How many integer solutions for $a$ and $b$ in $(ab)/(a+b)=3600$?
My attempt:
$(ab)/(a+b)=3600$
= $ab=3600(a+b)$ where $a+b\not=0$
= $ab=3600a+3600b$
=$ab-3600a-3600b$
=$(a-b)3600$
=$3600a-3600b$
=$(a-3600)(b-3600)$
=$ab-3600a-3600b+3600^2$ where $a+b\not=0$
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. I was trying this for a while now and I am not sure what to do. Can someone please help me with this solution?
Rewrite the equation as $ab-3600a-3600b=0$, and then as $(a-3600)(b-3600)=3600^2$.
So $a-3600$ must be a divisor $s$, positive or negative, of $3600^2$, and $b-3600$ must be $t$, where $t=\frac{3600^2}{s}$. Conversely, if we let $s$ be any divisor of $3600^2$, and let $t$ be as above, then $a=3600+s$, $b=3600+t$ is a solution.
Now we ask how many divisors $s$ of $3600^2$ are there? Note that $3600=2^4\cdot 3^2\cdot 5^2$, so $3600^2=2^8\cdot 3^4\cdot 5^4$.
Thus $3600^2$ has $(9+1)(4+1)(4+1)$ positive divisors, and an equal number of negative, for a total of $(2)(9)(5)(5)$.
Another way: The following start is less symmetrical, so I like it less. We can rewrite our equation as $$b=3600+\frac{3600^2}{a-3600}.$$ So again we want $a-3600$ to be a divisor of $3600^2$.