Philippine Mathematical Olympiad

Solution:
(ans. 12)
$$\begin{aligned} & m^{2} = 2^{8} + 2^{11} + 2^{n} \Rightarrow 2^{n} = m^{2} - 2^{8} - 2^{11} = m^{2} - 2^{8}(1 + 2^{3}) = \\ & m^{2} - (3 \cdot 2^{4})^{2} = (m - 3 \cdot 2^{4})(m + 3 \cdot 2^{4}) = (m - 48)(m + 48) \Rightarrow m - 48 = \\ & 2^{k},\ m + 48 = 2^{l},\ k + l = n \Rightarrow 2^{l} - 2^{k} = 96 \Rightarrow 2^{k}(2^{l - k} - 1) = 2^{5} \cdot 3 \Rightarrow \end{aligned}$$
$2^{l - k} - 1 = 3,\ 2^{k} = 2^{5}$ by unique factorization of integers $\Rightarrow l - k = 2$ and $l = 7 \Rightarrow n = 12$.

OR complete the square:
$$(2^{s})^{2} + 2(2^{6})(2^{4}) + (2^{4})^{2} = (2^{s} + 2^{4})^{2} \Rightarrow s = 6 \Rightarrow n = 2s = 12.$$