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

2 Answers
A. S. Adikesavan
Jan 20, 2017

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]}

Cesareo R.
Jan 20, 2017

#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#

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