Determine a value of $n$ for which the number $2^{8}+2^{11}+2^{n}$ is a perfect square.
Solution 1: Observe that $2^{8}+2^{11}+2^{n}=\left(2^{4}\right)^{2}+2 \times 2^{4} \times 2^{6}+\left(2^{\frac{n}{2}}\right)^{2}$. Therefore, for $n=12$, we have $2^{8}+2^{11}+2^{12}=\left(2^{4}+2^{6}\right)^{2}$. Thus, $n=12$ is a solution.

Solution 2: If $2^{8}+2^{11}+2^{n}=k^{2}$, then

$$
\begin{aligned}
2^{8}+2^{3} \times 2^{8}+2^{n} & =k^{2} \\
9 \times 2^{8}+2^{n} & =k^{2} \\
2^{n} & =k^{2}-\left(3 \times 2^{4}\right)^{2} \\
2^{n} & =\left(k-3 \times 2^{4}\right)\left(k+3 \times 2^{4}\right)
\end{aligned}
$$

Therefore, $\left(k-3 \times 2^{4}\right)$ and $\left(k+3 \times 2^{4}\right)$ are powers of 2, that is, $k+3 \times 2^{4}=2^{a}$ and $k-3 \times 2^{4}=2^{b}$, with $a+b=n$ and

$$
2^{a}-2^{b}=\left(k+3 \times 2^{4}\right)-\left(k-3 \times 2^{4}\right)=3 \times 2^{5}=96
$$

Let's examine the list of powers of 2:

$$
2,4,8,16,32,64,128,256, \ldots
$$

We observe that the difference between these powers is only 96 between $128=2^{7}$ and $32=2^{5}$. Therefore, $a=7$ and $b=5$. Thus, $n=7+5=12$ is the only solution.