<s>[INST] $ 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?

 [/INST] Your example is just a special case of this theorem,
Theorem
If $p\equiv 3\pmod 4$ Then whenever $p\mid x^2+y^2$, we have $p\mid x$ and $p\mid y$.
Proof
Assume that $p\mid x^2+y^2$ with $p\not\mid x,y$ then
$$x^2+y^2\equiv 0\pmod p$$
$$x^2\equiv -y^2\pmod p$$
Which means $$x^{2\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}y^{2\frac{p-1}{2}} \pmod p$$
$$x^{p-1}\equiv - y^{\frac{p-1}{2}}\pmod p$$
That’s because $(p-1)/2$ is odd, but let’s see what we’ve got, using Fermat’s little theorem $$1\equiv -1\pmod p$$
Hence $p=2$ but that’s clearly a contradiction. $\quad \quad \quad \quad \quad \quad\square$
</s>