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

I simplified the equation down to

$$(x+y)^2 = x^3 + y^3$$

And hence found out solutions $(0,1), (1,0), (1,2), (2,1), (2,2)$ but I dont think my approach is correct . Is further simplification required? Is there any other method to solve this? I am thankful to those who answer!

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  • $\begingroup$ How did you powers of 3 in your simplification, when there don't exist such in the given expression? $\endgroup$ Oct 6, 2017 at 14:45
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    $\begingroup$ You forgot at minimum $(0,0)$. $\endgroup$
    – Martigan
    Oct 6, 2017 at 14:51
  • $\begingroup$ when you bring your equation in standard polynomial form, you can observe that the Discriminant is negative, hence the conic section is an ellipse. It is always sensible to graph a closed curve as it gives you a finite amount of candidates you can check $\endgroup$
    – imranfat
    Oct 6, 2017 at 14:54

3 Answers 3

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Write $\Delta\geq0$.

It must help!

$$x^2-(y+1)x+y^2-y=0,$$ which gives $$(y+1)^2-4(y^2-y)\geq0$$ or $$3y^2-6y-1\leq0$$ or $$1-\frac{2}{\sqrt3}\leq y\leq1+\frac{2}{\sqrt3},$$ which gives $$0\leq y\leq2,$$ which gives all solutions:

For $y=0$ we get $x^2-x=0$, which gives $(0,0)$ and $(1,0)$.

For $y=1$ we get $x^2-2x=0$, which gives $(0,1)$ and $(2,1)$.

For $y=2$ we get $x^2-3x+2=0$, which gives $(2,2)$ and $(1,2).$

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$x+y=x^2-xy+y^2\to y^2-(x+1) y+x^2-x=0$

$y=\dfrac{1}{2} \left(1+x\pm\sqrt{-3 x^2+6 x+1}\right)$

discriminant must be positive $\Delta=-3 x^2+6 x+1\geq 0\to \dfrac{1}{3} \left(3-2 \sqrt{3}\right)\leq x\leq \dfrac{1}{3} \left(3+2 \sqrt{3}\right)$

which for integer $x$ means, $0\leq x \leq 2$

For $x=0$ we get $y=0;\;1$ solutions are $\color{red}{(0,0)\;(0;\;1)}$

for $x=1$ we have $y=0;\;y=2$ so $\color{red}{(1,0)\;(1;\;2)}$

for $x=2$ finally $y=1;\;y=2$ so $\color{red}{(2,1)\;(2;\;2)}$

hope this helps

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$x+y=x^2-xy+y^2 $.

Since this is symmetrical in $x$ and $y$, we can assume that $x \le y$.

If $x=y$, this becomes $2x = x^2$, so $x=0$ or $x=2$.

If $x < y$, then $2y \gt x+y =(x-y/2)^2+3y^2/4 \ge 3y^2/4 $.

There is no solution if $y \le 0$.

If $ > 0$, $2 \gt 3y/4 $ or $y \lt 8/3$ so $y \le 2$.

If $y=2$, then $x=0$ or $x=1$; of there, only $x=1$ works.

If $y=1$, then $x=0$ works.

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