For $\forall$ $m\in \mathbb{N}$ with $\pi (m)$ we denote the number of prime numbers that are no bigger than $m$. Find all pairs of natural numbers $(a,b)$ for which there exist polynomials $P,Q\in \mathbb{Z}[x]$ so that for 
$\forall$ $n\in \mathbb{N}$ the following equation is true:
$\frac{\pi (an)}{\pi (bn)} =\frac{P(n)}{Q(n)}$.

1. **Claim**: There exist infinitely many \( n \) such that both intervals \( (na, na+a] \) and \( (nb, nb+b] \) contain only composite numbers.

   **Proof**: For the sake of contradiction, suppose there exists \( N \in \mathbb{N} \), and for all \( n \ge N \) there is a prime either in \( (na, na+a] \) or in \( (nb, nb+b] \) (or in both). Then for any \( k \in \mathbb{N} \), in some of the intervals \( (Na, Na+ka] \) or \( (Nb, Nb+kb] \) there exist at least \( k/2 \) prime numbers. This implies that the prime numbers have a positive density, which contradicts the prime number theorem (PNT). \(\blacksquare\)

2. For those \( n \) satisfying the above claim, we have \( \pi((n+1)a) = \pi(na) \) and \( \pi((n+1)b) = \pi(nb) \). Hence,
   \[
   \frac{P(n)}{Q(n)} = \frac{P(n+1)}{Q(n+1)}
   \]
   for infinitely many natural numbers \( n \). This yields
   \[
   P(z)Q(z+1) = P(z+1)Q(z), \quad \forall z \in \mathbb{C} \quad \quad \quad (1)
   \]

3. If \( z_0 \in \mathbb{C} \) is a root of \( P(z) \) with maximal \( \mathrm{Re}(z) \), the above identity implies \( z_0 \) is also a root of \( Q(z) \). Note that \( P_1(z) := \frac{P(z)}{z - z_0} \) and \( Q_1(z) := \frac{Q(z)}{z - z_0} \) also satisfy \( (1) \). Thus, applying that operation several times, we get \( P(z) = cQ(z) \) for some constant \( c \in \mathbb{C} \).

4. By the prime number theorem (PNT), \( \pi(an) \sim \frac{an}{\ln(an)} \), so it easily follows that \( c = \frac{a}{b} \). Hence,
   \[
   \frac{\pi(an)}{\pi(bn)} = \frac{a}{b}
   \]

5. It gives
   \[
   \frac{\pi(an+a) - \pi(an)}{\pi(bn+b) - \pi(bn)} = \frac{a}{b}
   \]

6. Let \( n = kb! \). Then \( bn+2, bn+3, \dots, bn+b \) are all composite numbers. Consider the sequence \( kb \cdot b! + 1, k = 1, 2, \dots \). By Dirichlet's theorem, there exists \( k \in \mathbb{N} \) such that \( kb \cdot b! + 1 \) is a prime. For that \( k \), we get \( \pi(bn+b) - \pi(bn) = 1 \). It yields \( \frac{a}{b} \) is an integer. With the same trick, we can find \( n \) such that \( \pi(an+a) - \pi(an) = 1 \), which yields \( \frac{b}{a} \) is an integer.

7. Hence, the only possibility is \( a = b \). In this case, the condition of the statement obviously holds.

The final answer is \( \boxed{ (a, b) = (k, k) } \) for any \( k \in \mathbb{N} \).