Find all odd primes $\displaystyle p$ for which $\displaystyle \left(\frac{11}{p}\right) = 1$. $\displaystyle \left(\frac{11}{p}\right)$ is in Legendre Symbol.

$\displaystyle \left(\frac{11}{p}\right) = - \left( \frac{p}{11} \right)$ due to quadratic reciprocity. ($\displaystyle 11 \equiv 3 \pmod{4}$)

Since $\displaystyle 11$ is prime and $\displaystyle p$ is prime, we have $\displaystyle gcd(11, p) = 1$.

Apply Gauss' theorem.

$\displaystyle \frac{11 - 1}{2} = 5$.
Let $\displaystyle S = \{p, 2p, 3p, 4p, 5p \}$.
Let $\displaystyle k = \{ s | s > 5 \}$
Then $\displaystyle \left(\frac{p}{11}\right) = (-1)^k = -1$ iff $\displaystyle k$ is odd.

When $\displaystyle p = 3$, $\displaystyle k = 4$.
When $\displaystyle p = 5$, $\displaystyle k = 4$.
For all odd primes $\displaystyle p > 5$, $\displaystyle k = 5$.

So all odd primes $\displaystyle p > 5$ satisfy $\displaystyle \left(\frac{11}{p}\right) = 1$.

Is my attempt correct?