2 is a square modulo $p$ if and only if $p \equiv \pm 1 \pmod 8$ 
2 is a square modulo $p$ if and only if $p \equiv \pm 1 \pmod 8$

The indication for this exercise was to consider $\alpha$ such that $\alpha^8 = 1$ ($\alpha$ is eigth root of the unity) in a field extension of $\mathbb{F}_p$ and consider $y = \alpha + \alpha^{-1}$.
I have got the part that if $p \equiv \pm 1 \pmod 8$, then $2$ is a square modulo 8. I showed that $y^2 = 2$ (this is because $\alpha^8 = 1 \Rightarrow \alpha^4 = -1 \Rightarrow \alpha^2 = -\alpha^{-2}$. Then I used the proposition that if $\mathbb{K}$ is a field extension of $\mathbb{F}_p$, then $\mathbb{F_p} = \{x \in \mathbb{K} | x^p = x\}$ and I showed $y^p = y$ (easy in caracteristic $p$).
Edit 2: $y^p = y$ is shown as follows:
$$\begin{align*} y^p &= \alpha^p + \alpha^{-p} & \text{every other terms vanish as we work in car $p$} \\
 & =\alpha^{8n \pm 1} + \alpha^{-8n \mp 1} & p \equiv \pm 1 \pmod 8 \text{ by hypothesis}\\
&= (\alpha^8)^{n} \alpha^{\pm 1} + (\alpha^8)^{-n}\alpha^{\mp 1} & \alpha^8 = 1\\
&= \alpha^{\pm1} + \alpha^{\mp 1}\\
&= y.
\end{align*}$$
This shows that $2$ is a square modulo $p$, as $y^2 = 2$ and $y \in \mathbb{F}_p$.
Now I have some troubles for the other part, I don't know where to start and what to do. 
Edit: While searching on google for help, I fell on quadratic residues, and I don't know what they are, we did not see them in class.
 A: You have already done the case $p \equiv \pm 1 \pmod 8$, so let's look at the case $p \equiv \pm 3 \pmod 8$. The argument is very similar to the one you gave. Just like before, we let $\alpha$ be any primitive 8th root of unity, and set $y = \alpha + \alpha^{-1}$. In the same way as before, it follows that $y^2 = 2$, so $y$ and $-y$ are the two square roots of $2$ in an algebraic closure of $\mathbb{F}_p$. Therefore, in order to show that $2$ is not a square in $\mathbb{F}_p$, it is enough to show that $y$ is not in $\mathbb{F}_p$. 
To do this, let's calculate $y^p$. Just as before, we get
\begin{align}
y^p &= \alpha^p + \alpha^{-p} \\
    &= \alpha^{8n \pm 3} + \alpha^{-8n \mp 3} \\
    &= \alpha^{3}+\alpha^{-3}& \text{ (possibly with the terms revered, but addition is commutative) }\\
    &= \alpha^4 \cdot (\alpha^{-1} + \alpha^{-7})\\
    &= -1 \cdot (\alpha^{-1} + \alpha) \\&= -y
\end{align}
Since $y^p = -y \neq y$, we see that $y$ is not fixed under taking the $p$-th power, so $y \notin \mathbb{F}_p$. Thus, if $p \equiv \pm 3 \pmod 8$, then $2$ is not a square mod $8$. 
