Exercise 3.3.15* (Consecutive quadratic residues)

Proof. Since $p\ge 7$, $\left(\frac{2}{p}\right)\in \{-1,1\}$ and $\left(\frac{5}{p}\right)\in \{-1,1\}$.

• If $\left(\frac{2}{p}\right)=1$, then for $n=1$, $\left(\frac{n}{p}\right)=\left(\frac{n+1}{p}\right)=1$.
• If $\left(\frac{5}{p}\right)=1$, then for $n=4$, $\left(\frac{n}{p}\right)=\left(\frac{n+1}{p}\right)=1$.

• If no one of these two conditions is satisfied, then $\left(\frac{2}{p}\right)=\left(\frac{5}{p}\right)=-1$. Then

  $$\left(\frac{10}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{5}{p}\right)=1.$$

  Therefore, for $n=9$, $\left(\frac{n}{p}\right)=\left(\frac{n+1}{p}\right)=1$.

So $\left(\frac{n}{p}\right)=\left(\frac{n+1}{p}\right)=1$ for at least one number $n$ in the set $\{1,2,\dots ,9\}$. □