$p = x^2 + y^2$ with $p$ prime and $y$ even, show that $x$ and $y/2$ are quadratic residues mod $p$ 
Let $p$ be a prime which can be written as $p = x^2 + y^2$ for positive integers $x$ and $y$. Also assume that $y$ is an even integer.
Show that $x$ and $y/2$ are quadratic residues mod $p$.

My attempt:
The case $p = 2$ is trivial.
For $p > 2$ we know that $p \equiv 1$ mod $4$. The integer $y$ is even. Since $p = x^2+y^2$ and $y^2 \equiv 0 $ mod $4$, we have that $x^2 \equiv 1$ mod $4$.
From this point I would like to use the legendre symbol $\left(\frac{x}{p}\right)$ to get further, but I don't get anywhere. Any tips?
 A: Use properties of the Jacobi symbol.
Firstly, note that if $p>2$, then $p\equiv 1\pmod 4$. Since $x$ and $p$ are odd we have
$$
\left(\frac{x}{p}\right)=(-1)^{(x-1)(p-1)/4}\left(\frac{p}{x}\right)=\left(\frac{p}{x}\right)=\left(\frac{x^2+y^2}{x}\right)=\left(\frac{y^2}{x}\right)=1,
$$
so $x$ is a quadratic residue.
For $y$ we consider two cases: $y\equiv 2\pmod 4$ (meaning that $p\equiv 5\pmod 8$) or $y\equiv 0\pmod 4$ (meaning that $p\equiv 1\pmod 8$).
In the first case we can just do the same
$$
\left(\frac{y/2}{p}\right)=(-1)^{(y/2-1)(p-1)/4}\left(\frac{p}{y/2}\right)=\left(\frac{p}{y/2}\right)=\left(\frac{x^2+y^2}{y/2}\right)=\left(\frac{x^2}{y/2}\right)=1.
$$
For the second case let $y=2^kz$ where $z$ is odd and $k\ge 2$. Since $p\equiv 1\pmod 8$ we have $\left(\frac{2}{p}\right)=1$, so the previous argument can be modified in the following way:
$$
\left(\frac{y/2}{p}\right)=\left(\frac{2^{k-1}z}{p}\right)=\left(\frac{z}{p}\right)=(-1)^{(z-1)(p-1)/4}\left(\frac{p}{z}\right)=\left(\frac{p}{z}\right)=\left(\frac{x^2+y^2}{z}\right)=\left(\frac{x^2}{z}\right)=1.
$$
