For what positive integers k does the following property apply? Suppose you start at the origin $(0,0)$ and can move in the following ways:

*

*You can move horizontally by $k-1$: That is, $(x,y) \to (x\pm (k-1),y)$.


*You can move vertically by $k$: That is, $(x,y) \to (x,y\pm k)$.


*You can move diagonally by $k+1$: That is, $(x,y) \to (x\pm (k+1), y\pm(k+1))$ or $(x,y) \to (x\mp (k+1), y\pm(k+1))$.
The question is: For what $k\in\mathbb{Z}^{+}$ can you reach any point with integer coordinates by just these actions?
Observations:
-If $k$ is odd, the $x$-coordinate can change by $k-1$ or $k+1$, which are both even numbers. Thus, it is not possible for $x$ to be odd, which means  $(1,0)$ for instance can never be reached. It must therefore be necessary for $k$ to be even.
-In order to show that any point can be reached, it is sufficient to show that the points $(\pm 1,0)$ and $(0,\pm 1)$ can be reached by a combination of the actions 1,2 and 3. This is because if an arbitrary point $(X,Y)$ is to be reached, we can simply apply the action to get to $(\pm 1,0)$ $\vert X \vert$ times and the action to get to $(0,\pm 1)$ $\vert Y \vert$ times with the appropriate signs depending on which quadrant $(X,Y)$ lies in.
-If you move $k$ times to the right and $k-1$ times up, you reach $(k(k-1),(k-1)k)=(k^2-k,k^2-k)$.
 A: For any $k$, odd or even, you can "walk" the $y$-axis, viz:
$$0,0 \rightarrow k+1,k+1 \rightarrow k+1,1 \rightarrow 2,1$$
If you repeat that process a total of $k-1$ times, you get to point $2(k-1),k-1$, from which two $x$-axis moves and one $y$-axis move takes you to $0,-1$. Simply reflecting the diagonal moves (to the second quadrant) would allow you to arrive at $0,1$, and repeating this process allows you to arrive at any point along the $y$-axis direction, which remains true no matter what point on the $x$-axis you start from.
OP already showed that $k$ must be even.
For $k=2$, $k-1=1$ and that trivially permits walking along the $x$-axis in unit steps. For any other even number, find a number $m$ such that $m\equiv 0 \bmod (k-1)$ and $m\equiv \pm 1 \bmod (k+1)$. The Chinese Remainder Theorem assures that this is possible. Simply move $\frac{m}{k-1}$ times horizontally (always the same direction) and then $\frac{m\pm 1}{k+1}$ times (taking care to use the appropriate $\pm$ sign) diagonally towards the $y$-axis and the resulting $x$-coordinate will be $\pm 1$. Using the $y$-axis walk already described, you can arrive at $\pm 1,0$. Again, changing the direction of the horizontal moves will allow you to land on either $1,0$ or $-1,0$.
From there, as OP states, all things are possible.
Example: $k=4$. We want $m\equiv 0 \bmod 3$ and $m\equiv \pm 1 \bmod 5$. $m=21$ works. So we move $\frac{21}{3}=7$ times horizontally, then $\frac{20}{5}=4$ times diagonally, viz:
$$0,0 \rightarrow 21,0 \rightarrow 1,16$$
From there, we can  walk the $y$-axis in unit steps to $1,0$ if we so desire. Alternatively, we could alternate the vertical component of the diagonal moves to arrive at $1,0$ directly.
