$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $ 
Show that 
  $$ 7\mid x \text{ and } 7\mid  y \Longleftrightarrow   7\mid x^2+y^2  $$

Indeed,
First let's show 
$7\mid x \text{ and } 7\mid  y \Longrightarrow  7\mid x^2+y^2  $
we've  $7\mid x \implies 7\mid x^2$ the same for $7\mid  y \implies 7\mid  y^2$ then 
$ 7\mid x^2+y^2 $


*

*Am i right and can we write $a\mid x \implies a\mid x^P ,\ \forall p\in \mathbb{N}^*$


Now let's show 
$7\mid x^2+y^2 \Longrightarrow 7\mid x \text{ and } 7\mid  y$
$7\mid x^2+y^2  \Longleftrightarrow   x^2+y^2=0 \pmod 7 $
for 
\begin{array}{|c|c|c|c|c|} \hline
 x& 0 & 1 & 2& 3 & 4 & 5 & 6  \\ \hline
 x^2& 0 & 1 & 4& 2 & 2 & 4 & 1 &\pmod 7\\ \hline 
 y& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline
 y^2& 0 & 1 & 4& 2 & 2 & 4 & 1 & \pmod 7 \\ \hline
 \end{array}
which means we have one possibility that $x=y= 0 \pmod 7 $


*

*Am I right and are there other ways?


Let $x,y \in \mathbb{F}_p$ be with $x^2+y^2=0$. If $x=0$, then $y=0$. Now assume $x \neq 0$. Let $z:=y/x$, then $z^2=-1$. If $p=2$, this means $z = 1$. If $p > 2$, this means that $z$ has order $4$ in $\mathbb{F}_p^*$, which happens iff $4|p-1$ i.e. $p \equiv 1 \bmod 4$. Hence, for every odd prime $p$ with $p \not\equiv 1 \bmod 4$ the quadratic form $x^2+y^2=0$ has only the trivial solution.
(While your method for $p=7$ is fine, try it with $p=67$!)
