how to prove transitive relation for this one? $R=\{(x,y) \mid 3 |(x-y^2) \}$ while R is relation on set A=$\mathbb{Z}$
I need to check if it's reflective , symmetric , transitive .
I figured already that it's not reflective or symmetric with (2,1) or (10,2),(2,10)
and I don't how to check if it's transitive or not.
I think we need to prove that, we have $(x,y), (y,c)$ thus $(x,c) $ then $3|(x-y^2)  = 0$ and
3|$(y-c^2) = 0$ but $3|(x-c^2) \neq  0$

If $3|x-y^2$ and $3|y-z^2$ does it follow that $3|x-z^2$?
$3|x-y^2\implies z\equiv y^2 \pmod 3$ and $3|y-z^2\implies y\equiv z^2 \pmod 3$.
So if $x\equiv y^2$ and $y\equiv z^2$ then $x \equiv z^4\equiv \pmod 3$.
Must it follow that $z^4 \equiv z^2 \pmod 3$?
If $z\equiv 0$ then $z^2 \equiv z^4 \pmod 3$ and, yes,  $x\equiv y^2\pmod 3$ and $y\equiv 0^2\pmod 3\implies x\equiv 0\equiv z^2 \pmod 3$.
If $z \equiv \pm 1$ then $1\equiv z^2\equiv z^4 \pmod 3$ and yes $x\equiv y^2 \pmod 3$ and $y\equiv (\pm 1)^2 \equiv 1\pmod 3 \implies y^2 \equiv 1 \pmod 3$ and so $x\equiv y^2 \equiv 1\equiv z^2 \pmod 3$.
So yes, it is transitive.
......
Alternatively.  If $3|x-y^2$ then we have $x-y^2 = 3k$ for some $k$.  And $3|y-z^2$ then we have $y-z^2 = 3j$ for some $j$.  So $y=3j-z^2$.
So $x-y^2 = x-(9j^2 - 6jz^2 + z^4)= 3k$
So $x-z^2 = 3k + 9j^2 - (6j+1)z^2 + z^4=$
$3(k + 3j^2 -2jz^2) +z^4-z^2$.
so $3|x-z^2$ if and only if $z^4-z^2 = z^2(z^2 - 1)=z^2(z+1)(z-1)$ is divisible by $3$.
But either $z, z+1$ or $z-1$ must be divisible by $3$ as they are three consectutive integers and one of them must be divisible by $3$.
So $3|z^2(z+1)(z-1)$.
