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

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
$$y^2 = -\dfrac{a^4}4 + \dfrac{10}{16}a^4 =  \dfrac{6}{16}a^4$$
$$ y = \pm\dfrac{\sqrt{6}}{4}a^2$$
A: 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 $y^2$.
A: 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
A: 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 $a$,
$$\color{green}{x=\frac52\frac{a^2}4,\\y=\pm\sqrt6\frac{a^2}4}.$$
