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.

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 $y^2$.
