Path Contains Two Points Whose Coordinates Differ by Integers

Question

This question arised while studying topology, it seems intuitively reasonable, yet I do not know how to show this.

Let $\gamma : [0,1] \rightarrow \mathbb{R}^2 $ be a path (a contiunous map) from $(0,0)$ to $(4,6)$ (or to any pair $(n,m)$ with $n$ and $m$ are not coprime) in the plane. Then, there are two points on the path, $(x_t,y_t)$ and $(x_s,y_s)$ for different $t$ and $s$ (excluding the cases $s=0,t=1$ and $s=1,t=0$), such that $x_s−x_t$ and $y_s−y_t$ are both integers.

This problem can be reduced to showing that on any injective path from $(0,0)$ to $(4,6)$, there are two distinct points whose coordinates differ by integers since paths having self intersections obviously satisfy the condition.

I have seen a proof of this on the web, yet it was quite technical and it was done by considering several cases. I am looking for a more comprehensive and elegant proof if there is. Any kind of hint or reference is greatly appreciated.

Edit: The trivial case, considering the initial point and the final point is excluded.

Answer 1

Assume that $\gamma(s)-\gamma(t)$ is never on $\Bbb Z^2$ except at the trivial points and let us look for a contradiction.

Then we can define $f : [0 ; 1]^2 \to \Bbb R^2 \setminus \{ (0,0) \}$ given by $f(s,t) = \gamma(s) - \gamma(t) - (2,3)$.

We look at what $f$ does on the triangle of vertices $(0,0),(1,0),(1,1)$ :

First, $f(s,0) = \gamma(s) - (2,3)$, so it goes from $-(2,3)$ to $(2,3)$, then $f(1,s) = (4,6)-\gamma(s)-(2,3) = -f(s,0)$, so it goes from $(2,3)$ back to $-(2,3)$ using the same path but reflected through the origin, and finally $f(s,s) = -(2,3)$ is just a constant path.

Since $f$ is continuous on the interior of the triangle, this loop is clearly null-homotopic, so its winding number around $0$ has to be $0$.

On the other hand, the winding number from $-(2,3)$ to $(2,3)$ is $k+ \frac 12$ for some integer $k \in \Bbb Z$ (in a sense that can be made precise by identifying the plane to the complex numbers and computing $\frac 1 {2i\pi} \int_0^1 \frac {f'(s,0)} {f(s,0)} ds$), and the winding number on the return trip from $(2,3)$ back to $-(2,3)$ is the exact same $k+\frac 12$ because it is the same path but rotated by $\pi$ around $0$. So the winding number of $f$ around the triangle has winding number $2k+1$ for some $k \in \Bbb Z$.

Since $0$ isn't an odd integer, we have a contradiction.

Answer 2

I see, easier than indicated. As long as the curve is not too bad, the significance of the coprime condition is this: The path from beginning to end is a difference vector of (4,6). We are looking for a vector line that intersects the curve with a difference vector of (2,3), perhaps with other intersections in between.

Take all lines of the form $(A,0) + t (2,3).$ In a smooth portion of the curve, the distance between two intersections of the line with the curve is continuous. Find $A$ such that there are two intersections with $t$ values that differ by $1;$ put another way, the distance between the intersections is exactly $\sqrt{13}.$ We know we can do this as we know there is a larger length possible, namely $2 \sqrt {13}$

This picture shows a line where the distance is not quite equal to $\sqrt {13}$

I am not prepared to work out what happens if you begin with a Koch Snowflake curve. I am also not going to think about discontinuity when the curve has infinitely many points where this line is parallel. Those are some of the reasons the other proofs you have seen would need extra cases.