Let $p>11$ be a prime. Then there exists integer $a$ such that $a, a+1$ are quadratic residues modulo $p.$
May I know if my proof is correct? Please advise, thank you.
There are $\dfrac{p-1}{2}$ distinct quadratic residues, three of which are $1,4$ and $9.$ Suppose there is no such integer $a.$
If one quadratic residue is between $4$ and $9,$ the remaining $\dfrac{p-1}{2}-4$ quadratic residues are more than $9.$ It follows that the largest among the remaining $\dfrac{p-1}{2}-4$ quadratic residues is at least $9 + 2\times \Big(\dfrac{p-1}{2}-4\Big)=p \leq p-1.$(Contradiction)
If all the quadratic residues, except $1,4$ and $9$ are more than $9,$ we argue similarly and arrive at the contradiction: $p+2 \leq p-1.$