Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $

Question

The objective is to find all solutions to $$ x^3 - y^3 = 3(x^2 - y^2) $$ where $x,y \in \mathbb{Z}$.

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So far I've got one pair of solution. Try $(x, y)=(0,0)$: $$ 0^3-0^3=3(0^2-0^2) \\ 0=0 \qquad \text{equation satisfied}$$

Another try $ (x, y) = (x, x)$ then $$ x^3 - x^3 = 3(x^2 -x^2) \\ 0 = 0 \qquad \text{equation satisfied}$$

But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:

$$\begin{aligned} (x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \\ x^2+xy+y^2 &= 3(x+y) \end{aligned}\\ \begin{aligned} x^2+xy+y^2-3x-3y &= 0 \\ x^2+x(y-3)+y(y-3) &=0 \\ x^2 + (y-3)(x+y) &= 0 \end{aligned}$$

What now? Am I on the right track?

Answer 1

By symmetry, we can say that whenever $(x,y)$ is a solution, $(y,x)$ is also a solution. Also, one of them must be positive integer, as $x+y=\frac{x^2+xy+y^2}{3}>0$

Apply quadratic formula in $3$ specific cases, when $y=1,2,3.$ We get $(2,2), (-1,2), (0,3), (3,0), (2,-1)$ as the integer solutions in these cases. There is no other case, as otherwise $y>3$ or $x>3$ and Left Hand Side of the last equation you wrote will be strictly positive.

Hope it is helpful:)

Answer 2

$x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $x\ne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$. Solve the quadratic for $x$ to get $x=\frac{n+\sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $n\le 4$, and a perfect square to get an integer solution. The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$.

Answer 3

So first you factorize the expression: $$(x-y)(x^2+xy+y^2-3x-3y)=0$$ $$(x-y)(x^2+x(y-3)+(y^2-3y))=0$$ When the product of two expressions are $0$ either of them is $0$ So first factor is $(x-y)=0$ giving the solution pair $x=y$ so for all such pairs the equation will be 0. Second factor can be treated as a quadratic equation in terms of either $x$ or $y$, $$x^2+x(y-3)+(y^2-3y)=0$$ Where you consider each $y$ as a constant and solve the equation to get: $$x=\frac{-(y-3)\pm\sqrt{(y-3)^2-4(y^2-3y)}}{2}$$ $$x=\frac{-(y-3)\pm\sqrt{3(9-y^2)}}{2}$$ For $x$ to be real the quantity inside the square root should be positive, moreover you need the integer solutions for $x$ and $y$ so the only possibility that remains is if the discriminant becomes a perfect square, So $y=0,\pm1,\pm2,\pm3$ but you also want it to be a perfect square so only $y=\pm3$ which gives: $$x = 0,3 $$ Final solutions: All integers for which $x=y$, $(0,3)$, $(3,-3)$

You can also solve the equation for $y$, hope this helps .....