27. Find the least positive integer $n$ such that $2^{8}+2^{11}+2^{n}$ is a perfect square.
27. Answer: 12
Let $2^{8}+2^{11}+2^{n}=m^{2}$ and so
$$
2^{n}=m^{2}-2^{3}(1+8)=(m-48)(m+48) .
$$

If we let $2^{k}=m+48$, then $2^{n-k}=m-48$ and we have
$$
2^{k}-2^{n-k}=2^{n-k}\left(2^{2 k-n}-1\right)=96=2^{5} \times 3 \text {. }
$$

This means that $n-k=5$ and $2 k-n=2$, giving us $n=12$.