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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 $n=3$.

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  • $\begingroup$ What a very good and difficult question! It might be easy, as you propose, to find a counter example for the odd cases. Always feel free to use modular arithmetic when you are faced with integers (the top and bottom of the expression). $\endgroup$ – William Mar 20 '15 at 21:05
  • $\begingroup$ @William how so? $\endgroup$ – MCT Mar 21 '15 at 2:35
  • $\begingroup$ @Soke, I rescind my comment. Fermat's Last Theorem applies somehow but not sure exactly how. $\endgroup$ – William Mar 21 '15 at 3:04
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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!

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  • $\begingroup$ Also, see: en.wikipedia.org/wiki/Pell%27s_equation $\endgroup$ – David Scott Kirby Apr 15 '15 at 8:36
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    $\begingroup$ Could you explain where the 5 and 30 come from? $\endgroup$ – Peter Taylor Apr 15 '15 at 11:16
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    $\begingroup$ Trial and error -- simply plugging in digits until finding an integer solution. I suspect it's the first integer solution but I'm not sure. I'll expand the post with the steps I took. $\endgroup$ – David Scott Kirby Apr 15 '15 at 12:09
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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

$$ r = \left(2x/2u\right)\left(x^4+10x^2y^2+5y^4\right)/\left(u^4+10u^2v^2+5v^4\right)$$

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