# Find all ordered pair of integers for a diophantine equation

1. Sep 16, 2009

### choob

1. The problem statement, all variables and given/known data
find all orderde pairs of integers (x,y) such that x^2+y^2=4x+2y

2. Relevant equations

3. The attempt at a solution
rearrange to--> x^2=4x+2y-y^2
because x and y can only be integers, y(2-y) must be divisible by x
so y(2-y)>=x
y(2-y)=x(x-4)
x(x-4)>=x
x-4>=1
x>=5

however, if i test with the ordered pair (3, 3) the diophantine equation is true
x^2+y^2=4x+2y
9+9=12+6
18=18

2. Sep 16, 2009

### Dick

Complete the square in both x and y. I.e. write it as (x-a)^2+(y-b)^2=c. Figure out a, b and c. Now try and solve it again. You'll find it's pretty easy.

3. Sep 16, 2009

### ehild

y(y-2) does not need to be greater than x. It can be equal, and can be negative.
You have the equation in form y(2-y)=x(x-4). At what condition are both sides positive, 0, and negative?

ehild

4. Sep 19, 2009

### choob

@dick

if i complete the square i get (x-2)^2+(y-sqrt(2))^2=6

i have no idea what to do from there

@ehild

x(x-4)=0 when y=0 or y=2
y(2-y)=0 when x=0 or x=4

so (0, 0) is one pair

y(2-y)=x(x-4)

LS is +ve when 2>y>0
RS is +ve when x>4

LS is -ve when y>2 or y<0
RS is -ve when x<4

there are no integer solutions when ls and rs are +ve

Last edited: Sep 19, 2009
5. Sep 19, 2009

### Dick

When I complete the squares I get (x-2)^2+(y-1)^2=5. How did you get the sqrt(2) in there?