Question about finite fields I'm from Brazil, so my English is not the best, I'm sorry if there's something wrong. I'm using $|F_q|$ for the cardinal of the field.
I have $\mathbb{F}_q$ a finite field of characteristic $p$ and cardinal $q$, $p$ is a prime number and $q$ is a power of $p$.
It's a very long exercise, I already showed that for $p=2$ $\mathbb{F}_q=\mathbb{F}_q^2$ and for $p\neq2$ I show that $|\mathbb{F}_q^{*^2}|=\frac{q-1}{2}$ and $|\mathbb{F}_q^2|=\frac{q+1}{2}$.
Then I showed the properties:
$x\in \mathbb{F}_q^* \Longrightarrow x^{\frac{q-1}{2}}=\pm 1;$
$x \in \mathbb{F}_q^{*^2} \Longleftrightarrow x^{\frac{q-1}{2}}= 1;$
$-1\in \mathbb{F}_q^{*^2} \Longleftrightarrow q-1\equiv 0$ (mod $4$).
Now I'm having trouble to show this:
If $-1$ and $x$ are not squares of $\mathbb{F}_q^*$, then $-x$ is a square of $\mathbb{F}_q$.
And also, I'm having trouble to show that in the field $\mathbb{F}_p$, suppose that
$$A=\{ -x^2; x\in \mathbb{F}_p \} \quad \text{and} \quad B=\{ 1+y^2;y \in \mathbb{F}_p \}$$
I have to show that $|A|=|B|=\frac{p+1}{2}$, and deduce that exist $x$ and $y$ in $\mathbb{F}_p$ such that
$$1+x^2+y^2=0$$
and exist $a$ and $b$
$$p|1+a^2+b^2.$$
 A: By what you have proved already $(-1)^{(q-1)/2}= -1$. Therefore $(q-1)/2$ is odd. Also $x^{(q-1)/2}=-1$. But then $(-x)^{(q-1)/2}=(-1)^{(q-1)/2}(x)^{(q-1)/2}=1$, so, again by what you have proved, $-x$ is a square. QED
A proof of the rest of the question - for every odd prime power $q=p^n$.
The multiplicative group $\Bbb{F}_q^*$ is Abelian of order $q-1$. The map $x\mapsto x^2$ is a homomorphism. Its kernel is $\{x\mid x^2=1\}=\{-1,1\}$ of order 2 because $1\ne -1$ since $q$ is odd, so the image (the set of all non-zero squares) is of size $(q-1)/2$. Then the set $S$ of all (including zero) squares of $\Bbb{F}_q$ is of size $(q-1)/2+1=(q+1)/2$. Then the set $A$ and the set $B$ are of the same size: the set $A$ is obtained from $S$ by adding the $-$ sign to each element, the set $B$ is obtained by adding $1$ to each element of $S$. If $A$ and $B$ did not intersect then $A\cup B$ would contain $q+1$ elements $>|\Bbb{F}_q|$. Hence $A\cap B$ is not empty. Hence there exist $x, y$ such that $-x^2=1+y^2$ or $1+x^2+y^2=0$ which means $1+x^2+y^2\equiv 0\mod p$.
