Saudi Arabia Mathematical Competitions

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 $$\{\begin{array}{l}|x-y|=d\\ |x+y|=\frac{n}{d}\end{array}$$ 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 $x^2\equiv y^2(\mathrm{mod}4)$, that is $$x^2-y^2\equiv 0(\mathrm{mod}4),$$ giving $a_n=0$.

Case 3: $n=2^s(2k+1),s\ge 2$. The system (1) has no solutions when $d\mid 2k+1$ or $\frac{n}{d}|2k+1$. It follows that $$a_n=4(\tau (n)-2\tau (2k+1)).$$ Assume that $n=2^s{p}_1^{{\alpha }_1}\dots p_t^{{\alpha }_t}$, where $p_1,\dots ,p_t$ are odd primes. Then, $$\begin{array}{rl}{a}_n& =4\left((s+1)\left({\alpha }_1+1\right)\dots \left({\alpha }_t+1\right)-2\left({\alpha }_1+1\right)\dots \left({\alpha }_t+1\right)\right)\\ & =4(s-1)\left({\alpha }_1+1\right)\dots \left({\alpha }_t+1\right)\\ & =4\tau \left(\frac{n}{4}\right)\end{array}$$ Finally, we get $$a_n=\{\begin{array}{ll}4\tau (n)& \text{ if }n\text{ is odd }\\ 0& \text{ if }n=4k+2\\ 4\tau \left(\frac{n}{4}\right)& \text{ if }n=2^s(2k+1),s\ge 2.\end{array}$$

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.