<s>[INST] Solutions of $x^2 -xy + y^2 \equiv 0 \pmod{43}$ 
$x^2 -xy + y^2 \equiv 0 \pmod{43}$

We could transform this into

$\frac{x^3 + y^3}{x+y} \equiv 0 \pmod{43}$

and thus

$x^3 + y^3 \equiv 0 \pmod{43}$

From what I could manage so far, I know that if $x = 3k$ , $y= 3k+1$, as $43 = 3k+1$.
Also there are no solutions for $43$ itself, $84$ and $127$.
How should I proceed?
 [/INST] If we were working in $\mathbb{Q}$ we could transform your equation into:
$$x^2 - xy + y^2 = 0 \qquad \iff \qquad \left(x- \frac{1}{2}y\right)^2 + \frac{3}{4}y^2 = 0.$$
Fortunately, both $2$ and $4$ are invertible modulo $43$:
$$(x - (2^{-1})y)^2 + 3(4)^{-1}y^2 \equiv 0 \pmod{43} \quad \iff \quad (x+21y)^2 + 33y^2 \equiv 0 \pmod{43}.$$
Let $z = x + 21y$. We have the equation $z^2 + 33y^2 \equiv 0 \pmod {43}$, or better yet,
$$z^2 \equiv 10y^2 \pmod{43}.$$
Now $10$ is a quadratic residue modulo $43$, since $15^2 \equiv 10 \pmod{43}$, so
$$z^2 \equiv (15y)^2 \pmod{43}$$
which means
$$z \equiv \pm 15y\pmod{43} \qquad \iff \qquad x + 21y \equiv \pm 15y \pmod{43}.$$
This gives us two diferent families of solutions:
$$x \equiv 37y \pmod{43} \qquad \text{and} \qquad x \equiv 7y\pmod{43}.$$
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