About a month ago, I got the following :

For

every positive rational number$r$, there exists a set of fourpositive 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 findevery 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 $n=3$.