Which is the smallest positive integer $n$ for which $3^{2 n}-1$ is divisible by $2^{2010}$?
Solution. We will show that if $m$ is odd and $k$ is a positive integer, then $3^{2^{k} m}-1$ is divisible by exactly the $(k+2)$-th power of 2. If $k=1$, then

$$
\begin{gathered}
3^{2 m}-1=9^{m}-1=(8+1)^{m}-1= \\
=\left(8^{m}+\binom{m}{1} 8^{m-1}+\ldots+\binom{m}{m-2} 8^{2}+\binom{m}{m-1} 8+1\right)-1=8^{2} v+8 m
\end{gathered}
$$

which is indeed divisible by exactly the third power of 2. We can then prove by induction on $k$: if $k \geq 1$ and $3^{2^{k} m}-1=2^{k+2} \cdot(2 w+1)$, then

$$
\begin{aligned}
3^{2^{k+1} m}-1 & =\left(3^{2^{k} m}-1\right) \cdot\left(3^{2^{k} m}+1\right)=2^{k+2}(2 w+1) \cdot\left(2^{k+2}(2 w+1)+2\right)= \\
& =2^{k+2}(2 w+1) \cdot 2\left(2^{k+1}(2 w+1)+1\right)
\end{aligned}
$$

which is divisible by exactly the $(k+3)$-th power of 2.

According to the proven results, $3^{2 n}-1$ is divisible by $2^{2010}$ if and only if $2 n$ is divisible by $2^{2008}$, i.e., $n$ is divisible by $2^{2007}$. The smallest such positive integer is therefore $n=2^{2007}$.