Representing every positive rational number in the form of $(a^n+b^n)/(c^n+d^n)$ About a month ago, I got the following : 

For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that 
  $$r=\frac{a^\color{red}{3}+b^\color{red}{3}}{c^\color{red}{3}+d^\color{red}{3}}.$$
For $r=p/q$ where $p,q$ are positive integers, we can take 
  $$(a,b,c,d)=(3ps^3t+9qt^4,\ 3ps^3t-9qt^4,\ 9qst^3+ps^4,\ 9qst^3-ps^4)$$
  where $s,t$ are positive integers such that $3\lt r\cdot(s/t)^3\lt 9$.
For $r=2014/89$, for example, since we have $(2014/89)\cdot(2/3)^3\approx 6.7$, taking $(p,q,s,t)=(2014,89,2,3)$ gives us $$\frac{2014}{89}=\frac{209889^3+80127^3}{75478^3+11030^3}.$$

Then, I began to try to find every positive integer $n$ such that the following proposition is true : 
Proposition : For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that $$r=\frac{a^\color{red}{n}+b^\color{red}{n}}{c^\color{red}{n}+d^\color{red}{n}}.$$
The followings are what I've got. Let $r=p/q$ where $p,q$ are positive integers.


*

*For $n=1$, the proposition is true. We can take $(a,b,c,d)=(p,p,q,q)$.

*For $n=2$, the proposition is false. For example, no such sets exist for $r=7/3$.

*For even $n$, the proposition is false because the proposition is false for $n=2$.
However, I've been facing difficulty in the case of odd $n\ge 5$. I've tried to get a similar set of four positive integers $(a,b,c,d)$ as the set for $n=3$, but I have not been able to get any such set. So, here is my question.

Question : How can we find every odd number $n\color{red}{\ge 5}$ such that the following proposition is true?
Proposition : For every positive rational number $r$, there exists a set of four positive integers $(a,b,c,d)$ such that $$r=\frac{a^n+b^n}{c^n+d^n}.$$

Update : I posted this question on MO.
Added : Problem N2 of IMO 1999 Shortlist asks the case $n=3$.

Your solution for n=3 includes an implicit change of variables:
$$ \left(a,b,c,d\right)=\left(x+y,x-y,u+v,u-v\right) $$
$$ r = \left(2x/2u\right)\left(x^2+3y^2\right)/\left(u^2+3v^2\right)$$
at which point the substitution
$$ \left(x,y,u,v\right)=\left(3ps^3t,9qt^4,9qst^3,ps^4\right)$$
yields the desired result of $$r=p/q$$
A similar two-step substitution for 
$$n\ge 5$$
may simplify the search
for n=5, the substitution 
$$ \left(a,b,c,d\right)=\left(x+y,x-y,u+v,u-v\right) $$
yields
$$ r = \left(2x/2u\right)\left(x^4+10x^2y^2+5y^4\right)/\left(u^4+10u^2v^2+5v^4\right)$$
