Find all natural numbers $n$ for which $2^{8}+2^{11}+2^{n}$ is equal to the square of an integer.
The sum of the first two terms is a perfect square, indeed $2^{8}+2^{11}=2^{8}\left(1+2^{3}\right)=2^{8} \cdot 9=48^{2}$. Thus, if $k$ is an integer such that $2^{8}+2^{11}+2^{n}=k^{2}$, then

$$
2^{n}=k^{2}-48^{2}=(k-48)(k+48)
$$

We can assume that $k$ is positive, in which case the right side is a product of positive integers. If this product is a power of 2, then the factors are also powers of 2. Therefore, there exist non-negative integers $p$ and $q$ such that $pp$, so $2^{q-p}-1$ is odd, and in our case

$$
2^{p}=2^{5} \quad \text { and } \quad 2^{q-p}-1=3
$$

Thus, $p=5$ and $q=7$, so

$$
2^{n}=2^{p} \cdot 2^{q}=2^{12}
$$

The only solution to the problem is therefore $n=12$. In this case, $2^{8}+2^{11}+2^{12}=80^{2}$.