Integral Solutions of $x+y=x^2-xy+y^2$

Question

Find all integral solutions of $$x+y=x^2-xy+y^2$$

A modulo 2 analysis does not work here, only says cannot both be odd.

Answer 1

Solving $x+y=x^2-xy+y^2$ for $y$ gives you $$y=\frac{x+1\pm\sqrt{-3x^2+6x+1}}{2}.$$ Here, you have to have $$-3x^2+6x+1\ge 0\Rightarrow x=0,1,2.$$

Answer 2

Again consider only $x \neq 0 \neq y$.

Only a Hint this time: On the one hand the RHS is clearly larger than the LHS, when $xy < 0$. So we can assume $xy > 0$. Then we have $x+y-xy = (x-y)^2 \geq 0$. So the sum of two integers is as least as large as their product. Not too many choices there...

Answer 3

Note that: $$(x+y)(x^2 - xy + y^2) = x^3 + y^3$$ $$(x+y)^2 = x^3 + y^3$$

Now let's assume that $x\le y$ and we have:

$$(2y)^2 \ge (x+y)^2 = x^3 + y^3 \ge y^3 \implies 4y^2 \ge y^3 \implies 4\ge y$$

Obviously $y$ must be non-negative number, since otherwise RHS is negative, while LHS is non-negative in the second equation.