Find real solutions in $x$,$y$ for the system $\sqrt{x-y}+\sqrt{x+y}=a$ and $\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2.$ 
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.

$$\left\{
\begin{array}{l}
\sqrt{x-y}+\sqrt{x+y}=a \\ 
\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=a^2 \\ 
\end{array}
\right.$$

From first equation,
$$2x  + 2\sqrt{x^2 - y^2} = a^2 \tag 2$$
$$(x- a^2/2)^2 = x^2 - y^2$$
$$x^2 + \dfrac{a^4}4 - xa^2 = x^2 - y^2 $$
$$-\dfrac{a^4}4 + xa^2 = y^2 \tag 3$$

Add (2) to second equation
$$ x + \sqrt{x^2 + y^2} = \dfrac32 a^2$$
$$x^2 + \dfrac94a^4 - 3a^2x = x^2 + y^2 $$
$$ 3a^2 x +y^2 -\dfrac94a^4 = 0$$
Substituting from 3
$$ 3a^2 x -\dfrac{a^4}4 + xa^2  -\dfrac94a^4 = 0$$
$$ 4a^2 x   -\dfrac{10}4a^4 = 0$$
$$x = \dfrac{10}{16}a^2$$
Substituting back into 3
$$y^2 = -\dfrac{a^4}4 + \dfrac{10}{16}a^4 =  \dfrac{6}{16}a^4$$
$$ y = \pm\dfrac{\sqrt{6}}{4}a^2$$
