# Representing every positive rational number in the form of (an+bn)/(cn+dn)(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

For $r=p/q$ where $p,q$ are positive integers, we can take

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

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

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

Update : I posted this question on MO.

Added : Problem N2 of IMO 1999 Shortlist asks the case $n=3$.

For the above problem there are four sets of solutions (this is intuitive: for a, b, c, & d). In the case of positive rational r and any odd number n we can eliminate all but one of the solutions:

$d^n = 5 \wedge c^n = 1 \wedge a^n + b^n = 30 \wedge r = 5 \wedge a^n \in Z$

In the case of any odd number n≥3 we refer to the generating function:

$a^{2 n + 1} + b^{2 n + 1} = 30 \wedge c^{2 n + 1} = 1 \wedge d^{2 n + 1} = 5 \wedge r = 5 \wedge a^{2 n + 1} \in Z$

As well as the case of every odd number n≥5 (et. al):

$a^{2 n + 3} + b^{2 n + 3} = 30 \wedge c^{2 n + 3} = 1 \wedge d^{2 n + 3} = 5 \wedge r = 5 \wedge a^{2 n + 3} \in Z$

Quickly we discover that it doesn’t matter the value of n, as long as it’s odd and positive, leading to the generalization:

$r = -c_5-1 \wedge a^{2n+1} + b^{2n+1} = (c_1+c_4+1)(c_5+1) \wedge c^{2n+1}+c_3 = c_1+c_2+1 \wedge c_2+d^{2n+1} = c_3+c_4 \wedge (c_5 | c_4 | c_3 | c_2 | c_1 | a^{2n+1}) \in Z$

For all n:

$r = -c_5-1 \wedge$

$a^n + b^n = (c_1+c_4+1)(c_5+1) \wedge$

$c^n+c_3 = c_1+c_2+1 \wedge$

$c_2+d^n = c_3+c_4 \wedge$

$(c_5 | c_4 | c_3 | c_2 | c_1 | a^n) \in Z$

Note: this isn’t a complete answer so it might be more appropriate as a comment, but pending reputation I may as well take a naive crack at it. Excuse any abuse of notation or lack of comprehension–it’s been over a decade since I’ve had any formal mathematics. Lastly, I welcome criticism, especially if it’s informative and friendly!