Given that $(x,y)\in R$ (where $R$ is a relation) $\iff x^2+x=y^2+y$ is it true that for all $x\in \mathbb Z: (-x-1,x)\in R^2$?
Certainly $R$ contains the identity relation. In addition: $$ x^2+x = y^2 + y \implies x+y=-1, x-y \neq 0 $$ so $R=I_{\mathbb Z} \cup \begin{pmatrix}0&-1&1&-2&2&-3\dots\\-1&0&-2&1&-3&2 \dots\end{pmatrix}$
which essentially makes $R$ an equivalence relation.
Then $R^2=R$ so I guess the statement is false.
Exercise. Show R = $R^2.$
Thus your problem becomes showing
(-x-1,x) in R for all x in Z.
Does $(-x-1)^2 + (-x-1) = x^2 + x?