21. The numbers $m$ and $k$ satisfy the equations $2^{m}+2^{k}=p$ and $2^{m}-2^{k}=q$. What is the value of $2^{m+k}$ in terms of $p$ and $q$ ?
A $\frac{p^{2}-q^{2}}{4}$
B $\frac{p q}{2}$
C $p+q$
D $\frac{(p-q)^{2}}{4}$
$\mathrm{E} \frac{p+q}{p-q}$
Solution
A
We have
$$
2^{m}+2^{k}=p
$$
and
$$
2^{m}-2^{k}=q .
$$

Adding equations (1) and (2), we obtain
$$
2\left(2^{m}\right)=p+q
$$
and hence
$$
2^{m}=\frac{p+q}{2} .
$$

Subtracting equations (2) from equation (1), we obtain
$$
2\left(2^{k}\right)=p-q
$$
and hence
$$
2^{k}=\frac{p-q}{2} \text {. }
$$

Therefore, by (3) and (4)
$$
\begin{aligned}
2^{m+k} & =2^{m} \times 2^{k} \\
& =\left(\frac{p+q}{2}\right) \times\left(\frac{p-q}{2}\right) \\
& =\frac{(p+q)(p-q)}{4} \\
& =\frac{p^{2}-q^{2}}{4} .
\end{aligned}
$$