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?

Too long to write in the comment section.
Alternative proofs for CPCH's Case #2 where both $2$ and $3$ are nonresidues and $n\geqslant 11$.
Method A: We already have two consecutive nonresidues. Suppose we don't have two consecutive quadratic residues, then from $5$ to $p-1$ there are equal numbers of residues/nonresidues; and $(4/p)=1 \implies (5/p)=-1$, they must alternate with $-1$ and $1$, therefore $(9/p)=-1$, a contradiction.
Method B: $(2/p)(5/p)=(10/p)$, so at least one of them is equal to $1$, therefore, either $(1,2)$, or $(3,4)$, or $(9,10)$ are consecutive residues.
