<s>[INST] Solve $\frac{1}{x}+\frac{1}{y}= \frac{1}{2007}$ The number of positive integral pairs $(x<y)$ such that $\frac{1}{x}+\frac{1}{y}= \frac{1}{2007}$ 
The answer is 7 where as i am getting 6.
The ordered pair are (2676,8028),(2230,20070),(2016,449568),(2010,1344690),(2008,4030056)&(2008,4028049).
I cannot find my mistake.
 [/INST] $$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4\cdot 223^2=(x-2007)(y-2007)$$
\begin{align}3^4\cdot 223^2 &=(3^0) \cdot (3^4\cdot 223^2)\\
&=(3^1) \cdot (3^3\cdot 223^2)
\\
&=(3^2) \cdot (3^2\cdot 223^2)
\\
&=(3^3) \cdot (3^1\cdot 223^2)\\
&=(3^4) \cdot (3^0\cdot 223^2)\\
&=(3^0\cdot 223) \cdot (3^4\cdot 223)\\
&=(3^1\cdot 223) \cdot (3^3\cdot 223)
\end{align}
I hope you can recover $x$ and $y$ from here.
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