# One 3rd degree equation with two variables. Can it be a solution?

1. Feb 7, 2013

### Ceva

Hello this is my first post here. I found a problem in algebra which i cannot solve and don't even know if there are solutions as asked.
It is that one:

1. The problem statement, all variables and given/known data
Find all the integer solutions (x,y) that satisfy the equation:

2. Relevant equations
×3-727× = y3 - 727y

3. The attempt at a solution
x3 - y3 = 727x - 727y
(x-y)(x2+xy+y2)=727(x-y)

I didn't divided by (x-y) because i think i'll lose solutions. Also, x-y can be zero (i think).

So i finally have: (x-y)(x2+xy+y2-727)=0

How can i continue? Can i factorize more the second bracket? Or i can solve it graphically?

Also, x-y=0 doesn't mean that i have infinite solutions? If i set myself x and y Ε Z .

2. Feb 7, 2013

If x = y, then (x-y) = 0 and you have a set of solutions (a straight line).
If x <> y, then your other solutions satisfy x^2 + xy + y^2 = 727 (an ellipse).

3. Feb 7, 2013

### Ray Vickson

You can have x = y = any integer, so already there are infinitely many solutions. To look for solutions with x ≠ y, divide out your above expression by (x-y)---which IS allowed when x ≠ y-- to get x^2 + x*y + y^2 = 727. Using various methods you can find the maximum and minimum possible values of x, which are
$$x_{\min} = -\frac{2}{3}\sqrt{2181} \doteq -31.1342 \\ x_{\max} = \frac{2}{3}\sqrt{2181} \doteq 31.1342 ,$$
so the only integer values of x you need to look at are those from x = -31 to +31. For each x you can solve the quadratic equation to find the two values of y. You can just use a loop over x from -31 to + 31 and look at all the y-values to see if any of them come out as integers.

4. Feb 9, 2013

### Ceva

Okay, i found the solution. I solved with respect to y and what was over the root must be a square number in order to be an integer. From there it wasn't difficult to find the solutions. Thanks for your help!