Consecutive quadratic residues of primes that differ by 2 Show that if $p$ is prime and $p \ge 7$, then there are always two consecutive quadratic residues of $p$ that differ by 2.
I think that I am supposed to use the fact that at least one of $2, 5$ and $10$ is always a quadratic residue however I'm not sure exactly how to link this to the question.
 A: We will show for any prime $p\ge 7$ there are two consecutive positive quadratic residues  of $p$ that differ by $2$. The result is easy to verify for the prime $7$. So we suppose that $p\gt 7$. 
For brevity we write QR for quadratic residue  of $p$, and NR for quadratic non-residue  of $p$.
Let $q$ be the smallest positive NR of $p$. Note that $q$ is prime. If $q$ is odd, then $q-1$ and $q+1$ are consecutive QR that differ by $2$, since all their prime factors are $\lt q$.
So we are finished unless $2$ is a NR. Assume $2$ is a NR. If $3$ is  a QR, we are finished. So we may assume that $3$ is  a NR.  
Then $6$ is a QR, and therefore we are finished unless $5$ is a QR. If $7$ is a QR, we are finished, for $8$ is NR and $9$ is QR. 
So we have solved our problem unless $2$ is NR, $3$ is NR, $5$ is QR, and $7$ is NR.
But in that case $14$ is QR, $15$ is NR, and $16$ is QR. The result follows. 
A: Since $p$ is prime and $p \geq 7$, 1 and 4 must be quadratic residues of $p$, because $(\frac{2^2}{p}) = (\frac{1}{p})= 1$, where $()$ is Legendre symbol.
Let us assume that for all $p \geq 7$, it does not have two consecutive quadratic residues $a$ and $b$ such that $|a - b| = 2$. From the conclusion above, we know that 2, 3 and 6 cannot be quadratic residues of $p$, since $4 - 2 = 2, \space 6 - 4 = 2$ and $3 - 1 = 2$,  i.e., $(\frac{2}{p}) = (\frac{3}{p}) = (\frac{6}{p}) = -1$. Apparently we will soon get a contradiction: $(\frac{6}{p}) = (\frac{2}{p}) \cdot (\frac{3}{p}) = 1$, but $(\frac{6}{p})$ cannot be 1 and -1 simultaneously.
Q.E.D.
