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
**Short solution:**

WLOG, $a=2$ (as $x,y\propto a^2$).

By squaring the first equation and rearranging, you draw 
$$\sqrt{x^2-y^2}=2-x.$$

And from the second,
$$\sqrt{x^2+y^2}=4+\sqrt{x^2-y^2}=6-x.$$

Now by summing the squares of the LHS,

$$2x^2=(2-x)^2+(6-x)^2$$

which gives$$x=\frac52.$$

Then from the first identity

$$y=\pm\sqrt{x^2-(2-x)^2}=\pm\sqrt6.$$

For general $a$,

$$\color{green}{x=\frac52\frac{a^2}4,\\y=\pm\sqrt6\frac{a^2}4}.$$