Proving a "new" chess piece has an invalid moveset

Question

Imagine a new chess piece that starts in the center of an infinite board, let's call that point $(0,0)$. Our piece can only move in the following ways:

$(+2,-1)$, $(+1,-2)$, $(+1,+1)$, and $(-3,0)$ in relation to its current location.

In order to show this proposed piece is "invalid" I have to show that it cannot reach some squares. I have heard that it cannot reach $(0,2)$, specifically, and use that for the following proof.

First, consider each of the piece's coordinate motions as vectors in the plane: $\begin{bmatrix}2 \\ -1 \end{bmatrix}$,$\begin{bmatrix}1 \\ -2 \end{bmatrix}$,$\begin{bmatrix}1 \\ 1 \end{bmatrix}$,$\begin{bmatrix}-3 \\ 0 \end{bmatrix}$. We desire $x_1,x_2, x_3,x_4\in\mathbb{N}$ such that $x_1\begin{bmatrix}2 \\ -1 \end{bmatrix}+x_2\begin{bmatrix}1 \\ -2 \end{bmatrix}+x_3\begin{bmatrix}1 \\ 1 \end{bmatrix}+x_4\begin{bmatrix}-3 \\ 0 \end{bmatrix}=\begin{bmatrix}0 \\ 2 \end{bmatrix}$

When I solved this using an augmented matrix, I found that $x_1=-x_3+\frac{3}{2}x_4+\frac{2}{3}$ and $x_2=x_3+\frac{4}{3}$.

Therefore, no values of $x_3, x_4\in\mathbb{N}$ yield values of $x_1,x_2\in\mathbb{N}$.

My question: is my logic and setup correct? Less important: are my calculations correct (I have not done matrix algebra in a long time)?

Note: I have chosen to define $\mathbb{N}$ as including $0$.

Answer 1

1. The method is a good one, to prove that $(0,2)$ is unreachable.

2. I didn't check the calculations, but the result is plausible.

3. A simpler proof follows:

Claim: All reachable points $(x,y)$ satisfy the condition that $x-y$ is a multiple of $3$.

Proof: Note that $(0,0)$ satisfies the condition, as does each of the four possible moves. If $(x,y)$ satisfies the condition, as does $(x',y')$, then so does $(x+x', y+y')$. Hence every integer linear combination of the moves does as well.

Now, $(0,2)$ does not satisfy the above condition, so it is not reachable.

Answer 2

I'll write $x$ and $y$ for the usual coordinates on the plane. Unless I've overlooked something, none of the allowed moves can change the quantity $x-y$ modulo 3. Since that quantity is zero initially, it will remain zero after any number of moves.