For what value of $n$ will

$$
2^{11}+2^{8}+2^{n} \text { be a perfect square? }
$$
Two solutions emerged. In one, they sought the solution in the form $a^{2}+2 a b+b^{2}$ (with appropriate choices of $a, b$). Here, many overlooked the observation that if an expression is a perfect square, it can always be written in the form $(a+b)^{2}$.

Thus, if $2^{8}+2^{11}+2^{n}$ is exactly in the form $(a+b)^{2}$, then neither $a^{2}$ nor $b^{2}$ can be $2^{11}$, since it has an odd exponent. Therefore, the only possibility is that $2 a b=2^{11}$, from which $n=12$ follows.

The other type of solution was based on the following idea:

It is easy to verify that the expression cannot be a perfect square if $n<8$. Let's factor out $2^{8}$. Then $2^{8}\left(9+2^{n-8}\right)$ can only be $N^{2}$ (a perfect square) if $N^{2}-9=2^{n-8}$, from which we get $(N+3)(N-3)=2^{n-8}$. This cannot hold if $n-8$ is too large. Both $N-3$ and $N+3$ must be powers of two whose difference is 6, from which a small calculation yields $n=12$.