Find all functions $f$ defined on the integers greater than or equal to $1$ that satisfy:
(a) for all $n,f(n)$ is a positive integer.
(b) $f(n + m) =f(n)f(m)$ for all $m$ and $n$.
(c) There exists $n_0$ such that $f(f(n_0)) = [f(n_0)]^2$ .
1. **Using the functional equation \( f(n + m) = f(n)f(m) \):**

   Let's denote \( f(1) = a \). We will use induction to show that \( f(n) = a^n \) for all \( n \geq 1 \).

   - **Base case:** For \( n = 1 \), \( f(1) = a \).
   - **Inductive step:** Assume \( f(k) = a^k \) for some \( k \geq 1 \). Then for \( k + 1 \):
     \[
     f(k + 1) = f(k)f(1) = a^k \cdot a = a^{k+1}
     \]
     By induction, \( f(n) = a^n \) for all \( n \geq 1 \).

2. **Using the condition \( f(f(n_0)) = [f(n_0)]^2 \):**

   Given \( f(n) = a^n \), we need to find \( n_0 \) such that:
   \[
   f(f(n_0)) = [f(n_0)]^2
   \]
   Substituting \( f(n_0) = a^{n_0} \):
   \[
   f(a^{n_0}) = (a^{n_0})^2
   \]
   Since \( f(n) = a^n \), we have:
   \[
   a^{a^{n_0}} = a^{2n_0}
   \]
   This implies:
   \[
   a^{n_0} = 2n_0
   \]

3. **Solving \( a^{n_0} = 2n_0 \):**

   - If \( a = 1 \):
     \[
     1^{n_0} = 2n_0 \implies 1 = 2n_0 \implies n_0 = \frac{1}{2}
     \]
     This is not possible since \( n_0 \) must be an integer.

   - If \( a = 2 \):
     \[
     2^{n_0} = 2n_0 \implies 2^{n_0 - 1} = n_0
     \]
     We need to find integer solutions to \( 2^{n_0 - 1} = n_0 \):
     - For \( n_0 = 1 \):
       \[
       2^{1 - 1} = 1 \implies 1 = 1
       \]
       This is true.
     - For \( n_0 = 2 \):
       \[
       2^{2 - 1} = 2 \implies 2 = 2
       \]
       This is also true.
     - For \( n_0 \geq 3 \):
       \[
       2^{n_0 - 1} > n_0
       \]
       This can be shown by induction or by noting that \( 2^{n_0 - 1} \) grows exponentially while \( n_0 \) grows linearly.

4. **Conclusion:**

   The only possible values for \( a \) are 1 and 2. Therefore, the functions that satisfy all the given conditions are:
   \[
   f(n) = 1 \quad \text{and} \quad f(n) = 2^n
   \]

The final answer is \( \boxed{ f(n) = 1 } \) and \( f(n) = 2^n \).