Question
If x, y \in Z such that
x^2+y^2 =4
x^3+y^3 = 8
Find the value of 2x+y
If x, y \in Z such that
x^2+y^2 =4
x^3+y^3 = 8
Find the value of 2x+y
x^3+y^3 = (x+y)(x^2-xy+y^2)
Then,
Since
Let
Then
4m -mn -8 = 0 \\ m^2-2n-4 = 0
n = \dfrac{4m-8}{m}
m^2-2\cdot \dfrac{4m-8}{m} - 4 = 0
m^3-12m+16 = 0
m = 2
m^3-12m+16
=m^3-8-12m+24
=(m-2)(m^2+2m+4)-12(m-2)
=(m-2)(m^2+m-8)=0
Therefore,
m is the only integer root for the cubic equation.
Then,
n = \dfrac{4m-8}{m}
=\dfrac{4*2-8}{2} = 0
therefore, x=0,y=2 or x=2, y=0
2x+y = 2 or 4.