Question

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

Answer 1

The solution is the set of points (x, y ) on the graph. for `f( x, y )=x^3+3xy^2+2y^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]}

Answer 2

`x=673` and `y=672`

Explanation:

We will looking for solutions `x,y in ZZ`

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

`x^3(1-3lambda^2+2lambda^3)=2x^3(lambda-1)^2(lambda+1/2)=2(y-x)^2(y+x/2) = (x-y)^2(x+2y)=2017`

so

`x^3 - 3 x y^2 + 2 y^3=(x - y)^2 (x + 2 y)=2017`

but `2017` is a prime number so

`{((x-y)^2=1),( x + 2 y=2017):}`

or

`{((x-y)^2=2017),( x + 2 y=1):}`

The solution is given by the first system and is

`x=673` and `y=672`