assume you are a flea jumping on a grid (similar to $\mathbb{Z}^2$). You can do any jump that define an integer distance, but moreover it has to change both coordinates (for instance, the moves (3,4) or (5,12) are allowed, but not (1;0)).
The first question is to prove that in a finite number of jumps, you can reach any element of $\mathbb{Z}^2$. This is the easy part and let as an exercise for the reader :)
Now, you have done this easy part and you know that you can reach (1;0) in 3 jumps, which means that any point of coordinates (a;b) is reachable in at most 3(|a|+|b|) moves. Of course this bound is terrible. But here comes the real difficulty : is the number of moves needed to reach any point uniformly bounded ? Or, at least, can you do better than O(|n|) ? (say that n is max (|a|,|b|) for instance, it does not change anything as far as we are only studying what happens asymptotically).
So far, I guess one can use some Euclidean algorithm trick to obtain something in O(ln(|n|)), but I don't have any better idea. Is this problem (well)-known ? Studied somewhere ? Thanks in advance for any comment ;)
Here is an argument that shows we can always go from $(0,0)$ to $(a,b)$ in three jumps. This is best possible, since we cannot go from $(0,0)$ to $(0,1)$ in two jumps. (Proof: by the triangle inequality, the lengths of those jumps differ by less than $1$. But they can only be equal if we jump from $(0,0)$ to $(x,\frac12)$ to $(0,1)$, which is not on the grid.)
First, here is a scheme by which we may sometimes go from $(0,0)$ to $(a,b)$ in two jumps. First, jump from $(0,0)$ to $(3x,4x)$. Then, jump from $(3x,4x)$ to $(3x+4y,4x+3y)$. We solve for the $x$ and $y$ needed to get to $(a,b)$ and see that $$\begin{cases}3x+4y = a \\ 4x+3y =b\end{cases} \implies x = \frac{4b-3a}{7}, y = \frac{4a-3b}{7}.$$ This is not always valid, because $x$ and $y$ are not always integers. In fact, they are integers exactly when $a+b \equiv 0 \pmod 7$.
So instead of trying to get these jumps to go from $(0,0)$ to $(a,b)$, we will consider seven target locations: $(a + 7z, b + 24z)$ for $0 \le z \le 6$. As $z$ varies, the sum $(a+7z)+(b+24z)$ ranges over all possible values modulo $7$, so one of these target locations will be a point $(a',b')$ with $a'+b' \equiv 0 \pmod 7$, which means we can get to it with two jumps. And $(a',b')$ is an integer distance of $25z$ away from $(a,b)$, so we can end with a third jump to $(a,b)$.
In summary, we can get from $(0,0)$ to $(a,b)$ in three jumps: $$(0,0) \Rightarrow (3x,4x) \Rightarrow (3x+4y,4x+3y) = (a',b') \Rightarrow (a,b).$$