$2 \cdot 31$ If $x=am+bn$ satisfies the quadratic equation $(x+m)^{2}-(x+n)^{2}=(m-n)^{2}$, where $m$ and $n$ are unequal and non-zero constants, then
(A) $a=0, b$ has a unique non-zero value.
(B) $a=0, b$ has two non-zero values.
(C) $b=0, a$ has a unique non-zero value.
(D) $b=0, a$ has two non-zero values.
(E) $a$ and $b$ each have a unique non-zero value.
(11th American High School Mathematics Examination, 1960)
[Solution] By imitating problem $2 \cdot 30$, we get $x=-n$. Therefore, $a m+b n=-n$. Since $m, n$ are non-zero constants, then $a=0, b=-1$. Hence, the answer is $(A)$.