27. Find the positive integers $n$ so that $2^{8}+2^{11}+2^{n}$ is a perfect square.
(ans. 12 .
$$
\begin{array}{l}
m^{2}=2^{8}+2^{11}+2^{n} \Rightarrow 2^{n}=m^{2}-2^{8}-2^{11}=m^{2}-2^{8}\left(1+2^{3}\right)= \\
m^{2}-\left(3 \cdot 2^{4}\right)^{2}=\left(m-3 \cdot 2^{4}\right)\left(m+3 \cdot 2^{4}\right)=(m-48)(m+48) \Rightarrow m-48= \\
2^{k}, m+48=2^{l}, k+l=n \Rightarrow 2^{l}-2^{k}=96 \Rightarrow 2^{k}\left(2^{l-k}-1\right)=2^{5} \cdot 3 \Rightarrow
\end{array}
$$
$2^{l-k}-1=3,2^{k}=2^{5}$ by unique factorization of integers $\Rightarrow l-k=2$ and $l=7 \Rightarrow n=12$. OR complete the square: $\left(2^{s}\right)^{2}+2\left(2^{6}\right)\left(2^{4}\right)+\left(2^{4}\right)^{2}=$ $\left(2^{s}+2^{4}\right)^{2} \Rightarrow s=6 \Rightarrow n=2 s=12$. $)$