Find the number of functions $ f: \mathbb N\rightarrow\mathbb N$ which satisfying:
(i) $ f(1) \equal{} 1$
(ii) $ f(n)f(n \plus{} 2) \equal{} f^2(n \plus{} 1) \plus{} 1997$ for every natural numbers n.
1. **Initial Setup and Substitution**:
   Given the function \( f: \mathbb{N} \rightarrow \mathbb{N} \) satisfying:
   \[
   f(1) = 1
   \]
   and
   \[
   f(n)f(n+2) = f^2(n+1) + 1997 \quad \text{for all natural numbers } n,
   \]
   we can replace 1997 with a prime number \( p \). Let \( f(2) = m \), then:
   \[
   f(3) = m^2 + p.
   \]

2. **Finding \( f(4) \)**:
   Using the given functional equation for \( n = 2 \):
   \[
   f(2)f(4) = f^2(3) + p,
   \]
   substituting \( f(2) = m \) and \( f(3) = m^2 + p \):
   \[
   mf(4) = (m^2 + p)^2 + p.
   \]
   Solving for \( f(4) \):
   \[
   f(4) = \frac{(m^2 + p)^2 + p}{m} = m^3 + 2mp + \frac{p(p+1)}{m}.
   \]
   For \( f(4) \) to be an integer, \( m \) must divide \( p(p+1) \).

3. **General Form and Recurrence**:
   We have:
   \[
   f(n)f(n+2) = f^2(n+1) + p,
   \]
   and
   \[
   f(n+1)f(n+3) = f^2(n+2) + p.
   \]
   Subtracting these equations:
   \[
   f(n)f(n+2) - f(n+1)f(n+3) = f^2(n+1) + p - (f^2(n+2) + p),
   \]
   simplifying:
   \[
   f(n)f(n+2) - f(n+1)f(n+3) = f^2(n+1) - f^2(n+2).
   \]
   Dividing both sides by \( f(n+2)f(n+3) \):
   \[
   \frac{f(n)}{f(n+3)} = \frac{f(n+1)}{f(n+2)}.
   \]
   This implies:
   \[
   \frac{f(n+1)}{f(n+2)} = \frac{f(n)}{f(n+1)}.
   \]

4. **Finding the General Solution**:
   From the above, we get:
   \[
   \frac{f(2)}{f(n)} = \frac{f(2)}{f(3)} \cdot \frac{f(3)}{f(4)} \cdots \frac{f(n-1)}{f(n)} = \frac{f(1) + f(3)}{f(2) + f(4)} \cdot \frac{f(2) + f(4)}{f(3) + f(5)} \cdots \frac{f(n-2) + f(n)}{f(n-1) + f(n+1)}.
   \]
   Simplifying:
   \[
   \frac{f(2)}{f(n)} = \frac{f(1) + f(3)}{f(n-1) + f(n+1)}.
   \]
   This implies:
   \[
   f(n-1) + f(n+1) = \frac{f(1) + f(3)}{f(2)} f(n) = \frac{p+1}{m} f(n) + mf(n).
   \]

5. **Considering \( m \)**:
   If \( m = 1 \), then:
   \[
   f(n+1) = f(n) + 1.
   \]
   This sequence is valid and integer-valued.

   If \( m > 1 \), since \( p \) and \( p+1 \) are relatively prime, \( m \) must divide either \( p \) or \( p+1 \) but not both. If \( m \) divides \( p+1 \), it works. If \( m \) divides \( p \), then \( m = p \).

6. **Checking \( m = p \)**:
   For \( m = p \):
   \[
   f(1) = 1, \quad f(2) = p, \quad f(3) = p^2 + p = p(p+1),
   \]
   \[
   f(4) = p^3 + 2p^2 + p + 1.
   \]
   For \( f(5) \):
   \[
   f(5) = \frac{(p^3 + 2p^2 + p + 1)^2 + p}{p(p+1)}.
   \]
   The numerator is not divisible by \( p \), so \( f(5) \) is not an integer.

7. **Conclusion**:
   The only valid \( m \) is 1, leading to the sequence \( f(n) = n \).

The final answer is \( \boxed{1} \).