Question: > 
> Find all real solutions in $x$ and $y$, given $a$, to the system:
>  $$\left\{
> \begin{array}{l}
> \sqrt{x-y}+\sqrt{x+y}=a \\ 
> \sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\ 
> \end{array}
> \right. 
> $$
> 
> 
> 

From a math olympiad. Solutions presented: $(x,y)=(0.625 a^2,0.612372 a^2)$ and $(x,y)=(0.625 a^2,-0.612372 a^2)$. I tried first to make the substitution $u=x+y$ and $v=x-y$, noticing that $x^2+y^2=0.5((x+y)^2+(x-y)^2)$ but could not go far using that route. Then I moved to squaring both equations, hoping to get a solution, but without success. 

Hints and answers are appreciated. Sorry if this is a duplicate.

Answer: 
Square the first

$x-y+2\sqrt{x§2-y^2}+x+y=a^2$

$\sqrt{x^2-y^2}=\dfrac{a^2-2x}{2}$

Plug into the second

$\sqrt{x^2+y^2}=\dfrac{a^2-2x}{2}+a^2=\dfrac{3a^2-2x}{2}$

Square again

$x^2+y^2=\dfrac{9a^4-12a^2x+4x^2}{4}$

$x^2-y^2=\dfrac{a^4-4a^2x+4x^2}{4}$

Adding the last two equations we get

$2x^2=\dfrac{10a^4-16a^2x+8x^2}{4}\to x=\dfrac{5}{8}a^2$

Subtracting we get

$2y^2=\dfrac{8a^4-8a^2x}{4}\to y^2=a^4-a^x$

$y^2=a^4-\dfrac{5}{8}a^4=\dfrac{3}{8}a^4=\dfrac{6}{16}a^4$

$y=\pm \dfrac{\sqrt 6}{4}a^2$

so the solutions are

$x=\dfrac{5}{8}a^2;\;y=\pm \dfrac{\sqrt 6}{4}a^2$

I hope this helps
**Hint:**

Let $u=\sqrt{x-y},v=\sqrt{x+y}$. The system now reads

$$u+v=a,\\\sqrt{\frac{u^4+v^4}2}-uv=a^2$$

Raising the first equation to the fourth power,
$$a^4=u^4+v^4+4uv(u^2+v^2)+6u^2v^2.$$

Then using $u^4+v^4=2(a^2+uv)^2$ and $u^2+v^2=a^2-2uv$, you get an equation in $uv$, which simplifies:

$$a^4=2(a^2+uv)^2+4uv(a^2-2uv)+6u^2v^2.$$

> 
>  $uv=-\dfrac{a^2}8$.
> 
> 
> 

When $uv$ is known, $u^2+v^2=a^2-2uv$ gives you $2x$, and $x^2-u^2v^2$ gives you $y^2$.