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.

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
