Let $a$ be a positive integer such that $\gcd(an+1, 2n+1) = 1$ for all integer $n$.

a) Prove that $\gcd(a-2, 2n+1) = 1$ for all integer $n$.
b) Find all possible $a$.
Let \( a \) be a positive integer such that \(\gcd(an+1, 2n+1) = 1\) for all integers \( n \).

### Part (a)
Prove that \(\gcd(a-2, 2n+1) = 1\) for all integers \( n \).

1. Given \(\gcd(an+1, 2n+1) = 1\), we use the Euclidean algorithm to simplify this expression:
   \[
   \gcd(an+1, 2n+1) = \gcd(an+1 - (2n+1), 2n+1)
   \]
2. Simplify the expression inside the gcd:
   \[
   an+1 - (2n+1) = an + 1 - 2n - 1 = (a-2)n
   \]
   Therefore,
   \[
   \gcd(an+1, 2n+1) = \gcd((a-2)n, 2n+1)
   \]
3. Since \(\gcd(an+1, 2n+1) = 1\), it follows that:
   \[
   \gcd((a-2)n, 2n+1) = 1
   \]
4. Notice that \(\gcd((a-2)n, 2n+1) = 1\) implies \(\gcd(a-2, 2n+1) = 1\) because \(2n+1\) is odd and does not share any common factor with \(n\).

Thus, we have shown that:
\[
\gcd(a-2, 2n+1) = 1 \quad \text{for all integers } n.
\]

### Part (b)
Find all possible \( a \).

1. From part (a), we know that \(\gcd(a-2, 2n+1) = 1\) for all integers \( n \).
2. Consider \( n = \frac{p-1}{2} \) where \( p \) is an odd prime. Then \( 2n+1 = p \), and we have:
   \[
   \gcd(a-2, p) = 1
   \]
   for all odd primes \( p \).
3. Since \(\gcd(a-2, p) = 1\) for all odd primes \( p \), \( a-2 \) must be a power of 2. This is because any integer that is coprime with all odd primes must be of the form \( 2^m \) for some nonnegative integer \( m \).
4. Therefore, \( a-2 = 2^m \) for some nonnegative integer \( m \), which implies:
   \[
   a = 2 + 2^m
   \]
   Additionally, \( a = 1 \) is also a solution since \( \gcd(1-2, 2n+1) = \gcd(-1, 2n+1) = 1 \).

Thus, the possible values of \( a \) are:
\[
a = 1 \quad \text{or} \quad a = 2 + 2^m \text{ for any nonnegative integer } m.
\]

The final answer is \( \boxed{ a = 1 } \) or \( a = 2 + 2^m \text{ for any nonnegative integer } m \).