Let $P$ be the Petersen graph. Prove or disprove that there is a unique pair of non-negative integers (x,y) on the integer grid $\mathbb{Z}/\langle4\rangle \times\ \mathbb{Z}/\langle4\rangle$ such that no edge in $P$ is "accidentally incident" to an extra vertice. The graph $P$ has an accidental incidence iff $\exists v \in P \ \text{such that} \ \deg(v) \not= 3 $ Does changing the grid to $\mathbb{Z}/\langle4\rangle \times\ \mathbb{Z}/\langle3\rangle$ change the truth value of the claim? $\mathbb{Z}/\langle3\rangle \times \mathbb{Z}/\langle3\rangle$?
For example on the integer grid $\mathbb{Z}/\langle5\rangle \times\ \mathbb{Z}/\langle4\rangle$:
Petersen on $\mathbb{Z}/\langle5\rangle \times\ \mathbb{Z}/\langle4\rangle$
with points A(2, 2), B(0, 3), C(3, 2), D(1, 4), E(3, 3) F(2, 4), G(4, 4), H(4, 1), I(2, 0), J(0, 0)
However, an invalid assignment of vertices to the grid, or equivalently, a $P_0$ with an accidental incidence is as follows: (simply change A(2, 2) to A(0, 2))
with points A(0, 2), B(0, 3), C(3, 2), D(1, 4), E(3, 3) F(2, 4), G(4, 4), H(4, 1), I(2, 0), J(0, 0)
What I really want to know: Restrict your edges to only those created by straight lines: that is, polygons with one side who's total degree is 180˚. Is there a grid with a valid permutation (without "accidental incidence") that works? If so, what is the smallest one?