The number of positive integer pairs (x,y) such that 1x+1y=12007,x<y is
x+yxy=12007
⇒xy−2007(x+y)=0
Adding 20072 to both sides, we get
xy−2007(x+y)+20072=20072
⇒(x−2007)(y−2007)=20072
Let x−2007=A and y−2007=B
The equation becomes AB=20072
Number of solutions of above equation is equal to number of factors of 20072
20072=34×2232
Hence, number of factors of 20072 is (4+1)(2+1)=15
In one case A=B=2007
Of the remaining 14 cases, half of the case A>B and remaining half A<B
Accordingly we get 7 cases, where x<y.