Show whether $R=\{(x,y):x^2+2y=8\}$ is reflexive, symmetric, or transitive

Question

Let $R$ be the relation on the set of integers $R=\{(x,y):x^2+2y=8\}$.

Reflexivity:

$R$ is reflexive since $2\in\mathbb{Z}$ and $2^2+2\cdot2=8$, so $(2,2)\in R$.

Symmetry:

For $(0,4)\in R$ since $0^2+2\cdot4=8$, but $(4,0)\notin R$ since $4^2+2\cdot0\neq8$. Thus, $R$ is not symmetric.

I can only prove reflexive and symmetric. I have no idea how to do transitivity.

Answer 1

No, $R$ is not reflexive. After all $0\not\mathrel{R}0$.

Indeed, it is not symmetric, for the reason that you mentioned.

And it is not transitive: $6\mathrel{R}(-14)$ and $(-14)\mathrel{R}(-94)$, but $6\not\mathrel R(-94)$.