Show that if $p\equiv 3 \pmod 4$ is a prime, exactly one between $2$ and $-2$ is a quadratic residue modulo $p$.
The "most obvious" solution is the following: since $\displaystyle \left(\frac{-1}{p} \right)=-1$ and Legendre's symbol is multiplicative, we conclude that exactly one between $\displaystyle \left(\frac{-2}{p} \right)$ and $\displaystyle \left(\frac{2}{p} \right)$ is $1$.
A friend of mine told me that we can show this fact also considering the kernel of the homomorphism $\varphi: (\mathbb{Z}/p\mathbb{Z})^*\to (\mathbb{Z}/p\mathbb{Z})^*$ such that $\varphi (x)=x^2$ (or something like that...). It seems very interesting but I can't figure it out how this may be helpful to the solution. Do you have any idea?
Suppose $\;\left(\frac2p\right)=1\;$ :
$$\phi(-2)=\phi(-1)\phi(2)=(-1)\cdot \phi(2)=-1\implies\left(\frac{-2}{\;\;p}\right)\neq1$$
and now suppose $\;\left(\frac{-2}{\;\;p}\right)=1\;$ :
$$\phi(2)=\phi((-1)(-2))=\phi(-1)\phi(-2)=(-1)\cdot \phi(-2)=-1\implies\left(\frac2p\right)\neq1$$
In both cases above we use, of course, that
$$\left(\frac{-1}{\;\;p}\right)=1\iff p=1\pmod 4\implies\left(\frac{-1}{\;\;p}\right)=-1\;\;\text{when}\;\; p=3\pmod 4$$