# Integer solutions for x^3 - 3 x y^2 + 2 y^3=2017 ?

Jan 20, 2017

The solution is the set of points (x, y ) on the graph. for $f \left(x , y\right) = {x}^{3} + 3 x {y}^{2} + 2 {y}^{3} - 2017 = 0$. See the asymptotes-inclusive graphical depiction for f ( x, y, ) = 0.

#### Explanation:

The x-intercept is (2017)^(1/3)=12.635, nearly.

y-intercept ; (2017/2)^(1/3)=10.028, nearly.

The graph depicts the solution set {( x, y ) }. It seems that there are

asymptotes given by (x-y)(x+2y)=0, in proximity.

graph{(x^3-3xy^2+2y^3-2017)(x-y)(x+2y) =0 [-49.5, 49.5, -24.75, 24.75]}

Jan 20, 2017

$x = 673$ and $y = 672$

#### Explanation:

We will looking for solutions $x , y \in \mathbb{Z}$

This is a homogeneous equation. So doing $y = \lambda x$ we get

${x}^{3} \left(1 - 3 {\lambda}^{2} + 2 {\lambda}^{3}\right) = 2 {x}^{3} {\left(\lambda - 1\right)}^{2} \left(\lambda + \frac{1}{2}\right) = 2 {\left(y - x\right)}^{2} \left(y + \frac{x}{2}\right) = {\left(x - y\right)}^{2} \left(x + 2 y\right) = 2017$

so

${x}^{3} - 3 x {y}^{2} + 2 {y}^{3} = {\left(x - y\right)}^{2} \left(x + 2 y\right) = 2017$

but $2017$ is a prime number so

$\left\{\begin{matrix}{\left(x - y\right)}^{2} = 1 \\ x + 2 y = 2017\end{matrix}\right.$

or

$\left\{\begin{matrix}{\left(x - y\right)}^{2} = 2017 \\ x + 2 y = 1\end{matrix}\right.$

The solution is given by the first system and is

$x = 673$ and $y = 672$