0

In the equation $x^3+y^3=(x+y)^2$, solve for $x$ and $y$ in integers.

So far, I factorized and cancelled out a copy of $(x+y)$ on both sides, leaving me with $x+y=x^2-xy+y^2$. Then, I added $xy+1$ on both sides, and got $(x+1)(y+1)=x^2+y^2+1$. I can't continue from this point.

player3236
  • 16,413
xrider1000
  • 385
  • 1
    When you cancelled $x+y$ from both sides, you lost all solutions of the form $y=-x$. – TonyK Feb 16 '21 at 15:04
  • 1
    https://math.stackexchange.com/questions/3805246/find-all-pairs-of-integers-x-y-such-that-x3y3-xy2?noredirect=1 may help. – Tan Feb 16 '21 at 15:04
  • @Tan: The accepted solution at your link is missing the solution $(x,y)=(2.2)$. – TonyK Feb 16 '21 at 15:07
  • Maybe, starting from $x+y=x^2-xy+y^2$, you should add $3xy$ to both sides to conclude that $(x+y) + 3xy = (x+y)^2$. This way you know that $3xy$ is dividable by $x+y$. Since we can assume that $(x,y) = 1$ this would lead to $3|(x+y)$ leading to four options on $x+y$, that is $x+y = 1,-1,3,-3$ – Salcio Feb 16 '21 at 15:08
  • @TonyK It seems that they found it but forgot to write at the end. – Tan Feb 16 '21 at 15:10

2 Answers2

0

The simplified expression that you obtained defines an ellipse:

enter image description here

Additionally, the line $y=-x$ also satisfies the relation and so, apart from the solutions already found on the elipse ($(0,0), (1,0), (0,1),(2,1),(1,2)$ and $(2,2)$), you also get all points of the form $(n,-n), n \in \mathbb{Z}$.

PierreCarre
  • 20,974
  • 1
  • 18
  • 34
0

Let's write everything on one side and factor (and not discard infinite families of solutions). \begin{align*} x^3 + y^3 - (x+y)^2 &= 0 \\ (x+y)(y^2 - xy + x^2 -y -x) &= 0 \end{align*}

A product is zero when any factor is zero, so, from $x+y = 0$, we find the infinite family of solutions $$ y = -x \text{.} $$ Now we attack $y^2 - xy + x^2 -y -x = 0$. Note that this is the equation of an ellipse with major axis parallel to the line $y = x$ and centered at $(1,1)$. If you sketch that ellipse, you will find that it is in the box $\left[\frac{1}{3}(3 - 2 \sqrt{3}),\frac{1}{3}(3 + 2 \sqrt{3})\right] \times{}$ $\left[\frac{1}{3}(3 - 2 \sqrt{3}),\frac{1}{3}(3 + 2 \sqrt{3})\right]$. So we check the nine integer points in that box. (We could exclude $(1,1)$ since it is the center, but the automatic/unthinking way to proceed is to check every point in the box). \begin{align*} &x & &y & y^2 - xy + &x^2 -y -x & \\ \hline &0 & &0 & 0& &\checkmark \\ &0 & &1 & 0& &\checkmark \\ &0 & &2 & 2& & \\ &1 & &0 & 0& &\checkmark \\ &1 & &1 & -1& & \\ &1 & &2 & 0& &\checkmark \\ &2 & &0 & 2& & \\ &2 & &1 & 0& &\checkmark \\ &2 & &2 & 0& &\checkmark \\ \end{align*}

So the full solution set is $\{(x,y) \mid y = -x\} \cup {}$ $\{(0,0)$, $(0,1)$, $(1,0)$, $(1,2)$, $(2,1)$, $(2,2)\}$.

Eric Towers
  • 67,037