If $a$ and $b$ are positive integers such that $a^2-b^4= 2009$, find $a+b$.
1. We start with the given equation:
   \[
   a^2 - b^4 = 2009
   \]

2. We use the difference of squares factorization:
   \[
   a^2 - b^4 = (a + b^2)(a - b^2)
   \]
   Let \( x = a + b^2 \) and \( y = a - b^2 \). Then we have:
   \[
   xy = 2009
   \]

3. The prime factorization of \( 2009 \) is:
   \[
   2009 = 7^2 \cdot 41
   \]

4. We need to find pairs \((x, y)\) such that \( xy = 2009 \) and both \( x \) and \( y \) are positive integers. The possible pairs are:
   \[
   (2009, 1), (287, 7), (49, 41)
   \]

5. For each pair, we check if \( 2b^2 = x - y \) results in \( b \) being an integer:
   - For \( (x, y) = (2009, 1) \):
     \[
     2b^2 = 2009 - 1 = 2008 \implies b^2 = 1004 \implies b = \sqrt{1004}
     \]
     Since \( \sqrt{1004} \) is not an integer, this pair is not valid.

   - For \( (x, y) = (287, 7) \):
     \[
     2b^2 = 287 - 7 = 280 \implies b^2 = 140 \implies b = \sqrt{140}
     \]
     Since \( \sqrt{140} \) is not an integer, this pair is not valid.

   - For \( (x, y) = (49, 41) \):
     \[
     2b^2 = 49 - 41 = 8 \implies b^2 = 4 \implies b = 2
     \]
     Since \( b = 2 \) is an integer, this pair is valid.

6. Now, we find \( a \) using \( x = a + b^2 \) and \( y = a - b^2 \):
   \[
   a + b^2 = 49 \quad \text{and} \quad a - b^2 = 41
   \]
   Substituting \( b = 2 \):
   \[
   a + 4 = 49 \implies a = 45
   \]
   \[
   a - 4 = 41 \implies a = 45
   \]

7. Therefore, \( a + b = 45 + 2 = 47 \).

The final answer is \( \boxed{47} \).