Problem 8.3. Find the least positive integer $m$ such that $2^{2000}$ divides $2003^{m}-$ 1.

Ivan Tonov
8.3. Set $m=2^{k} p$, where $p>1$ is an odd integer. Then

$$
2003^{m}-1=2003^{2^{k} p}-1=\left(2003^{2^{k}}-1\right) K
$$

where $K$ is a sum of $p$ even integers and 1 ; in particular, $K$ is odd. Hence $m$ must be of the form $m=2^{k}$. In this case we have that

$$
\begin{aligned}
2003^{2^{k}}-1 & =\left(2003^{2^{k-1}}-1\right)\left(2003^{2^{k-1}}+1\right) \\
& =\left(2003^{2^{k-1}}+1\right)\left(2003^{2^{k-2}}+1\right) \ldots(2003+1)(2003-1)
\end{aligned}
$$

Since $2003^{2^{i}}+1 \equiv 2(\bmod 4)$, then $2^{k+2}$ divides $2003^{2^{k}}-1$ but $2^{k+3}$ does not (use that $2003+1 \equiv 4(\bmod 8)$ and $2003-1 \equiv 2(\bmod 4))$. Therefore $k+2=2000$, i.e. $k=1998$. Thus the desired number is equal to $2^{1998}$.