Find the largest positive integer $k$ such that $\phi ( \sigma ( 2^k)) = 2^k$. ($\phi(n)$ denotes the number of positive integers that are smaller than $n$ and relatively prime to $n$, and $\sigma(n)$ denotes the sum of divisors of $n$). As a hint, you are given that $641|2^{32}+1$.
1. We need to find the largest positive integer \( k \) such that \( \phi(\sigma(2^k)) = 2^k \). Here, \(\phi(n)\) denotes Euler's totient function, and \(\sigma(n)\) denotes the sum of divisors of \( n \).

2. First, we calculate \(\sigma(2^k)\):
   \[
   \sigma(2^k) = 1 + 2 + 2^2 + \cdots + 2^k = \sum_{i=0}^k 2^i = 2^{k+1} - 1
   \]
   This is a geometric series sum.

3. Next, we need to find \(\phi(2^{k+1} - 1)\) and set it equal to \(2^k\):
   \[
   \phi(2^{k+1} - 1) = 2^k
   \]

4. Suppose \(2^{k+1} - 1 = \prod_{i=1}^m p_i^{e_i}\), where \(p_i\) are distinct primes. Then:
   \[
   \phi(2^{k+1} - 1) = (2^{k+1} - 1) \prod_{i=1}^m \left(1 - \frac{1}{p_i}\right)
   \]
   Setting this equal to \(2^k\), we get:
   \[
   2^k = (2^{k+1} - 1) \prod_{i=1}^m \left(1 - \frac{1}{p_i}\right)
   \]

5. Since \(2^k\) has no odd factors, it follows that:
   \[
   \prod_{i=1}^m p_i \ge 2^{k+1} - 1
   \]
   But by definition:
   \[
   \prod_{i=1}^m p_i \le 2^{k+1} - 1
   \]
   Therefore:
   \[
   \prod_{i=1}^m p_i = 2^{k+1} - 1
   \]

6. Then:
   \[
   2^k = \prod_{i=1}^m (p_i - 1)
   \]
   This implies that \(p_i = 2^{j_i} + 1\) for some integers \(j_i\).

7. We need to solve:
   \[
   \prod_{i=1}^m (2^{j_i} + 1) = 2^{k+1} - 1
   \]
   Since \(2^{j_i + 1}\) is prime, \(j_i\) cannot have any odd factors, so \(j_i = 2^{l_i}\) for integers \(l_i\).

8. Multiplying both sides by \(\prod_{i=1}^m (2^{2^{l_i}} - 1)\), we get:
   \[
   \prod_{i=1}^m (2^{2^{l_i+1}} - 1) = (2^{k+1} - 1) \prod_{i=1}^m (2^{2^{l_i}} - 1)
   \]

9. By the Euclidean Algorithm, \(\text{gcd}(2^{2^{l_i}} - 1, 2^{k+1} - 1) \neq 1\) for some \(i\). This is true if and only if \(k\) is a multiple of \(2^{l_i}\) or vice versa, and \(k+1\) has no odd prime factors. Thus, \(k = 2^r - 1\) for some \(r\).

10. If \(k + 1 = 2^r\), then \(2^{2^i} + 1 \mid 2^{k+1} - 1\) and \(\text{gcd}(2^{2^i} + 1, 2^{2^j} + 1) = 1\) for \(0 \le i, j \le r - 1\). Expanding and using a telescoping difference of squares product gives:
    \[
    \left[\frac{1}{2^1 - 1}\right] \cdot (2^{k+1} - 1) \prod_{i=0}^{r-1} \left(\frac{2^{2^i}}{2^{2^i} + 1}\right) = 2^{2^r - 1} = 2^k
    \]
    This holds true if \(2^{2^i} + 1\) is prime for all \(1 \le i \le r - 1\). Therefore, \(r \le 5\) and \(k \le 31\).

11. Finally, we verify that \(k = 31\) works:
    \[
    \sigma(2^{31}) = 2^{32} - 1
    \]
    \[
    \phi(2^{32} - 1) = 2^{31}
    \]
    This satisfies the condition.

The final answer is \( \boxed{31} \)