# 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.

• How can $1/2008+1/4030056$ and $1/2008+1/4028049$ both be $1/2007$? – hmakholm left over Monica Jul 24 '18 at 9:43
• There are solutions with $x=2034, 2088$ (and $4014$, but you may not accept that) – Henry Jul 24 '18 at 9:57
• How can we find your mistake if you don't tell us how you calculated them? – miracle173 Jul 24 '18 at 9:59
• I checked in excel – Samar Imam Zaidi Jul 24 '18 at 10:28

$$2007x + 2007 y = xy$$

$$0=xy-2007-2007y$$

$$2007^2=xy-2007x-2007y+2007^2$$

$$(2007^2)=(x-2007)(y-2007)$$

$$3^4\cdot 223^2=(x-2007)(y-2007)$$

\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.

• There is also $(3^2\cdot 223) \cdot (3^2\cdot 223)$ though this may be excluded by $x \lt y$ – Henry Jul 24 '18 at 9:59
• @Henry: “may be” excluded?? – TonyK Jul 30 '18 at 13:02
• @TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4\cdot 223^2$ – Henry Jul 30 '18 at 15:25
• I would prefer “is”. – TonyK Jul 30 '18 at 15:59

Write the equations as,

$$(x+y)2007=xy$$

$$xy-2007x-2007y+2007^2=2007^2$$

$$(x-2007)(y-2007)=2007^2$$

also, $2007=3^2.223$

Can you continue?