# Equation with two variables (integers)

1. Apr 24, 2012

### staples

1. The problem statement, all variables and given/known data
Solve the equation (x & y are integers):
$(x^3+4)(xy^2-x^2y+3y^2-12)=x^6$

2. Relevant equations
3. The attempt at a solution

$$xy^2-x^2y+3y^2-12=\frac{x^6}{x^3+4} \\ xy^2-x^2y+3y^2-12=x^3-4 + \frac{16}{x^3+4} \\ 16 \geq x^3+4 \\ x^3 \leq 12$$

That's all I can think of to do. I've tried expanding it and it doesn't seem to help. Any hints, please? Thanks.

2. Apr 24, 2012

### Staff: Mentor

Hi staples! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
Continuing on with your approach, we see that with all other terms being integers, then 16/(x3 +4) must also be a whole number, +ve or -ve, so that narrows it down to just a few possibilities for you to try for x candidates. Then try these one at a time to see whether you can find any corresponding y solution/s.

Last edited by a moderator: May 5, 2017
3. Apr 25, 2012

Thanks :)

4. May 2, 2012

### tiny-tim

ooh, nice welcome-smilie, NascentOxygen!

(is that a self-portrait? )

Last edited by a moderator: May 6, 2017
5. May 4, 2012

### Staff: Mentor

Indeed. It's the spitting image.