Diophantine equation $x^2 + xy + y^2 = \left({{x+y}\over{3}} + 1\right)^3$.

Question

Solve in integers the equation$$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3.$$

Answer 1

Hint $1$ :

Let $t=\frac{x+y}{3}\in \mathbb Z$ the equation becomes: $$y^2-3ty+(3t)^2-(t+1)^3=0 $$

a quadratic equation on $y$ which is soluble up to the condition $4t+1$ is a square.

Hint $2$: $\Delta_y=(t-2)^2(4t+1)$

Solutions $(t,y)=(a^2+a,-a^3+3a+1),(a^2+a,a^3+3a^2-1)$ for any parameter $a$

And it's your turn to do some work!

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Edit

Because my answer is down-voted and I don't see the reasons, I will write a full solution based on my little hints below:

Claim If $(x,y)$ are integer solutions to the equation : $$x^2 + xy + y^2 = \left({{x+y}\over3} + 1\right)^3\tag 1$$ then $(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}$

Proof

Let $(x,y)$ be a solution to the given equation, and let $t=\frac{x+y}{3}\in \Bbb Z$ so that: $$y^2-3ty+(3t)^2-(t+1)^3=0 \tag 2$$ and this is a quadratic equation on $y$ so let's compute its discriminant: $$\Delta_y=(3t)^2-4\left((3t)^2-(t+1)^3\right)=4(t+1)^3-27t^2=(t-2)^2(4t+1)\tag 3$$ so this equation has an integer solution $y$ only if $\Delta_y$ is a square of an integer, and this is equivalent to $4t+1$ is a square of an integer which implies that $t=a^2+a$ for some integer $a$. and because the equation $(2)$ is quadratic and we know the value of its discriminant we can find the two solutions on $y$: $$\begin{align}y&=&\frac12\left(3t-(t-2)\sqrt{4t+1}\right)&=-a^3+3a+1\tag 4\\ y&=&\frac12\left(3t+(t-2)\sqrt{4t+1}\right)&=a^3+3a^2-1 \tag 5\end{align}$$ now that we found the value of $t$ and $y$ we can find the value of $x=3t-y$ which gives: $$\left\{\begin{matrix} x=a^3+3a^2-1&\text{ and }& y=-a^3+3a+1 &\text{ or }\tag 6\\ x=-a^3+3a+1&\text{ and } & y=a^3+3a^2-1 \end{matrix}\right.$$

Finally: $$(x,y)\text{ or } (y,x) \in\left\{\left(-a^3+3a+1,a^3+3a^2-1\right)\big | a\in \Bbb Z \right\}\tag 7$$

Answer 2

Setting $x+y=3t$ and $xy=s$, we obtain that $$9t^2-s = (t+1)^3 \implies s = -t^3+6t^2-3t-1$$ $x$ and $y$ satisfying the quadratic $a^2 -3ta + s =0$. This means $9t^2-4s = k^2$. Eliminating $s$, we obtain $$4t^3-15t^2+12t+4 = k^2 \implies 64t^3 - 240t^2 + 192t + 64 = (4k)^2$$ $$(4t-5)^3 - 108t + 189 = (8k)^2 \implies (4t-5)^3 - 27(4t-5) + 54 = (4k)^2$$ Hence, it boils down to solving for integer points on the elliptic curve $$Y^2 = X^3 - 27X+54 = (X-3)^2(X+6)$$ Hence, we need $X+6=m^2$, i.e., $4t+1=m^2 \implies t = \dfrac{m^2-1}4$. This means $m$ has to be odd. Choosing $m=2p+1$, we obtain $$t = \dfrac{4p^2+4p}4 = p^2 + p$$ This gives us that $$Y = m(m^2-9) = (2p+1)(4p^2+4p-8) \implies k=(2p+1)(p+2)(p-1)$$ This gives us $x,y = \dfrac{3t \pm k}2 = t + \dfrac{t \pm k}2$.

Hence, to summarize,

1. Pick any $p \in \mathbb{Z}$.
2. Set $t=p(p+1)$ and $k = (2p+1)(p+2)(p-1)$
3. This gives us $\boxed{\color{blue}{x,y = t + \dfrac{t \pm k}2}}$

If we only want unordered pairs, it suffice to vary $p$ just over $\mathbb{N}$. The first few values by varying $p$ from $0$ to $5$ are

\begin{array}{|c|c|c|} \hline p & x & y\\ \hline 0 & -1 & 1\\ 1 & 3 & 3\\ 2 & 19 & -1\\ 3 & 53 & -17\\ 4 & 111 & -51\\ 5 & 199 & -109\\ \vdots & \vdots & \vdots\\ \hline \end{array}