2.134 Let positive integers $a, m, n$ satisfy $\sqrt{a^{2}-4 \sqrt{2}}=\sqrt{m}-\sqrt{n}$. Then the values of $a, m, n$ 
(A) have one set.
(B) have two sets.
(C) have more than two sets.
(D) do not exist.
(China Junior High School Mathematics League, 1996)
[Solution] Squaring both sides of the given equation, we get
$$
a^{2}-4 \sqrt{2}=m+n-2 \sqrt{m n} .
$$

Since $a, m, n$ are natural numbers, $a^{2}-4 \sqrt{2}$ is an irrational number. Therefore,
$$
\left\{\begin{array} { l } 
{ \sqrt { m n } = \sqrt { 8 } , } \\
{ m + n = a ^ { 2 } , }
\end{array} \text { i.e., } \quad \left\{\begin{array}{l}
m n=8, \\
m+n=a^{2} .
\end{array}\right.\right.
$$

Given that $m>n$, the only solution is $m=8, n=1, a=3$. Therefore, the answer is $(A)$.