Find all pair of integers $(m,n)$ and $m \ge  n$ such that there exist a positive integer $s$ and
a) Product of all divisor of $sm, sn$ are equal.
b) Number of divisors of $sm,sn$ are equal.
1. **Understanding the product of all divisors**:
   - The product of all divisors of a number \( t \) is given by \( t^{\frac{\tau(t)}{2}} \), where \( \tau(t) \) denotes the number of divisors of \( t \).
   - Given that the product of all divisors of \( sm \) and \( sn \) are equal, we have:
     \[
     (sm)^{\frac{\tau(sm)}{2}} = (sn)^{\frac{\tau(sn)}{2}}
     \]
   - This implies:
     \[
     sm^{\frac{\tau(sm)}{2}} = sn^{\frac{\tau(sn)}{2}}
     \]

2. **Equal number of divisors**:
   - The number of divisors of \( sm \) and \( sn \) are equal, i.e., \( \tau(sm) = \tau(sn) \).
   - Let \( \tau(sm) = \tau(sn) = \tau \).

3. **Prime factorization and exponents**:
   - Since \( sm \) and \( sn \) share the same set of prime divisors, we can write:
     \[
     sm = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \quad \text{and} \quad sn = p_1^{f_1} p_2^{f_2} \cdots p_k^{f_k}
     \]
   - Given \( \tau(sm) = \tau(sn) \), we have:
     \[
     \prod_{i=1}^k (e_i + 1) = \prod_{i=1}^k (f_i + 1)
     \]

4. **Equality of exponents**:
   - Suppose \( sm \neq sn \). Then there exists at least one prime \( p \) such that \( v_p(sm) \neq v_p(sn) \).
   - Without loss of generality, assume \( v_p(sm) < v_p(sn) \). This implies:
     \[
     \frac{\tau(sm)}{2} v_p(sm) < \frac{\tau(sn)}{2} v_p(sn)
     \]
   - Since \( \tau(sm) = \tau(sn) \), we get:
     \[
     v_p(sm) < v_p(sn)
     \]
   - This must hold for all primes dividing \( sm \) and \( sn \).

5. **Contradiction**:
   - If \( v_p(sm) < v_p(sn) \) for all primes \( p \), then:
     \[
     \tau(sm) = \prod_{i=1}^k (e_i + 1) < \prod_{i=1}^k (f_i + 1) = \tau(sn)
     \]
   - This contradicts the assumption that \( \tau(sm) = \tau(sn) \).

6. **Conclusion**:
   - Therefore, \( sm = sn \) must hold, implying \( m = n \).

The final answer is \( \boxed{ (m, n) } \) such that \( m = n \).