How prove this frog can finite steps jump the point $(\frac{1}{5},\frac{1}{17})$ 
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.

 A: Valid jumps move the frog by
$$\tag1(\Delta_x,\Delta_y)=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right) $$
with $t\in\mathbb Q$.
Or equivalently
$$\tag2(\Delta_x,\Delta_y)=\left(\frac{a^2-b^2}{a^2+b^2},\frac{2ab}{a^2+b^2}\right) $$
with $a,b\in\mathbb Z$, $(a,b)\ne(0,0)$ and wlog $\gcd(a,b)=1$.
Let $v(x)$ denote $2$adic valuation, i.e. $v(2^k\frac uv)=k$ if $u,v$ are odd, and $v(0)=\infty$.
Then one of the coordinates in $(1)$ (or $(2)$) has valuation $0$ and the other has valuation $\ge 2$, i.e. either $v(\Delta_x)=0, v(\Delta_v)\ge 2$ or $v(\Delta_y)=0, v(\Delta_v)\ge 0$. The sum of rationals with nonnegative valuation is again of nonnegative valuation. Therefore no sum of jump coordinates can have valuation $v(\frac14)=-2$. This shows that the second target cannot be reached. 
For the first opart, note that the set of reachable points is a subgroup of $\mathbb Q^2$ and contains $(\frac35,\frac45)$ and reflections and rotations thereof. Try to combine these into $(\frac15,0)$.
Similarly, from $(\frac1{17},\frac4{17})$ and its symmetric variants, try to combine $(0,\frac1{17})$. Together this should give you a jump sequence to $(\frac15,\frac1{17})$.
A: For part (1), observe that 
$$\left(\frac{3}{5},\frac{4}{5}\right) + \left(\frac{3}{5},-\frac{4}{5}\right) + (-1, 0)
+ \left(\frac{15}{17},-\frac{8}{17}\right) + \left(-\frac{15}{17},-\frac{8}{17}\right) 
+ (0, 1)
= \left(\frac{1}{5},\frac{1}{17}\right).$$
A: 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.  $\dfrac{a}{2b + 1} + \dfrac{c}{2d + 1} = \dfrac{e}{2(2bd + b + d) + 1}$
A: very easy:  The even or odd quality of the The denominator is invariant, so 
If  this frog can jump  to a point $(\dfrac{a}{b},\dfrac{c}{d})$, then $b,d$ can not be even numbers
