Ramification and Quadratic Reciprocity Law

I have a question regarding the follow problem:

Show that the prime number 27644437 splits completely in $L = \mathbb{Q}(\sqrt{55})$.

From what I understand. This deals with ramification.

\begin{eqnarray} \sum_1^{r}e_if_i = n &, &\text{where }n = [L:K] \end{eqnarray}

For our prime number, $p$, to split completely into $L$, then $e_i=f_i =1$ for all $i$. Now in order for it to be completely split, it can not be ramified. Being ramified will require at least one of the $e_i>1$. However, we do not need to worry about this situation in this problem because $p\not| \text{disc}(L)$.

Minimal Polynomial of $L$ is $x^2-55$. Since it is not ramified, this will only leave two possibilities: inert or split. Since we are dealing with a second degree polynomial and we already ruled out the possibility of being ramified, then it must be completely split. Thus, inert or completely split.

This is where it gets complicated. If I want to show it is completely split I need to find at least one integer such that $x^2 -55 (\text{mod}p) = 0$.

I do not need to find the second integer because $p$ does not ramify $L$. Looking over so many numbers just to verify it meets this condition is ridiculous. There has to be another approach.

I asked my professor and he gave me a hint suggesting I use the Quadratic Reciprocity Law. However, I do not know how to apply it correctly in this case. I would assume looking at the Quadratic Residue symbol formed by the discriminate of $L$ and the prime number $p$. This will either result in an answer of $0, 1, \text{or, } -1$; however, I am not sure what the value would mean.

Thank You for your time, and thank you in advance for any feedback.

$p$ splits completely iff $55$ is a quadratic residue mod $p$. Now use quadratic reciprocity.

Thus we want to find

$$\left(\frac{55}{27644437}\right)$$

Now $27644437\equiv 1 \pmod 4$ so we have $$\left(\frac{55}{27644437}\right)= \left(\frac{27644437}{55}\right)= \left(\frac{7}{55}\right)=-\left(\frac{55}{7}\right)=$$ $$-\left(\frac{6}{7}\right) =-\left(\frac{2}{7}\right)\left(\frac{3}{7}\right) =\left(\frac{2}{7}\right)\left(\frac{7}{3}\right) =\left(\frac{2}{7}\right)\left(\frac{1}{3}\right)=\left(\frac{2}{7}\right)=1$$

So its a quadratic residue and it splits completely.

For what it may be worth, mostly because $p \equiv 1 \pmod 4,$ $p \equiv 2 \pmod 5,$ $p \equiv 7 \pmod {11},$there really is an expression $$p = 2 x^2 + 2 x y - 27 y^2$$ in integers, or $$2p = u^2 - 55 v^2.$$ Yep, $$\color{red}{ x=23908, \; \; \; y=-5603 }$$

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