Find all positive integers $ a$ and $ b$ such that $ \frac{a^{4}\plus{}a^{3}\plus{}1}{a^{2}b^{2}\plus{}ab^{2}\plus{}1}$ is an integer.
To find all positive integers \( a \) and \( b \) such that \( \frac{a^4 + a^3 + 1}{a^2 b^2 + ab^2 + 1} \) is an integer, we start by setting:

\[
\frac{a^4 + a^3 + 1}{a^2 b^2 + ab^2 + 1} = k
\]

where \( k \) is a positive integer. This implies:

\[
a^4 + a^3 + 1 = k(a^2 b^2 + ab^2 + 1)
\]

Expanding and rearranging terms, we get:

\[
a^4 + a^3 + 1 = k a^2 b^2 + k ab^2 + k
\]

\[
a^4 + a^3 + 1 - k a^2 b^2 - k ab^2 - k = 0
\]

We need to find conditions under which this equation holds for positive integers \( a \) and \( b \).

1. **Case \( k = 1 \):**

   If \( k = 1 \), the equation simplifies to:

   \[
   a^4 + a^3 + 1 = a^2 b^2 + ab^2 + 1
   \]

   Subtracting 1 from both sides, we get:

   \[
   a^4 + a^3 = a^2 b^2 + ab^2
   \]

   Factoring out common terms:

   \[
   a^3(a + 1) = b^2 a(a + 1)
   \]

   Since \( a \neq 0 \) and \( a + 1 \neq 0 \), we can divide both sides by \( a(a + 1) \):

   \[
   a^2 = b^2
   \]

   Taking the square root of both sides, we get:

   \[
   a = b
   \]

2. **Case \( k > 1 \):**

   If \( k > 1 \), we analyze the inequality:

   \[
   (a^2 + a)(a^2 - k b^2) = k - 1 \geq 0
   \]

   For the product \((a^2 + a)(a^2 - k b^2)\) to be non-negative, we consider the following:

   - \( a^2 + a > 0 \) for all positive \( a \).
   - \( a^2 - k b^2 \geq 0 \) implies \( a^2 \geq k b^2 \).

   Since \( k > 1 \), this implies \( a^2 > b^2 \). However, if \( a^2 > k b^2 \), then \( b^2 \) must be very small compared to \( a^2 \). The only integer solution that fits this condition is \( b = 0 \), but since \( b \) must be positive, this case does not provide any valid solutions.

Therefore, the only solution is when \( k = 1 \) and \( a = b \).

The final answer is \( \boxed{ a = b } \).