Prove that: $xy = 96$ has $24$ possible integer solutions.

Question

According to wolframalpha the equation $xy = 96$ has $24$ possible integer solutions

How is possible to determine this number of solutions?

For Example

Answer 1

Since $96=2^5\cdot 3$ it is well known that this number has $(5+1)(1+1)=12$ factors, the trivial ones $1$ and $96$ included. When we consider $xy$ and $yx$ as different solutions we have the $24$ solutions of the statement however the trivial $1\cdot96$ and $96\cdot1$ should be included as two solutions for this.

Answer 2

Hint. Note that $96=2^5\cdot 3$. So $x=\pm 2^a\cdot 3^{b}$ with $a\in\{0,1,2,3,4,5\}$ and $b\in\{0,1\}$. Then $y=96/x$.

Answer 3

Number of factors of a number which is of the form ${p_1}^{n_1}\times{p_2}^{n_2}\times{p_3}^a_{n_3}.......\times{p_n}^{n_n}$ are given by $({n_1}+1)({n_2}+1)({n_3}+1).......({n_n}+1)$. Use this and you will get that 96 have $6\times 2=12 $ factors {since $96=2^5\times 3^1$}.

Since $ xy=96$, so we get $(x,y)$ have $24$ solutions because $(x,y)$ and $(y,x)$ are not same or $(x,y)$ is an order pair.