Question: > 
> 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.

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

---

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

$$y^2 = -\dfrac{a^4}4 + \dfrac{10}{16}a^4 = \dfrac{6}{16}a^4$$

$$ y = \pm\dfrac{\sqrt{6}}{4}a^2$$