Find two positive integers $a,b$ such that $a | b^2, b^2 | a^3, a^3 | b^4, b^4 | a^5$, but $a^5$  does not divide $b^6$
1. Let \( a = p^x \) and \( b = p^y \), where \( p \) is a prime number and \( x, y \) are nonnegative integers.
2. We need to satisfy the following conditions:
   \[
   a \mid b^2 \implies p^x \mid p^{2y} \implies x \leq 2y
   \]
   \[
   b^2 \mid a^3 \implies p^{2y} \mid p^{3x} \implies 2y \leq 3x
   \]
   \[
   a^3 \mid b^4 \implies p^{3x} \mid p^{4y} \implies 3x \leq 4y
   \]
   \[
   b^4 \mid a^5 \implies p^{4y} \mid p^{5x} \implies 4y \leq 5x
   \]
   \[
   a^5 \nmid b^6 \implies p^{5x} \nmid p^{6y} \implies 5x > 6y
   \]
3. From the inequalities, we have:
   \[
   x \leq 2y
   \]
   \[
   2y \leq 3x \implies y \leq \frac{3}{2}x
   \]
   \[
   3x \leq 4y \implies x \leq \frac{4}{3}y
   \]
   \[
   4y \leq 5x \implies y \leq \frac{5}{4}x
   \]
   \[
   5x > 6y \implies y < \frac{5}{6}x
   \]
4. Combining these inequalities, we get:
   \[
   \frac{3}{4}x \leq y < \frac{5}{6}x
   \]
5. We need to find integers \( x \) and \( y \) that satisfy this range. For example, let \( x = 12 \):
   \[
   \frac{3}{4} \cdot 12 = 9 \quad \text{and} \quad \frac{5}{6} \cdot 12 = 10
   \]
   Thus, \( 9 \leq y < 10 \). The only integer \( y \) in this range is \( y = 9 \).

6. Therefore, one possible solution is:
   \[
   a = 2^{12}, \quad b = 2^9
   \]

The final answer is \( \boxed{ a = 2^{12}, b = 2^9 } \)