1. Calculate the remainder obtained when dividing $2^{4 n+2}+2^{4}$ by $2^{2 n+1}+2^{n+1}+1$ for every natural number $n \geq 2$.
Solution. We have:

$$
\begin{aligned}
2^{4 n+2}+2^{4} & =\left(2^{2 n+1}\right)^{2}+16=\left(2^{2 n+1}\right)^{2}+2 \cdot 2^{2 n+1}+1-2 \cdot 2^{2 n+1}+15 \\
& =\left(2^{2 n+1}+1\right)^{2}-2^{2 n+2}+15=\left(2^{2 n+1}+1\right)^{2}-\left(2^{n+1}\right)^{2}+15 \\
& =\left(2^{2 n+1}+2^{n+1}+1\right)\left(2^{2 n+1}-2^{n+1}+1\right)+15
\end{aligned}
$$

Now, since for $n \geq 2$ we have $2^{2 n+1}+2^{n+1}+1 \geq 2^{5}+2^{3}+1=41$, it follows that when dividing $2^{4 n+2}+2^{4}$ by $2^{2 n+1}+2^{n+1}+1$, we get a quotient of $2^{2 n+1}-2^{n+1}+1$ and a remainder of 15.