For any positive integer $n$, let $a_n$ be the number of pairs $(x,y)$ of integers satisfying $|x^2-y^2| = n$. 
(a) Find $a_{1432}$ and $a_{1433}$. 
(b) Find $a_n$ .
1. **Expressing the given equation in a factored form:**
   \[
   |x^2 - y^2| = n \implies (x+y)(x-y) = n
   \]
   This means we need to find pairs \((a, b)\) such that \(ab = n\).

2. **Considering the cases for \(a_{1432}\):**
   \[
   1432 = 2^3 \cdot 179
   \]
   We need to find pairs \((a, b)\) such that \(ab = 1432\) and \(a + b\) is even. This is because:
   \[
   x = \frac{a+b}{2}, \quad y = \frac{a-b}{2}
   \]
   Both \(x\) and \(y\) must be integers, so \(a + b\) and \(a - b\) must be even.

3. **Finding valid pairs \((a, b)\):**
   - The factors of \(1432\) are \(1, 2, 4, 8, 179, 358, 716, 1432\).
   - We need to pair these factors such that their sum is even.
   - Valid pairs are \((2, 716), (4, 358), (8, 179)\) and their reverses.

4. **Counting the solutions:**
   - Each valid pair \((a, b)\) gives two solutions: \((x, y)\) and \((y, x)\).
   - There are 4 valid pairs, each contributing 2 solutions.
   - Therefore, \(a_{1432} = 4 \times 2 = 8\).

5. **Considering the cases for \(a_{1433}\):**
   \[
   1433 \text{ is prime}
   \]
   The only pairs \((a, b)\) such that \(ab = 1433\) are \((1, 1433)\) and \((1433, 1)\).

6. **Counting the solutions:**
   - For \((1, 1433)\), we have:
     \[
     x+y = 1433, \quad x-y = 1 \implies x = \frac{1434}{2}, \quad y = \frac{1432}{2}
     \]
     - This gives one solution.
   - For \((1433, 1)\), we have:
     \[
     x+y = 1, \quad x-y = 1433 \implies x = \frac{1434}{2}, \quad y = \frac{-1432}{2}
     \]
     - This gives one solution.
   - Each pair contributes 2 solutions (one for \(x^2 - y^2 = 1433\) and one for \(y^2 - x^2 = 1433\)).

7. **Summarizing the solutions:**
   - Therefore, \(a_{1433} = 2 + 2 = 4\).

The final answer is \( \boxed{ a_{1432} = 8 } \) and \(a_{1433} = 4\).