TITLE: How prove this frog can finite steps jump the point $(\frac{1}{5},\frac{1}{17})$
QUESTION [6 upvotes]: A frog starts at $A(0,0)$ and jumps repeatedly, such that jump covers a distance of exactly $1$, and  such each point jumped to has rational coordinates. Show that:
(1): This frog can jump with finitely many jumps to the point $(\frac{1}{5},\frac{1}{17})$
(2): This frog can't jump with finitely many jumps to the point $(0,\dfrac{1}{4})$

I think this problem is interesting. And note
$$1=\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2$$
so I want to first jump to $(\frac{3}{5},\frac{4}{5})$, but then I don't know how to continue.

REPLY [2 votes]: It's obvious you can jump a distance of 1 in the x or y direction.  You can obviously use two $\left(\dfrac35 , \dfrac45\right)$ jumps (say up and to the right, then up and to the left) to move $\dfrac65$ or $\dfrac85$ along an axis.  Combine this with unit jumps and you can reach any coordinate divisible by 5.
Likewise, you can use two $\left(\dfrac{15}{17}, \dfrac{8}{17}\right)$ to move $\dfrac{16}{17}$ along an axis, which when combined with a unit jump allows you to move $\dfrac{1}{17}$ in any direction.
Generally, if $(a,b,c)$ is a primitive pythagorean triple then you can jump $\left(\dfrac{a}{c}, \dfrac{b}{c}\right)$ then $\left(\dfrac{a}{c}, -\dfrac{b}{c}\right)$ to wind up at $\left(\dfrac{2a}{c}, 0\right)$.  Since $c$ is odd and $\gcd(2a,c) = 1$, there exist integer $x, y$ such that $2ax + cy = 1$, so $x$ double-diagonal steps to the right and $-y$ unit steps to the left leave you at $\left(\dfrac{2ax + cy}{c}, 0\right) = \left(\dfrac{1}{c}, 0\right)$.
For part 2, others have had the right idea about even and odd denominators.  $\dfrac{a}{2b + 1} + \dfrac{c}{2d + 1} = \dfrac{e}{2(2bd + b + d) + 1}$