How many ordered pairs of integers (x, y) satisfy x^2 + 2y = xy?

Question

Can you solve the problem below:

How many ordered pairs of integers (x, y) satisfy x^2 + 2y = xy?

Answer 1

x^2 = xy - 2y

 

x^2 = y(x-2)

 

x^2/(x-2) = y

 

So any integer can be selected for x except x=2.

Those than do not result in an integer for y are 

ruled out 

 

For example consider the following table

 

X           Y

-2        4/-4 = -1  ---> solution (-2,-1)

-1        -1/3 <---- not a solution

0          0   ---> solution (0,0)

1         1/-1 = -1 ---> solution (1,-1)

3          9/1 = 9 ---> solution (3,9)

4      16/2 = 8 ---> solution (4,8)

 

you can now extend this table in both the negative and positive

directions to generate as many solutions as you like.

Answer 2

x² + 2y = xy

x² = -2y + xy

x² = y(-2 + x)

x² / (x - 2) = y

 

Unlimited number of solutions. 

Root at 0. 

Vertical asymptote at x = 2