Integer solutions of an equation that is set to a number How many integer solutions for $a$ and $b$ in $(ab)/(a+b)=3600$?


My attempt:

$(ab)/(a+b)=3600$

= $ab=3600(a+b)$

= $ab=3600a+3600b$

=$ab=3600a=3600b$

Dividing $3600b$ on both sides

=$a(1-3600)/3600$

I am not really sure if this is correct. I was just doing this for fun after I was reading this in some mathematical book. It just explained about a few things not much. Can someone please help me with this? I was trying this for a while now and wanted to know what it would look like.
HINT : Note that 
$$\begin{align}\frac{ab}{a+b}=3600&amp;\iff ab=3600(a+b)\ \ \text{and}\ \ a+b\not=0\\&amp;\iff (a-3600)(b-3600)=3600^2\ \ \text{and}\ \ a+b\not=0.\end{align}$$
Can you take it from here?
We get $ab=3600(a+b)$ or $-3600a+ab-3600b=0$. We can factor $-3600a+ab$ as $a(-3600+b)$. We get $$a(-3600+b)-3600b=0$$. Adding $3600^2$ to both sides gives $$a(-3600+b)+3600(3600)-3600b=3600^2$$ or equivalently, $$a(-3600+b)-3600(-3600)-3600b=3600^2$$. 

This factors as $a(-3600+b)-3600(-3600+b)$ or equivanlently, $$(a-3600)(-3600+b)=3600^2$$ .

The prime factorization of $3600^2$ is $(2^4 \cdot 3^2 \cdot 5^2)^2$ or $2^8 \cdot 3^4 \cdot 5^4$. There are $225$ factors of $3600^2$.

$(a-3600)(-3600+b)$ can each be factors of $3600^2$ and since there are $225$ factors, there are 225 integer solutions for positive $a$ and $b$.

There are also $225$ integer solutions for negative $a$ and $b$. 

Now we must deal with the condition where $a+b \neq 0$. We can let $a=-b$ and plug this into our equation to find our wrong solutions.

We get $$(-b-3600)(b-3600)=3600^2$$. 

$(-b-3600)$ must equal $\pm 3600$. $3600$ factors as $2^4 \cdot 3^2 \cdot 5^2$ so there are $45$ solutions for positive $b$ and $45$ solutions for negative $b$. So there are $90$ wrong solutions.

In total, we have $225+225-90=\boxed{360}$ integer values for $a$ and $b$.