Quadratic Reciprocity: Determine if 11 is a quadratic residue $\mod p $for primes of the form: 44k+5?

Question

I'm currently studying for exams and this has me stuck. A sample question from a past paper states:

Use the quadratic reciprocity theorem to determine whether $11$ is a quadratic residue $\mod p$ for primes of the form:

(i) $44k+5$

(ii)$44k+7$ etc...

Neither my books or my notes have any specific proofs related to this sort of mod p, so I would be very grateful if someone could go through the proof for the first value 44k+5. Step by step is vital.

My own progress

I do not know if its correct, but so far I have tried using what i think is the right formula: First, get ${1\over 2(p-1)} = 22k+2$ Second, list of numbers making up $44k+4$. This is the part confusing me. How the hell do I list numbers from 1 to $44k+4$ with $22k+2$ as the midway point? The book somewhat skips over this, assuming you just know how.

I have tried to make this question as clear as possible, but it is based on my own (very bad!) understanding of the question. Please feel free to ask for clarification if needed and I will supply it.

Answer 1

Assuming that $44k+5$ is prime and observing that it is $\equiv1\bmod4$, $$ \left(\frac{11}{44k+5}\right)= \left(\frac{44k+5}{11}\right)=\left(\frac{5}{11}\right)=1 $$ since $5\equiv4^2\bmod11$.

Answer 2

Quadratic reciprocity states that if $p$ and $q$ are odd primes, then $\left(\frac{p}{q}\right)=\left(\frac{q}{p}\right)(-1)^{(p-1)(q-1)/4}$.

Now, let $p=11$ and $q=44k+5$ (where this is a prime number). We have

$$\left(\frac{11}{44k+5}\right)=\left(\frac{44k+5}{11}\right)(-1)^{5(22k+2)}=\left(\frac{5}{11}\right)=\left(\frac{11}{5}\right)=\left(\frac{1}{5}\right)=1.$$

Therefore, $11$ is a square modulo a prime congruent to $5$ mod $44$.

---

It is worth going back and seeing why we needed it to be $44$ in the question. When we applied quadratic reciprocity, we had two simplifications that were made. The first was that $44k+5 \equiv 5 \pmod{11}$. This allowed us to simplify $\left(\frac{44k+5}{11}\right)$. For this to happen, we needed that $44$ is divisible by $11$. The second simplification needed was to simplify $(-1)^{\frac{11-1}{2}\frac{(44k+5)-1}{2}}=(-1)^{\frac{(44k+5)-1}{2}}$. For this term to not depend on the choice of $k$, we need $44k+5 \pmod 4$ not to depend on $k$. The reason this happens is because $44$ is divisible by $4$.

In summary, the reason they chose $44$ is because for the answer to not depend on the choice of $k$, they needed a coefficient which was a multiple of both $4$ and $11$. The least common multiple of $4$ and $11$ is $44$.