Does $p=x^2+4y^2$ imply that $x$ is a quadratic residue mod $p$?

Question

Does $p=x^2+4y^2$ imply that $x$ is a quadratic residue mod $p$?

I'm stuck on this problem. My attempt:

We know that since $p$ is a sum of squares $p\equiv 1 (4)$. This means that $\left(\frac{-1}{p}\right)=1$. Hence we have $$ x^2\equiv -4y^2\equiv (k2y)^2 (p) $$ for some $k$ with $k^2\equiv-1(p)$. And hence $$ x\equiv \pm (k2y) (p) $$ So we can write $$ \left(\frac{x}{p}\right)=\left(\frac{k}{p}\right)\left(\frac{2}{p}\right)\left(\frac{y}{p}\right) $$ And since $$ \left(\frac{2}{p}\right)=\left(\frac{k}{p}\right)=\begin{cases} 1&\text{ if }p\equiv 1(8)\\ -1&\text{ if }p\equiv 5(8)\\ \end{cases} $$ We can reduce to $$ \left(\frac{x}{p}\right)=\left(\frac{y}{p}\right) $$ i.e. either both $x$ and $y$ are quadratic residues or neither is. Which might be interesting or it might not be; I'm stuck here.

Any input would be appreciated, either telling me where to go from here, or using completely different arguments.

Answer 1

You already proved that $p=x^2+4y^2$ implies $p\equiv 1\pmod{4}$.

Then, by quadratic reciprocity: $$ \left(\frac{x}{p}\right)=\left(\frac{x}{x^2+4y^2}\right)=\left(\frac{x^2+4y^2}{x}\right)=\left(\frac{4y^2}{x}\right) = 1 $$ since $4y^2 = (2y)^2$.

Answer 2

We work with Legendre symbols only. Note that $x$ is odd. Then $x$ can be expressed as $$x=(\pm 1)(q_1q_2\cdots q_k),\tag{1}$$ where the $q_i$ are odd primes. We have $(-1/p)=1$. Also, for any $i$, we have $(q_i/p)=(p/q_i)$. But since $q_i$ divides $x$, it follows that $(p/q_i)=((x^2+4y^2)/q_i)=(4y^2/q_i)=1$.

Now we can use the representation (1) to conclude that $(x/p)=1$.