Please help with finding all integers to an equation

[Peter159]

How can i find all integers to this equation?

3(x+y)-xy=0

I already found just by tring that if x=y then its 6 and x=12 ,y=4 (x=4 and y=12)

I think that there are more, but how can i find all of them?

PS. the first task was: 1/x + 1/y = 1/3 and then i had to find all integers.

Thank you for your help!

Sorry for my bad english.

 

[Dick]

I would write it in the form x=3*y/(y-3). For large values of y that's a little larger than 3 and not an integer. So once you tried a y value large enough that 3*y/(y-3)<4, there aren't any larger solutions. Then I'd just try all y values less than that to see if x is also an integer. Do you need negative solutions as well?

 

[snipez90]

$$\frac{1}{x} + \frac{1}{y} = \frac{1}{3} \Rightarrow 3x + 3y = xy \Rightarrow xy - 3x - 3y + 9 = 9 \Rightarrow (x-3)(y-3) = 9$$.

The number 9 has three positive divisors: 1, 3, and 9. You just have to solve each of the systems of equations to get all the solutions.

 

[Dick]

$$\frac{1}{x} + \frac{1}{y} = \frac{1}{3} \Rightarrow 3x + 3y = xy \Rightarrow xy - 3x - 3y + 9 = 9 \Rightarrow (x-3)(y-3) = 9$$.

The number 9 has three positive divisors: 1, 3, and 9. You just have to solve each of the systems of equations to get all the solutions.

Very clever.

 

[Peter159]

Ok, thanks Dick and Snipez90 for your help!

I have one question. xy-3x-3y+9=9 > (x-3)(y-3) How did you factor that one?

Do you need negative solutions as well?

No, but thanks for asking.

EDIT:OH, No need for help on this question any more. Just figured it out how to do that :D

 

[NoMoreExams]

By inspection I would imagine, you have a cross term xy so you know that if you will have a product of linear factors it will be of the form (x-alpha)*(y-beta) [to give you that xy term] and then the choices for alpha and beta are obvious once you write out the product.