue$\pmod{p}$. in other words, $\big(\frac{n^2}{p}\big)=1$ (I used Legendre's symbol here)

Lets now start the proof!

Consider the residues$\pmod{p}$: $(1,2,...,p-1)$. Assign $1$ to a residue $k$ if it is a quadratic residue, and $0$ if it is not a quadratic residue. We will now get a new sequence of $1$s and $0$s out of the originl sequence.
As an example, for $p=7$ we would get $(1,2,3,4,5,6)\mapsto(1,1,0,1,0,0)$.
Observe that to prove your question, we must show that for any prime $p\geq 5$, in this sequence, there are $2$ consecutive $1$s and $2$ consecutive $0$s (at least).
Because we have a sequence of $p-1$ $1$s and $0$s and there are exactly $\frac{p-1}{2}$ $1$s and $\frac{p-1}{2}$ $0$s $($this is an implication of lemma $(1)$, because there are $\frac{p-1}{2}$ quadratic residues and $\frac{p-1}{2}$ non-quadratic residues$)$ we can easily see that if there are 2 consecutive $0$s, then there must be $2$ consecutive $1$s too.
Suppose for this prime $p$ there are no $2$ consecutive $1$s or $0$s. Then, using lemma $(2)$, because $1$ is a quadratic residue, our sequence should look like this: $(1,0,1,0,......,0)$, in other words all odd numbers from $1$ to $p-1$ are quadratic residues and all even numbers from $1$ to $p-1$ are non-quadratid residues. But using lemma $(2)$ again, we know $4$ is a quadratic residue, thus leding in a contradiction for $p\geq 5$ $(p$ must be $\geq 5$ so $4\leq p-1$).
So for $p\geq 5$ we will always find $2$ consecutive $1$s and $2$ consecutive $0$s.
