If $$x, y \in Z$$ such that
$$x^2+y^2 =4$$
$$x^3+y^3 = 8$$
Find the value of $$2x+y$$

Question

If $$x, y \in Z$$ such that 
$$x^2+y^2 =4$$
$$x^3+y^3 = 8$$
Find the value of $$2x+y$$

Uniteasy Admin · Accepted Answer

$$x^3+y^3 = (x+y)(x^2-xy+y^2)$$
Then,
$$8 = (x+y)(4-xy)$$n
Since
$$x^2+y^2 =(x+y)^2-2xy=4$$n
Let 
$$m = x+y$$n
$$n = xy$$n
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 $$4$$.

Question

Answer