Existence of an integer $i$ such that $i$ and $i+1$ are both squares mod $p$

Question

Let $p$ be a prime $\ne 2,5$

Prove $\exists i\le 9$ such that $(\frac{i}{p})=(\frac{i+1}{p})=1$ where $(.)$ is the Legendre symbol.

I tried to use the Euler formula $(\frac{a}{p})\equiv a^{\frac{p-1}{2}} \mod p$ but it leads nowhere.

Thank you for any hints.

Edit

We should also have $p\ne 3$ for the statement above to be correct as 2 is not a quadratic residue modulo $3$.

Answer 1

Note: I assume you intend $i\in \{1,2,\cdots, 9\}$, else $i=0$ is a cheap example.

Hint: argue that at least one of $2,5,10$ is a quadratic residue $\pmod p$.