Tenth Philippine Mathematical Olympiad

Solution:
$(4,16)$ and $(16,4)$

After rewriting the first equation into $x + y = 28 - \sqrt{x y}$, we square to get
$$x^{2} + x y + y^{2} = 784 - 56 \sqrt{x y}$$
Using the second given equation, the last equation becomes
$$336 = 784 - 56 \sqrt{x y} \quad \text{or} \quad \sqrt{x y} = 8$$
Substituting this last equation to the first given equation, we get $y = 20 - x$, and the second given equation then becomes
$$x^{2} + (20 - x)^{2} + 64 = 336$$
which yields $x = 4$ and $x = 16$. Knowing that $x y = 64$, the solutions are the ordered pairs $(4,16)$ and $(16,4)$.