1. If the natural numbers $a, x, y$ satisfy $\sqrt{a-2 \sqrt{6}}=\sqrt{x} -\sqrt{y}$, then the maximum value of $a$ is $\qquad$.
$$
=, 1, a_{\max }=7 \text {. }
$$

Squaring both sides of the known equation yields $a-2 \sqrt{6}=x+y-$ number, it must be that $x+y=\alpha$ and $x y=6$. From the known equation, we also know that $x>y$, so it can only be $x=6, y=1$ or $x=3, y=2$. Therefore, $a=7$ or 2, the maximum value is 7.