A number theory problem by Bhaskar Arya

Whenever you see this kind of problem,this thought should come to your mind: $(x-2007)(y-2007)$ as this simplifies to $x*y-2007(x+y)+2007^{2}$ because from the given data we have: $x*y-2007(x+y)=0$ $\Longrightarrow(x-2007)(y-2007)=2007^{2}.$ Now, $2007=3^{2}*223\Longrightarrow\ 2007^{2}=3^{4}*223^{2}$ The total number of factors is equal to $5*3=15.$ A new technique that I found today was that to find the number of un-ordered pairs of factors,if you have the total number of factors you divide the total number of factors by 2 and take the smallest integer that is greater than it.In mathematical terms: $\lceil\dfrac{15}{2}=8$ Does somebody have the proof of this?

$\frac {1}{x} + \frac {1}{y} = \frac {1}{2007}$

$2007x + 2007y = xy$

$2007x= y (x-2007)$

$y= \frac {2007x}{x-2007}$

Since $x,y$ are natural no.s. so using the relation we can put the values of $x$ to get the value of $y$ .

We will finally get 8 unordered pairs that are

$x=2008,\ y=4030056$

$x=2010,\ y=1344690$

$x=2016,\ y=449568$

$x=2034,\ y=151194$

$x=2088,\ y=51736$

$x=2230,\ y=20070$

$x=2676,\ y=80208$

$x=4014,\ y=4014$

Number of ordered solutions= number of factors of (2007)^2=15 for one value x=y and 14 ordered pairs= 7 unordered pairs therefore total number of solutions = 8

multiplying both sides with 2007xy $2007(x+y)=xy\longrightarrow xy-2007(x+y)+2007^2=2007^2$ $(x-2007)(y-2007)=2007^2$ lets see the number of factors f(2007^2) $2007^2=3^4*233^2$ $f(2007^2)=(4+1)(2+1)=15$ the number of order pairs are 15 and $no.\quad of\quad unordered\quad pair=\lceil \dfrac{no.\quad of\quad ordered\quad pair}{2}\rceil=\boxed{8}$

There are 8-

2008, 4030056

2010,1344690

2016,449568

2034,151194

2088,51736

2230,20070

2676,8028

4014,4014

I used a c++ program for it.