[b]a)[/b] Show that if two non-negative integers $ p,q $ satisfy the property that both $ \sqrt{2p-q} $ and $ \sqrt{2p+q} $ are non-negative integers, then $ q $ is even.

[b]b)[/b] Determine how many natural numbers $ m $ are there such that $ \sqrt{2m-4030} $ and $ \sqrt{2m+4030} $ are both natural.
### Part A:
1. Given that \( \sqrt{2p - q} \) and \( \sqrt{2p + q} \) are non-negative integers, let:
   \[
   \sqrt{2p - q} = n \quad \text{and} \quad \sqrt{2p + q} = m
   \]
   where \( n \) and \( m \) are non-negative integers.

2. Squaring both sides, we get:
   \[
   2p - q = n^2 \quad \text{and} \quad 2p + q = m^2
   \]

3. Adding these two equations:
   \[
   (2p - q) + (2p + q) = n^2 + m^2 \implies 4p = n^2 + m^2 \implies p = \frac{n^2 + m^2}{4}
   \]

4. Subtracting the first equation from the second:
   \[
   (2p + q) - (2p - q) = m^2 - n^2 \implies 2q = m^2 - n^2 \implies q = \frac{m^2 - n^2}{2}
   \]

5. For \( q \) to be an integer, \( m^2 - n^2 \) must be even. Since \( m^2 \) and \( n^2 \) are both perfect squares, they are congruent to either 0 or 1 modulo 4:
   - If \( m \) and \( n \) are both even, \( m^2 \equiv 0 \pmod{4} \) and \( n^2 \equiv 0 \pmod{4} \), so \( m^2 - n^2 \equiv 0 \pmod{4} \).
   - If \( m \) and \( n \) are both odd, \( m^2 \equiv 1 \pmod{4} \) and \( n^2 \equiv 1 \pmod{4} \), so \( m^2 - n^2 \equiv 0 \pmod{4} \).

6. In both cases, \( m^2 - n^2 \) is divisible by 4, and thus \( q = \frac{m^2 - n^2}{2} \) is even.

\(\blacksquare\)

### Part B:
1. Given \( \sqrt{2m - 4030} \) and \( \sqrt{2m + 4030} \) are both natural numbers, let:
   \[
   \sqrt{2m - 4030} = n \quad \text{and} \quad \sqrt{2m + 4030} = k
   \]
   where \( n \) and \( k \) are natural numbers.

2. Squaring both sides, we get:
   \[
   2m - 4030 = n^2 \quad \text{and} \quad 2m + 4030 = k^2
   \]

3. Adding these two equations:
   \[
   (2m - 4030) + (2m + 4030) = n^2 + k^2 \implies 4m = n^2 + k^2 \implies m = \frac{n^2 + k^2}{4}
   \]

4. Subtracting the first equation from the second:
   \[
   (2m + 4030) - (2m - 4030) = k^2 - n^2 \implies 8060 = k^2 - n^2 \implies 8060 = (k - n)(k + n)
   \]

5. We need to find pairs \((k, n)\) such that \( k - n \) and \( k + n \) are factors of 8060. The factor pairs of 8060 are:
   \[
   (1, 8060), (2, 4030), (4, 2015), (5, 1612), (10, 806), (20, 403), (31, 260), (62, 130), (65, 124), (124, 65), (130, 62), (260, 31), (403, 20), (806, 10), (1612, 5), (2015, 4), (4030, 2), (8060, 1)
   \]

6. For each pair \((a, b)\), solve:
   \[
   k - n = a \quad \text{and} \quad k + n = b
   \]
   Adding and subtracting these equations:
   \[
   2k = a + b \implies k = \frac{a + b}{2}
   \]
   \[
   2n = b - a \implies n = \frac{b - a}{2}
   \]

7. Both \( k \) and \( n \) must be integers, so \( a + b \) and \( b - a \) must be even. Checking the factor pairs, we find:
   \[
   (2, 4030), (10, 806), (62, 130), (130, 62), (806, 10), (4030, 2)
   \]

8. For each valid pair, calculate \( m \):
   \[
   m = \frac{n^2 + k^2}{4}
   \]

9. The valid pairs yield the following values of \( m \):
   \[
   m = 2593, 12097, 81217, 2030113
   \]

Thus, there are 4 natural numbers \( m \) that satisfy the given conditions.

\(\boxed{4}\)