Let $m,n$ be natural numbers and let $d = gcd(m,n)$. Let $x = 2^{m} -1$ and $y= 2^n +1$
(a) If $\frac{m}{d}$ is odd, prove that $gcd(x,y) = 1$
(b) If $\frac{m}{d}$ is even, Find $gcd(x,y)$
### Part (a)
Given: \( m, n \) are natural numbers, \( d = \gcd(m, n) \), \( x = 2^m - 1 \), and \( y = 2^n + 1 \). We need to prove that if \( \frac{m}{d} \) is odd, then \( \gcd(x, y) = 1 \).

1. Let \( a = 2^d \), \( k = \frac{m}{d} \), and \( l = \frac{n}{d} \). Then we have:
   \[
   x = 2^m - 1 = a^k - 1 \quad \text{and} \quad y = 2^n + 1 = a^l + 1
   \]

2. We use the property of the greatest common divisor:
   \[
   \gcd(2^m - 1, 2^{2n} - 1) = \gcd(a^k - 1, a^{2l} - 1)
   \]

3. By the Euclidean algorithm, we know:
   \[
   \gcd(a^k - 1, a^{2l} - 1) = a^{\gcd(k, 2l)} - 1
   \]

4. Since \( k \) is odd, \( \gcd(k, 2l) = 1 \). Therefore:
   \[
   \gcd(a^k - 1, a^{2l} - 1) = a - 1
   \]

5. Now, we need to find \( \gcd(a^k - 1, a^l + 1) \). Since \( a = 2^d \), we have:
   \[
   \gcd(2^m - 1, 2^n + 1) = \gcd(a^k - 1, a^l + 1)
   \]

6. We know that \( \gcd(a - 1, a^l + 1) = 1 \) because \( a - 1 \) and \( a^l + 1 \) are coprime when \( a = 2^d \) and \( l \) is odd.

Thus, we conclude:
\[
\gcd(2^m - 1, 2^n + 1) = 1
\]

### Part (b)
Given: \( m, n \) are natural numbers, \( d = \gcd(m, n) \), \( x = 2^m - 1 \), and \( y = 2^n + 1 \). We need to find \( \gcd(x, y) \) if \( \frac{m}{d} \) is even.

1. Let \( a = 2^d \), \( k = \frac{m}{d} \), and \( l = \frac{n}{d} \). Then we have:
   \[
   x = 2^m - 1 = a^k - 1 \quad \text{and} \quad y = 2^n + 1 = a^l + 1
   \]

2. We use the property of the greatest common divisor:
   \[
   \gcd(2^m - 1, 2^{2n} - 1) = \gcd(a^k - 1, a^{2l} - 1)
   \]

3. By the Euclidean algorithm, we know:
   \[
   \gcd(a^k - 1, a^{2l} - 1) = a^{\gcd(k, 2l)} - 1
   \]

4. Since \( k \) is even, \( \gcd(k, 2l) = 2 \). Therefore:
   \[
   \gcd(a^k - 1, a^{2l} - 1) = a^2 - 1
   \]

5. Now, we need to find \( \gcd(a^k - 1, a^l + 1) \). Since \( a = 2^d \), we have:
   \[
   \gcd(2^m - 1, 2^n + 1) = \gcd(a^k - 1, a^l + 1)
   \]

6. We know that \( a^2 - 1 = (a - 1)(a + 1) \). Since \( k \) is even, \( a^k - 1 \) is divisible by \( a^2 - 1 \). Therefore:
   \[
   \gcd(a^k - 1, a^l + 1) \mid a^2 - 1
   \]

7. It is easy to see that \( a + 1 \) divides both \( a^k - 1 \) and \( a^l + 1 \). Also, \( a - 1 \) does not divide \( a^l + 1 \) because \( a - 1 = 2^d - 1 \) and \( 2^d - 1 \mid 2 \) implies \( d = 1 \), which does not change our gcd.

Thus, we conclude:
\[
\gcd(2^m - 1, 2^n + 1) = a + 1 = 2^d + 1
\]

The final answer is \( \boxed{2^d + 1} \)