The radius $r$ of a circle with center at the origin is an odd integer. 
There is a point ($p^m, q^n$) on the circle, with $p,q$ prime numbers and $m,n$ positive integers. 
Determine $r$.
1. Given that the radius \( r \) of a circle with center at the origin is an odd integer, and there is a point \((p^m, q^n)\) on the circle, where \( p \) and \( q \) are prime numbers and \( m \) and \( n \) are positive integers, we need to determine \( r \).

2. Since the point \((p^m, q^n)\) lies on the circle, it satisfies the equation of the circle:
   \[
   p^{2m} + q^{2n} = r^2
   \]
   Given that \( r \) is an odd integer, \( r^2 \) is also odd. For the sum \( p^{2m} + q^{2n} \) to be odd, one of \( p \) or \( q \) must be 2 (since the square of an even number is even, and the square of an odd number is odd).

3. Without loss of generality, assume \( p = 2 \). Then the equation becomes:
   \[
   2^{2m} + q^{2n} = r^2
   \]

4. Rearrange the equation to:
   \[
   q^{2n} = r^2 - 2^{2m}
   \]
   Notice that \( r^2 - 2^{2m} \) must be a perfect power of the prime \( q \).

5. Factorize the right-hand side:
   \[
   q^{2n} = (r - 2^m)(r + 2^m)
   \]
   Since \( q \) is a prime number, and \( r \) is odd, \( r - 2^m \) and \( r + 2^m \) must be powers of \( q \).

6. Let \( r - 2^m = q^k \) and \( r + 2^m = q^{2n-k} \) for some integer \( k \). Then:
   \[
   2^{m+1} = q^{2n-k} - q^k
   \]
   Since \( q \) is a prime number greater than 2, the only way for the right-hand side to be a power of 2 is if \( k = 0 \). Thus:
   \[
   q^{2n} = 2^{m+1} + 1
   \]

7. This implies:
   \[
   (q^n - 1)(q^n + 1) = 2^{m+1}
   \]
   Since \( q \) is a prime number, \( q^n - 1 \) and \( q^n + 1 \) are consecutive even numbers. The only powers of 2 that are consecutive are 2 and 4. Therefore:
   \[
   q^n - 1 = 2 \quad \text{and} \quad q^n + 1 = 4
   \]
   Solving these, we get:
   \[
   q^n = 3
   \]
   Since \( q \) is a prime number, \( q = 3 \) and \( n = 1 \).

8. Substituting \( q = 3 \) and \( n = 1 \) back into the equation:
   \[
   3^2 = 2^{m+1} + 1
   \]
   \[
   9 = 2^{m+1} + 1
   \]
   \[
   2^{m+1} = 8
   \]
   \[
   m + 1 = 3
   \]
   \[
   m = 2
   \]

9. Finally, substituting \( m = 2 \) back into the expression for \( r \):
   \[
   r = 2^m + q^k = 2^2 + 1 = 4 + 1 = 5
   \]

The final answer is \( \boxed{5} \).