Find all values of the positive integer $ m$ such that there exists polynomials $ P(x),Q(x),R(x,y)$ with real coefficient satisfying the condition: For every real numbers $ a,b$ which satisfying $ a^m-b^2=0$, we always have that $ P(R(a,b))=a$ and $ Q(R(a,b))=b$.
1. **Case 1: \( m = 2k \) (even \( m \))**

   Suppose \( m = 2k \). Let \( R(x, x^k) = L(x) \). We need to satisfy the conditions:
   \[
   P(R(a, b)) = a \quad \text{and} \quad Q(R(a, b)) = b
   \]
   for every \( a, b \) such that \( a^m - b^2 = 0 \). This implies \( b = \pm a^k \).

   Since \( P(R(x, x^k)) = x \), and knowing that \( P \) and \( R \) are not constant polynomials, we have:
   \[
   \deg(P) = \deg(L) = 1
   \]
   Let \( L(x) = ux + v \). Then:
   \[
   P(x) = \frac{x - v}{u}
   \]
   and also:
   \[
   Q(x) = P(x)^k
   \]
   Therefore, \( b = a^k \) or \( b = -a^k \).

2. **Case 2: \( m = 2k + 1 \) (odd \( m \))**

   Suppose \( m = 2k + 1 \). Let \( R(x^2, x^m) = L(x) \). We need to satisfy:
   \[
   P(R(x^2, x^m)) = x^2
   \]
   If \( \deg(L) = 2 \), then:
   \[
   Q(L(x)) = x^m
   \]
   Since \( m \) is odd, this leads to a contradiction because the degrees do not match. Therefore, \( \deg(L) = 1 \).

   However, in \( R(x^2, x^m) \), the minimum power of \( x \) is \( \min(2, m) \). For \( \deg(L) = 1 \), we must have \( m = 1 \).

   For \( m = 1 \):
   \[
   R(a, b) = b
   \]
   and:
   \[
   P(a) = a^2, \quad Q(a) = a
   \]

Thus, the only value of \( m \) that satisfies the given conditions is \( m = 1 \).

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