8.15 If $2^{a}+2^{b}=3^{c}+3^{d}$, then the number of negative integers among $a, b, c, d$ is
(A) 4.
(B) 3.
(C) 2.
(D) 1.
(E) 0.
(14th American High School Mathematics Examination, 1963)
［Solution］We discuss the following cases:
(1) If only one of $a, b, c, d$ is negative, then one side of the equation is a fraction, and the other side is an integer, which is not appropriate;
(2) If exactly two of the four numbers are negative:
(i) If $a\beta$ the equation can be transformed into:
$$
\left(2^{\alpha-\beta}+1\right) 3^{\gamma+\delta}=\left(3^{\gamma}+3^{\delta}\right) 2^{\alpha},
$$

the left side of the above equation is odd, while the right side is even, which is not appropriate;
(ii) If $\alpha=\beta$, then the previous equation becomes:
$$
2^{\alpha+1} 3^{\gamma+\delta}=\left(3^{\gamma}+3^{\delta}\right) 2^{2 \alpha} \text {. }
$$

When $\alpha=1$, it becomes $3^{\gamma} \cdot 3^{\delta}=3^{\gamma}+3^{\delta}$, which is not true for all $\gamma, \delta$;
When $\alpha>1$, then $\alpha-1>0$, the previous equation becomes:
$$
3^{\gamma+\delta}=2^{\alpha-1}\left(3^{\gamma}+3^{\delta}\right) \text {. }
$$

The left side of the above equation is odd, while the right side is even, which is also not appropriate.
In summary, the number of negative numbers among $a, b, c, d$ can only be 0.
Therefore, the answer is $(E)$.