$ 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?


$►$ If $x=7x_1$ and $y=7y_1$ then $x^2+y^2=7(x_1^2+y_1^2$).
$►$ If $x^2+y^2\equiv0\pmod7\iff x^2\equiv -y^2\pmod7$ then because of $\mathbb F_7^2=\{1,4,2,0\}$ and $(-1)\mathbb F_7^2=\{6,3,5,0\}$ the only possibility for
$x^2\equiv -y^2\pmod7$ is that both $x^2$ and $y^2$ are equal to $0$ modulo $7$ so $x$ and $y$ are equal to $0$ modulo $7$ (since $7$ is prime). Then $x\equiv y\equiv0\pmod7$.
