Let us consider the following cases.

Case 1: $n$ is odd. The equation $|x^{2}-y^{2}|=n$ is equivalent to
$$|x-y||x+y|=n,$$
hence $|x-y|=d$ and $|x+y|=\frac{n}{d}$, where $d$ is a divisor of $n$. The system
$$\left\{\begin{array}{l} |x-y|=d \\ |x+y|=\frac{n}{d} \end{array}\right.$$
has four solutions, hence we have $a_{n}=4 \tau(n)$, where $\tau(n)$ is the number of divisors of $n$.

Case 2: $n=4k+2$. In this case $x^{2} \equiv y^{2} (\bmod 4)$, that is
$$x^{2}-y^{2} \equiv 0 \quad(\bmod 4),$$
giving $a_{n}=0$.

Case 3: $n=2^{s}(2k+1),\ s \geq 2$. The system (1) has no solutions when $d \mid 2k+1$ or $\left.\frac{n}{d} \right\rvert\, 2k+1$. It follows that
$$a_{n}=4(\tau(n)-2 \tau(2k+1)) .$$
Assume that $n=2^{s} p_{1}^{\alpha_{1}} \ldots p_{t}^{\alpha_{t}}$, where $p_{1}, \ldots, p_{t}$ are odd primes. Then,
$$\begin{aligned} a_{n} & =4\left((s+1)\left(\alpha_{1}+1\right) \ldots\left(\alpha_{t}+1\right)-2\left(\alpha_{1}+1\right) \ldots\left(\alpha_{t}+1\right)\right) \\ & =4(s-1)\left(\alpha_{1}+1\right) \ldots\left(\alpha_{t}+1\right) \\ & =4 \tau\left(\frac{n}{4}\right) \end{aligned}$$
Finally, we get
$$a_{n}= \begin{cases}4 \tau(n) & \text{ if } n \text{ is odd } \\ 0 & \text{ if } n=4k+2 \\ 4 \tau\left(\frac{n}{4}\right) & \text{ if } n=2^{s}(2k+1),\ s \geq 2 .\end{cases}$$

For $n=1432=2^{3} \cdot 179$, we have $a_{1432}=4 \tau(2 \cdot 179)=16$. For $n=1433$, we have $a_{1433}=4 \tau(1433)=8$, since $1433$ is a prime.