3.4 The non-zero real numbers $(a, b)$ that satisfy the equation $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$, such $(a, b)$
(A) None.
(B) 1.
(C) 2.
(D) There is one pair for each $b \neq 0$.
(E) There are two pairs for each $b \neq 0$.
(31st American High School Mathematics Examination, 1980)
[Solution] The given equation can be transformed into
$$
\frac{a+b}{a b}=\frac{1}{a+b} \text { or }(a+b)^{2}=a b \text {, }
$$

which is equivalent to
$$
a^{2}+a b+b^{2}=0 .
$$

Since $(a, b)$ satisfies the final equation, we have
$$
a=\frac{-b \pm \sqrt{-3 b^{2}}}{2},
$$

Noting that $b \neq 0$, there is no pair of real numbers that satisfies the original equation. Therefore, the answer is $(A)$.