$2 \cdot 101$ satisfies the number of non-negative integer pairs $(a, b)$ for $|a-b|+a b=1$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) 4 .
(Chinese Junior High School Mathematics League, 1992)
［Solution］Given $a b \geqslant 0$, and $|a-b| \geqslant 0$, then we have $\left\{\begin{array}{l}|a-b|=1, \\ a b=0,\end{array} \quad\right.$ and $\quad\left\{\begin{array}{l}|a-b|=0, \\ a b=1 .\end{array}\right.$ Solving these, we get $\left\{\begin{array}{l}a=0, \\ b=1 ;\end{array} \quad\left\{\begin{array}{l}a=1, \\ b=0 ;\end{array}\right.\right.$ and $\quad\left\{\begin{array}{l}a=1, \\ b=1 .\end{array}\right.$ Therefore, the answer is $(C)$.