<s>[INST] 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.
 [/INST] 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 $a$,
$$\color{green}{x=\frac52\frac{a^2}4,\\y=\pm\sqrt6\frac{a^2}4}.$$
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