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Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.

I think we'll have to use number theory to do it. Simply solving the equations won't do.

If we divide the second equation by the first, we get: $$x^2 - xy + y^2 = 1 + z$$

Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies $x + y = 1 - z \ge 1 - z^2 = x^3 + y^3$. This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive.

I have tried but am not able to proceed further. Can you help me with this?

Thanks.

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    $\begingroup$ When you divide, you actually discard a whole family of solutions, i.e. if $x+y=0$ and $z=1$, the three numbers solve the equation. $\endgroup$ – 5xum May 30 '16 at 10:05
  • $\begingroup$ @5xum Thanks. I didn't see that coming. $\endgroup$ – TheRandomGuy May 30 '16 at 10:07
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Substituting the first in the second gives $$x^3+y^3+(x+y)^2-2(x+y)=0$$ so $x+y=0$ (giving $z=1$) or $$x^2-xy+y^2+x+y-2=0,$$ that is, $$\left(x+\frac12-\frac y2\right)^2+\frac34(y+1)^2-3=0$$ so $(y+1)^2\leq4$, which leaves to check $y\in\{-3,-2,-1,0,1\}$.

All solutions are given by $$\begin{align*}(x,y,z)\in\{&(a,-a),\;a\in\mathbb Z,&&(z=1)\\ &(-2,-3),&&(z=6)\\ &(-3,-2),(0,-2),&&(z=6,z=3)\\ &(-2,0),(1,0),&&(z=3,z=0)\\ &(0,1)&&(z=0)\}\end{align*}$$


Perhaps some explanation how I got $\left(x+\frac12-\frac y2\right)^2+\frac34(y+1)^2-3=0$. This is called Completing the square:

Starting from $x^2-xy+y^2+x+y-2=0$ we first get rid of the linear term in $x$. Using $x^2+x=(x+\frac12)^2-\frac14$ we find: $$\left(x+\frac12\right)^2-\frac14-xy+y^2+y-2=0.$$ Let $X=x+\frac12$. We have $$X^2-\frac14-Xy+\frac y2+y^2+y-2=0.$$ Now we want to get rid of the mixed term (for the moment we don't care about additional terms in $y$ or constant terms). Using $X^2-Xy=(X-\frac y2)^2-\frac{y^2}4$ we find: $$\left(X-\frac y2\right)^2-\frac{y^2}4-\frac14+\frac y2+y^2+y-2=0.$$ Now we're left only with terms in $y$: $\frac34y^2+\frac32y-\frac94$. Using $y^2+2y=(y+1)^2-1$ we find: $$\frac34y^2+\frac32y-\frac94=\frac34(y+1)^2-\frac34-\frac94.$$ So finally, $$\left(X-\frac y2\right)^2+\frac34(y+1)^2-3=0;\qquad X=x+\frac12.$$


Note: Using this technique, any (inhomogeneous) binary quadratic equation $$ax^2+bxy+cy^2+\text{linear and constant terms}=0$$ with nonzero discriminant $D=b^2-4ac$ can be rewritten in the form $$U^2-DV^2=c$$ where $U$ is a linear (better: affine) function of $x$ and $y$, and $V$ is an affine function of $y$. If $D<0$ (as was the case here), the equation clearly has only finitely many solutions. It can be shown that if $D>0$ it has either $0$ or $\infty$ solutions (in that case we call it a Pell-type-equation or something).

If $D=0$ things get ugly.

Geometrically, these correspond to finding integer points on an ellipse if $D<0$, a hyperbola if $D>0$ and a parabola or a union of at most two lines if $D=0$.

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  • $\begingroup$ I still don't get why $(y+1)^2 \le 4$. Could you please explain? $\endgroup$ – TheRandomGuy May 30 '16 at 10:22
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    $\begingroup$ @Dhruv Because $\frac34(y+1)^2\leq\left(x+\frac12-\frac y2\right)^2+\frac34(y+1)^2=3$. $\endgroup$ – Bart Michels May 30 '16 at 10:23
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    $\begingroup$ Better to say that ${3 \over 4}(y+1)^2-3\le0$. This implies ${1 \over 4}(y+1)^2\le1$ and, finally, $(y+1)^2\le4$ $\endgroup$ – N74 May 30 '16 at 10:27
  • $\begingroup$ @barto Thanks. I got it now. But how did you arrive upon the $(x + 1/2 - y/2)..$ polynomial from the first one. I mean what motivated you to do so? $\endgroup$ – TheRandomGuy May 30 '16 at 10:28
  • $\begingroup$ Many, many thanks. $\endgroup$ – TheRandomGuy May 30 '16 at 10:34
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A general identity that is nice to know is $$x^3+y^3=(x+y)(x^2-xy+y^2).$$ Given that $x+y=1-z$ and $x^3+y^3=1-z^2$ we have $z=1-x-y$ and hence $$x^3+y^3=1-z^2=(1-z)(1+z)=(x+y)(2-x-y).$$ This means that either $x+y=0$, so $y=-x$ and $z=1$, or $$x^2-xy+y^2=2-x-y.$$ This is a quadratic in $x$, and quadratics are easy. Applying the quadratic formula yields $$x=\frac{y-1\pm\sqrt{(1-y)^2-4(y^2+y-2)}}{2}=\frac{1}{2}\left(y-1\pm\sqrt{-3(y+3)(y-1)}\right),$$ and for these $x$ to be integers we need $-3(y+3)(y-1)$ to be a square, so certainly this must be nonnegative. It follows that $$y\in\{-3,-2,-1,0,1\},$$ and these few cases can be checked by hand. For $(x,y,z)$ we find the solutions $$(-2,-3,6),\ (-3,-2,6),\ (0,-2,3),\ (-2,0,3),\ (1,0,3),\ (0,1,3),$$ and $(x,-x,1)$ for all $x\in\Bbb{Z}$ from before.

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  • $\begingroup$ I think this is exactly what OP did when he divided the two equations... The equation you have is almost equivalent to $x^2-xy+y^2=1-z$... $\endgroup$ – 5xum May 30 '16 at 10:09
  • $\begingroup$ Many, many thanks. $\endgroup$ – TheRandomGuy May 30 '16 at 10:34
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$$y=-x, z=1$$

$${y = \frac{-1 + x - \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x + \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$

$${y = \frac{-1 + x + \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x - \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$

All integer solutions:

$x=-3, y=-2, z=6$

$x=-2, y=-3, z=6$

$x=-2, y=0, z=3$

$x=0, y=-2, z=3$

$x=0, y=1, z=0$

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    $\begingroup$ How can you claim 'all'? $\endgroup$ – TheRandomGuy May 30 '16 at 10:11

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