Making an Integer
Given that and , what is the number of ordered pairs of integers such that is an integer?
Inspired by : 1998 IMO Problem Number 4
The answer is 6.
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1 solution
We have x y 2 + y + 1 1 ∣ x 2 y + x + y , hence x y 2 + y + 1 1 ∣ y ( x 2 y + x + y ) − x ( x y 2 + y + 1 1 ) = y 2 − 1 1 x Now if y 2 − 1 1 x = 0 we get a family of solutions, where ( x , y ) = ( 1 1 k 2 , 1 1 k ) .
If y 2 − 1 1 x > 0 , the relation x y 2 + y + 1 1 ∣ y 2 − 1 1 x cannot be true. Because y 2 − 1 1 x < x y 2 + y + 1 1 .
If y 2 − 1 1 x < 0 , we must have 1 1 x − y 2 ≥ x y 2 + y + 1 1 or y 2 < 1 1 (compare the coefficient of x on two sides of inequality). Thus, in this case y < 4 . It turns out that y = 1 and y = 2 work. For y = 1 we get x + 1 2 ∣ 1 1 x − 1 or x + 1 2 ∣ 1 1 ( x + 1 2 ) − 1 1 x + 1 = 1 3 3 , which gives x = 7 and x = 1 2 1 . For y = 2 we get 4 x + 1 3 ∣ 1 1 x − 4 or 4 x + 1 3 ∣ 1 1 ( 4 x + 1 3 ) − 4 ( 1 1 x − 4 ) = 1 5 9 , which gives x = 1 0 .
So we get 6 solutions with 0 < x < 1 2 2 and 0 < y < 1 2 2 and they are ( 1 1 , 1 1 ) , ( 4 4 , 2 2 ) , ( 9 9 , 3 3 ) , ( 7 , 1 ) , ( 1 2 1 , 1 ) , ( 1 0 , 2 )