2. determine the largest natural number $n$ so that

$$
4^{995}+4^{1500}+4^{n}
$$

is a square number.
## 1st solution

Answer: $n=2004$

We show more generally: If $a\left(2^{n-a}\right)^{2}$ applies, but on the other hand, because of $2 b-2 a<n-a+1$ it also applies

$$
\begin{aligned}
A & =1+2^{2 b-2 a}+\left(2^{n-a}\right)^{2} \\
& <1+2 \cdot 2^{n-a}+\left(2^{n-a}\right)^{2}=\left(2^{n-a}+1\right)^{2}
\end{aligned}
$$

This means that $A$ lies between two consecutive square numbers and cannot itself be a square number.