<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] The polynomial $p(T)=T^2-T+1$ is the sixth cyclotomic polynomial. Meaning that its zeros have order six.
Because $43-1=6\cdot7$ an element $a$ of the cyclic group $\Bbb{Z}_{43}^*$ is of order six if and only if it is a seventh power but is not a 14th or a 21st power.
Let's just list seventh powers and do a bit of screening:
$2^7=128\equiv-1$, no good. $3^7\equiv37$, this works because its square is $37^2\equiv-7$ and its cube is $\equiv-1$ so it has order six. Therefore $37$ is one zero of $p(T)$. The sum of the zeros is, by Vieta, equal to $1$, so the other zero is $1-37=7$.
Therefore 
$$p(T)=(T-7)(T-37).$$
This settles your question. Whenever $x=ay$ we have 
$$x^2-xy+y^2=y^2(a^2-a+1)=y^2p(a).$$
The solutions are thus $x=7y$ and $x=37y$, $y$ can be anything.
These are all the solutions because if we write $a=x/y$ the above calculation shows that $p(a)=0$. If $y=0$ then the only solution is obviously $x=0$.
</s>